ContemporaryAbstract Algebra

SEVENTH EDITION

Joseph A. GallianUniversity of Minnesota Duluth

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Contemporary Abstract Algebra,Seventh EditionJoseph A. Gallian

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Contents

Preface xi

PART 1 Integers and Equivalence Relations 1

0 Preliminaries 3Properties of Integers 3 | Modular Arithmetic 7 |

Mathematical Induction 12 | Equivalence Relations 15 |

Functions (Mappings) 18

Exercises 21

Computer Exercises 25

PART 2 Groups 27

1 Introduction to Groups 29Symmetries of a Square 29 | The Dihedral Groups 32

Exercises 35

Biography of Niels Abel 39

2 Groups 40Definition and Examples of Groups 40 | Elementary

Properties of Groups 48 | Historical Note 51

Exercises 52

Computer Exercises 55

3 Finite Groups; Subgroups 57Terminology and Notation 57 | Subgroup Tests 58 |

Examples of Subgroups 61

Exercises 64

Computer Exercises 70

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iv Contents

4 Cyclic Groups 72Properties of Cyclic Groups 72 | Classification of Subgroups

of Cyclic Groups 77

Exercises 81

Computer Exercises 86

Biography of J. J. Sylvester 89

Supplementary Exercises for Chapters 1–4 91

5 Permutation Groups 95Definition and Notation 95 | Cycle Notation 98 | Properties of

Permutations 100 | A Check Digit Scheme Based on D5 110

Exercises 113

Computer Exercises 118

Biography of Augustin Cauchy 121

6 Isomorphisms 122Motivation 122 | Definition and Examples 122 | Cayley’s

Theorem 126 | Properties of Isomorphisms 128 |

Automorphisms 129

Exercises 133

Computer Exercise 136

Biography of Arthur Cayley 137

7 Cosets and Lagrange’s Theorem 138Properties of Cosets 138 | Lagrange’s Theorem and

Consequences 141 | An Application of Cosets to Permutation

Groups 145 | The Rotation Group of a Cube and a Soccer Ball 146

Exercises 149

Computer Exercise 153

Biography of Joseph Lagrange 154

8 External Direct Products 155Definition and Examples 155 | Properties of External Direct

Products 156 | The Group of Units Modulo n as an External Direct

Product 159 | Applications 161

Exercises 167

Computer Exercises 170

Biography of Leonard Adleman 173

Supplementary Exercises for Chapters 5–8 174

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Contents v

9 Normal Subgroups and Factor Groups 178Normal Subgroups 178 | Factor Groups 180 | Applications of

Factor Groups 185 | Internal Direct Products 188

Exercises 193

Biography of Évariste Galois 199

10 Group hom*omorphisms 200Definition and Examples 200 | Properties of hom*omorphisms

202 | The First Isomorphism Theorem 206

Exercises 211

Computer Exercise 216

Biography of Camille Jordan 217

11 Fundamental Theorem of Finite Abelian Groups 218The Fundamental Theorem 218 | The Isomorphism Classes of

Abelian Groups 218 | Proof of the Fundamental Theorem 223

Exercises 226

Computer Exercises 228

Supplementary Exercises for Chapters 9–11 230

PART 3 Rings 235

12 Introduction to Rings 237Motivation and Definition 237 | Examples of Rings 238 |

Properties of Rings 239 | Subrings 240

Exercises 242

Computer Exercises 245

Biography of I. N. Herstein 248

13 Integral Domains 249Definition and Examples 249 | Fields 250 | Characteristic of a

Ring 252

Exercises 255

Computer Exercises 259

Biography of Nathan Jacobson 261

14 Ideals and Factor Rings 262Ideals 262 | Factor Rings 263 | Prime Ideals and Maximal

Ideals 267

Exercises 269

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vi Contents

Computer Exercises 273

Biography of Richard Dedekind 274

Biography of Emmy Noether 275

Supplementary Exercises for Chapters 12–14 276

15 Ring hom*omorphisms 280Definition and Examples 280 | Properties of Ring hom*omorphisms

283 | The Field of Quotients 285

Exercises 287

16 Polynomial Rings 293Notation and Terminology 293 | The Division Algorithm and

Consequences 296

Exercises 300

Biography of Saunders Mac Lane 304

17 Factorization of Polynomials 305Reducibility Tests 305 | Irreducibility Tests 308 | Unique

Factorization in Z[x] 313 | Weird Dice: An Application of Unique

Factorization 314

Exercises 316

Computer Exercises 319

Biography of Serge Lang 321

18 Divisibility in Integral Domains 322Irreducibles, Primes 322 | Historical Discussion of Fermat’s Last

Theorem 325 | Unique Factorization Domains 328 | Euclidean

Domains 331

Exercises 335

Computer Exercise 337

Biography of Sophie Germain 339

Biography of Andrew Wiles 340

Supplementary Exercises for Chapters 15–18 341

PART 4 Fields 343

19 Vector Spaces 345Definition and Examples 345 | Subspaces 346 | Linear

Independence 347

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Contents vii

Exercises 349

Biography of Emil Artin 352

Biography of Olga Taussky-Todd 353

20 Extension Fields 354The Fundamental Theorem of Field Theory 354 | Splitting

Fields 356 | Zeros of an Irreducible Polynomial 362

Exercises 366

Biography of Leopold Kronecker 369

21 Algebraic Extensions 370Characterization of Extensions 370 | Finite Extensions 372 |

Properties of Algebraic Extensions 376 |

Exercises 378

Biography of Irving Kaplansky 381

22 Finite Fields 382Classification of Finite Fields 382 | Structure of Finite Fields 383 |

Subfields of a Finite Field 387

Exercises 389

Computer Exercises 391

Biography of L. E. Dickson 392

23 Geometric Constructions 393Historical Discussion of Geometric Constructions 393 |

Constructible Numbers 394 | Angle-Trisectors and

Circle-Squarers 396

Exercises 396

Supplementary Exercises for Chapters 19–23 399

PART 5 Special Topics 401

24 Sylow Theorems 403Conjugacy Classes 403 | The Class Equation 404 | The

Probability That Two Elements Commute 405 | The Sylow

Theorems 406 | Applications of Sylow Theorems 411

Exercises 414

Computer Exercise 418

Biography of Ludwig Sylow 419

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viii Contents

25 Finite Simple Groups 420Historical Background 420 | Nonsimplicity Tests 425 |

The Simplicity of A5 429 | The Fields Medal 430 |

The Cole Prize 430 |

Exercises 431

Computer Exercises 432

Biography of Michael Aschbacher 434

Biography of Daniel Gorenstein 435

Biography of John Thompson 436

26 Generators and Relations 437Motivation 437 | Definitions and Notation 438 | Free

Group 439 | Generators and Relations 440 | Classification of

Groups of Order Up to 15 444 | Characterization of Dihedral

Groups 446 | Realizing the Dihedral Groups with Mirrors 447

Exercises 449

Biography of Marshall Hall, Jr. 452

27 Symmetry Groups 453Isometries 453 | Classification of Finite Plane Symmetry

Groups 455 | Classification of Finite Groups of Rotations in R3 456

Exercises 458

28 Frieze Groups and Crystallographic Groups 461The Frieze Groups 461 | The Crystallographic Groups 467 |

Identification of Plane Periodic Patterns 473

Exercises 479

Biography of M. C. Escher 484

Biography of George Pólya 485

Biography of John H. Conway 486

29 Symmetry and Counting 487Motivation 487 | Burnside’s Theorem 488 | Applications 490 |

Group Action 493

Exercises 494

Biography of William Burnside 497

30 Cayley Digraphs of Groups 498Motivation 498 | The Cayley Digraph of a Group 498 |

Hamiltonian Circuits and Paths 502 | Some Applications 508

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Contents ix

Exercises 511

Biography of William Rowan Hamilton 516

Biography of Paul Erdös 517

31 Introduction to Algebraic Coding Theory 518Motivation 518 | Linear Codes 523 | Parity-Check Matrix

Decoding 528 | Coset Decoding 531 | Historical Note: The

Ubiquitous Reed-Solomon Codes 535

Exercises 537

Biography of Richard W. Hamming 542

Biography of Jessie MacWilliams 543

Biography of Vera Pless 544

32 An Introduction to Galois Theory 545Fundamental Theorem of Galois Theory 545 | Solvability of

Polynomials by Radicals 552 | Insolvability of a Quintic 556

Exercises 557

Biography of Philip Hall 560

33 Cyclotomic Extensions 561Motivation 561 | Cyclotomic Polynomials 562 |

The Constructible Regular n-gons 566

Exercises 568

Computer Exercise 569

Biography of Carl Friedrich Gauss 570

Biography of Manjul Bhargava 571

Supplementary Exercises for Chapters 24–33 572

Selected Answers A1

Text Credits A40

Photo Credits A42

Index of Mathematicians A43

Index of Terms A45

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Preface

Dear Sir or Madam, will you read my book, it took me years to write, will youtake a look?

JOHN LENNON AND PAUL MCCARTNEY, Paperback Writer, single

Although I wrote the first edition of this book more than twenty yearsago, my goals for it remain the same. I want students to receive a solidintroduction to the traditional topics. I want readers to come away withthe view that abstract algebra is a contemporary subject—that its con-cepts and methodologies are being used by working mathematicians,computer scientists, physicists, and chemists. I want students to enjoyreading the book. To this end, I have included lines from popular songs,poems, quotations, biographies, historical notes, dozens of photographs,hundreds of figures, numerous tables and charts, and reproductions ofstamps and currency that honor mathematicians. I want students to beable to do computations and to write proofs. Accordingly, I haveincluded an abundance of exercises to develop both skills.

Changes for the seventh edition include 120 new exercises, newtheorems and examples, and a freshening of the quotations and biogra-phies. I have also expanded the supplemental material for abstract alge-bra available at my website.

These changes accentuate and enhance the hallmark features thathave made previous editions of the book a comprehensive, lively, andengaging introduction to the subject:

• Extensive coverage of groups, rings, and fields, plus a variety ofnon-traditional special topics

• A good mixture of now more than 1750 computational and theoreti-cal exercises appearing in each chapter and in SupplementaryExercise sets that synthesize concepts from multiple chapters

• Worked-out examples—now totaling 275—providing thoroughpractice for key concepts

• Computer exercises performed using interactive software availableon my website

xi

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xii Preface

• A large number of applications from scientific and computing fields,as well as from everyday life

• Numerous historical notes and biographies that illuminate the peo-ple and events behind the mathematics

• Annotated suggested readings and media for interesting furtherexploration of topics.

My website—accessible at www.d.umn.edu/~jgallian or throughCengage’s book companion site at www.cengage.com/math/gallian—offers a wealth of additional online resources supporting the book,including:

• True/false questions• Flash cards• Essays on learning abstract algebra, doing proofs, and reasons why

abstract algebra is a valuable subject to learn• Links to abstract algebra-related websites and software packages• . . . and much, much more.

Additionally, Cengage offers the following student and instructorancillaries to accompany the book:

• A Student Solutions Manual, available for purchase separately, withworked-out solutions to the odd-numbered exercises in the book(ISBN-13: 978-0-547-16539-4; ISBN-10: 0-547-16539-0)

• An online laboratory manual, written by Julianne Rainbolt and me,with exercises designed to be done with the free computer algebrasystem software GAP

• An online Instructor’s Solutions Manual with solutions to the even-numbered exercises in the book and additional test questions andsolutions

• Online instructor answer keys to the book’s computer exercises andthe exercises in the GAP lab manual.

Connie Day was the copyeditor and Robert Messer was the accuracyreviewer. I am grateful to each of them for their careful reading of themanuscript. I also wish to express my appreciation to Janine Tangney,Daniel Seibert, and Molly Taylor from Cengage Learning, as well asTamela Ambush and the Cengage production staff.

I greatly valued the thoughtful input of the following people, whokindly served as reviewers for the seventh edition:

Rebecca Berg, Bowie State University; Monte Boisen, University ofIdaho; Tara Brendle, Louisiana State University; Jeff Clark, ElonUniversity; Carl Eckberg, San Diego State University; Tom Farmer,Miami University; Yuval Flicker, Ohio State University; Ed Hinson,

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Preface xiii

University of New Hampshire; Gizem Karaali, Pomona College; MohanShrikhande, Central Michigan University; Ernie Stitzinger, NorthCarolina State University.

Over the years, many faculty and students have kindly sent me valu-able comments and suggestions. They have helped to make each editionbetter. I owe thanks to my UMD colleague Robert McFarland for giv-ing me numerous exercises and comments that have been included inthis edition. Please send any comments and suggestions you have to meat [emailprotected].

Joseph A. Gallian

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1

P A R T 1

Integers andEquivalence Relations

For online student resources, visit this textbook’s website athttp://college.hmco.com/PIC/gallian7e

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3

Preliminaries

The whole of science is nothing more than a refinement of everyday thinking.

ALBERT EINSTEIN, Physics and Reality

Properties of IntegersMuch of abstract algebra involves properties of integers and sets. In thischapter we collect the properties we need for future reference.

An important property of the integers, which we will often use, is theso-called Well Ordering Principle. Since this property cannot be provedfrom the usual properties of arithmetic, we will take it as an axiom.

Well Ordering Principle

The concept of divisibility plays a fundamental role in the theory ofnumbers. We say a nonzero integer t is a divisor of an integer s if thereis an integer u such that s 5 tu. In this case, we write t | s (read “tdivides s”). When t is not a divisor of s, we write t B s. A prime is apositive integer greater than 1 whose only positive divisors are 1 anditself. We say an integer s is a multiple of an integer t if there is an in-teger u such that s 5 tu.

As our first application of the Well Ordering Principle, we establisha fundamental property of integers that we will use often.

Theorem 0.1 Division Algorithm

PROOF We begin with the existence portion of the theorem. Considerthe set S 5 {a 2 bk | k is an integer and a 2 bk $ 0}. If 0 [ S, then b

Let a and b be integers with b . 0. Then there exist unique integers qand r with the property that a 5 bq 1 r, where 0 # r , b.

Every nonempty set of positive integers contains a smallest member.

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4 Integers and Equivalence Relations

divides a and we may obtain the desired result with q 5 a/b and r 5 0.Now assume 0 n S. Since S is nonempty [if a . 0, a 2 b ? 0 [ S; if a ,0, a 2 b(2a) 5 a(1 2 2b) [ S; a � 0 since 0 n S], we may apply theWell Ordering Principle to conclude that S has a smallest member, sayr 5 a 2 bq. Then a 5 bq 1 r and r $ 0, so all that remains to beproved is that r , b.

If r $ b, then a 2 b(q 1 1) 5 a 2 bq 2 b 5 r 2 b $ 0, so thata 2 b(q 1 1) [ S. But a 2 b(q 1 1) , a 2 bq, and a 2 bq is thesmallest member of S. So, r , b.

To establish the uniqueness of q and r, let us suppose that there areintegers q, q9, r, and r9 such that

a 5 bq 1 r, 0 # r , b and a 5 bq9 1 r9, 0 # r9 , b.

For convenience, we may also suppose that r9 $ r. Then bq 1 r 5bq9 1 r9 and b(q 2 q9) 5 r9 2 r. So, b divides r9 2 r and 0 # r9 2 r #r9 , b. It follows that r9 2 r 5 0, and therefore r9 5 r and q 5 q9.

The integer q in the division algorithm is called the quotient upon di-viding a by b; the integer r is called the remainder upon dividing a by b.

EXAMPLE 1 For a 5 17 and b 5 5, the division algorithm gives17 5 5 ? 3 1 2; for a 5 223 and b 5 6, the division algorithm gives223 5 6(24) 1 1.

Several states use linear functions to encode the month and date ofbirth into a three-digit number that is incorporated into driver’s li-cense numbers. If the encoding function is known, the division algo-rithm can be used to recapture the month and date of birth from thethree-digit number. For instance, the last three digits of a Florida maledriver’s license number are those given by the formula 40(m 2 1) 1 b,where m is the number of the month of birth and b is the day of birth.Thus, since 177 5 40 ? 4 1 17, a person with these last three digitswas born on May 17. For New York licenses issued prior toSeptember of 1992, the last two digits indicate the year of birth, andthe three preceding digits code the month and date of birth. For amale driver, these three digits are 63m 1 2b, where m denotes thenumber of the month of birth and b is the date of birth. So, since 701 563 ? 11 1 2 ? 4, a license that ends with 70174 indicates that theholder is a male born on November 4, 1974. (In cases where the for-mula for the driver’s license number yields the same result for two ormore people, a “tie-breaking” digit is inserted before the two digits

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0 | Preliminaries 5

for the year of birth.) Incidentally, Wisconsin uses the same methodas Florida to encode birth information, but the numbers immediatelyprecede the last pair of digits.

Definitions Greatest Common Divisor, Relatively Prime Integers

The greatest common divisor of two nonzero integers a and b is thelargest of all common divisors of a and b. We denote this integer bygcd(a, b). When gcd(a, b) 5 1, we say a and b are relatively prime.

The following property of the greatest common divisor of two inte-gers plays a critical role in abstract algebra. The proof provides an ap-plication of the division algorithm and our second application of theWell Ordering Principle.

Theorem 0.2 GCD Is a Linear Combination

PROOF Consider the set S 5 {am 1 bn | m, n are integers and am 1 bn . 0}. Since S is obviously nonempty (if some choice of mand n makes am 1 bn , 0, then replace m and n by 2m and 2n), theWell Ordering Principle asserts that S has a smallest member, say,d 5 as 1 bt. We claim that d 5 gcd(a, b). To verify this claim, use thedivision algorithm to write a 5 dq 1 r, where 0 # r , d. If r . 0,then r 5 a 2 dq 5 a 2 (as 1 bt)q 5 a 2 asq 2 btq 5 a(1 2 sq) 1b(2tq) [ S, contradicting the fact that d is the smallest member of S.So, r 5 0 and d divides a. Analogously (or, better yet, by symmetry),d divides b as well. This proves that d is a common divisor of a and b.Now suppose d9 is another common divisor of a and b and write a 5d9h and b 5 d9k. Then d 5 as 1 bt 5 (d9h)s 1 (d9k)t 5 d9(hs 1 kt),so that d9 is a divisor of d. Thus, among all common divisors of a andb, d is the greatest.

The special case of Theorem 0.2 when a and b are relatively prime isso important in abstract algebra that we single it out as a corollary.

Corollary

If a and b are relatively prime, than there exist integers s and t suchthat as 1 bt 5 1.

For any nonzero integers a and b, there exist integers s and t such thatgcd(a, b) 5 as 1 bt. Moreover, gcd(a, b) is the smallest positive integerof the form as 1 bt.

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EXAMPLE 2 gcd(4, 15) 5 1; gcd(4, 10) 5 2; gcd(22 ? 32 ? 5, 2 ? 33 ?72) 5 2 ? 32. Note that 4 and 15 are relatively prime, whereas 4 and 10 arenot. Also, 4 ? 4 1 15(21) 5 1 and 4(22) 1 10 ? 1 5 2.

The next lemma is frequently used. It appeared in Euclid’s Elements.

Euclid’s Lemma p | ab Implies p | a or p | b

PROOF Suppose p is a prime that divides ab but does not divide a. Wemust show that p divides b. Since p does not divide a, there are integers s and t such that 1 5 as 1 pt. Then b 5 abs 1 ptb, and sincep divides the right-hand side of this equation, p also divides b.

Note that Euclid’s Lemma may fail when p is not a prime, since 6 | (4 ? 3) but 6 B 4 and 6 B 3.

Our next property shows that the primes are the building blocks forall integers. We will often use this property without explicitly saying so.

Theorem 0.3 Fundamental Theorem of Arithmetic

We will prove the existence portion of Theorem 0.3 later in thischapter. The uniqueness portion is a consequence of Euclid’s Lemma(Exercise 27).

Another concept that frequently arises is that of the least commonmultiple of two integers.

Definition Least Common Multiple

The least common multiple of two nonzero integers a and b is thesmallest positive integer that is a multiple of both a and b. We will de-note this integer by lcm(a, b).

We leave it as an exercise (Exercise 12) to prove that every commonmultiple of a and b is a multiple of lcm(a, b).

Every integer greater than 1 is a prime or a product of primes. Thisproduct is unique, except for the order in which the factors appear.That is, if n 5 p1p2

. . . pr and n 5 q1q2. . . qs, where the p’s and q’s

are primes, then r 5 s and, after renumbering the q’s, we have pi 5 qifor all i.

If p is a prime that divides ab, then p divides a or p divides b.

6 Integers and Equivalence Relations

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0 | Preliminaries 7

EXAMPLE 3 lcm(4, 6) 5 12; lcm(4, 8) 5 8; lcm(10, 12) 5 60;lcm(6, 5) 5 30; lcm(22 ? 32 ? 5, 2 ? 33 ? 72) 5 22 ? 33 ? 5 ? 72.

Modular ArithmeticAnother application of the division algorithm that will be important tous is modular arithmetic. Modular arithmetic is an abstraction of amethod of counting that you often use. For example, if it is nowSeptember, what month will it be 25 months from now? Of course, theanswer is October, but the interesting fact is that you didn’t arrive at theanswer by starting with September and counting off 25 months.Instead, without even thinking about it, you simply observed that25 5 2 ? 12 1 1, and you added 1 month to September. Similarly, if itis now Wednesday, you know that in 23 days it will be Friday. Thistime, you arrived at your answer by noting that 23 5 7 ? 3 1 2, so youadded 2 days to Wednesday instead of counting off 23 days. If yourelectricity is off for 26 hours, you must advance your clock 2 hours,since 26 5 2 ? 12 1 2. Surprisingly, this simple idea has numerous im-portant applications in mathematics and computer science. You will seea few of them in this section. The following notation is convenient.

When a 5 qn 1 r, where q is the quotient and r is the remainderupon dividing a by n, we write a mod n 5 r. Thus,

3 mod 2 5 1 since 3 5 1 ? 2 1 1,6 mod 2 5 0 since 6 5 3 ? 2 1 0,

11 mod 3 5 2 since 11 5 3 ? 3 1 2,62 mod 85 5 62 since 62 5 0 ? 85 1 62,

22 mod 15 5 13 since 22 5 (21)15 1 13.

In general, if a and b are integers and n is a positive integer, then a mod n 5 b mod n if and only if n divides a 2 b (Exercise 9).

In our applications, we will use addition and multiplication mod n.When you wish to compute ab mod n or (a 1 b) mod n, and a or b isgreater than n, it is easier to “mod first.” For example, to compute(27 ? 36) mod 11, we note that 27 mod 11 5 5 and 36 mod 11 5 3, so(27 ? 36) mod 11 5 (5 ? 3) mod 11 5 4. (See Exercise 11.)

Modular arithmetic is often used in assigning an extra digit to identi-fication numbers for the purpose of detecting forgery or errors. We pre-sent two such applications.

EXAMPLE 4 The United States Postal Service money order shownin Figure 0.1 has an identification number consisting of 10 digits togetherwith an extra digit called a check. The check digit is the 10-digit numbermodulo 9. Thus, the number 3953988164 has the check digit 2, since

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Figure 0.1

3953988164 mod 9 5 2.† If the number 39539881642 were incorrectlyentered into a computer (programmed to calculate the check digit) as,say, 39559881642 (an error in the fourth position), the machine wouldcalculate the check digit as 4, whereas the entered check digit would be2. Thus the error would be detected.

EXAMPLE 5 Airline companies, United Parcel Service, and therental car companies Avis and National use the modulo 7 values ofidentification numbers to assign check digits. Thus, the identificationnumber 00121373147367 (see Figure 0.2) has the check digit 3 appended

Figure 0.2

8 Integers and Equivalence Relations

†The value of N mod 9 is easy to compute with a calculator. If N 5 9q 1 r, where r isthe remainder upon dividing N by 9, then on a calculator screen N 4 9 appears asq.rrrrr . . . , so the first decimal digit is the check digit. For example, 3953988164 4 9 5439332018.222, so 2 is the check digit. If N has too many digits for your calculator, re-place N by the sum of its digits and divide that number by 9. Thus, 3953988164 mod 9 556 mod 9 5 2. The value of 3953988164 mod 9 can also be computed by searchingGoogle for 3953988164 mod 9.

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0 | Preliminaries 9

Figure 0.3

to it because 121373147367 mod 7 5 3. Similarly, the UPS pickuprecord number 768113999, shown in Figure 0.3, has the check digit 2appended to it.

The methods used by the Postal Service and the airline companies donot detect all single-digit errors (see Exercises 35 and 39). However, detec-tion of all single-digit errors, as well as nearly all errors involving the trans-position of two adjacent digits, is easily achieved. One method that doesthis is the one used to assign the so-called Universal Product Code (UPC)to most retail items (see Figure 0.4). A UPC identification number has 12digits. The first six digits identify the manufacturer, the next five identifythe product, and the last is a check. (For many items, the 12th digit is notprinted, but it is always bar-coded.) In Figure 0.4, the check digit is 8.

Figure 0.4

To explain how the check digit is calculated, it is convenient to intro-duce the dot product notation for two k-tuples:

(a1, a2, . . . , ak) ? (w1, w2, . . . , wk) 5 a1w1 1 a2w2 1 ? ? ? 1 akwk.

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An item with the UPC identification number a1a2 ??? a12 satisfies thecondition

(a1, a2, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 0.

To verify that the number in Figure 0.4 satisfies the condition above, wecalculate

(0 ? 3 1 2 ? 1 1 1 ? 3 1 0 ? 1 1 0 ? 3 1 0 ? 1 1 6 ? 3 1 5 ? 11 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1) mod 10 5 90 mod 10 5 0.

The fixed k-tuple used in the calculation of check digits is called theweighting vector.

Now suppose a single error is made in entering the number inFigure 0.4 into a computer. Say, for instance, that 021000958978 isentered (notice that the seventh digit is incorrect). Then the computercalculates

0 ? 3 1 2 ? 1 1 1 ? 3 1 0 ? 1 1 0 ? 3 1 0 ? 1 1 9 ? 3 1 5 ? 1 1 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1 5 99.

Since 99 mod 10 � 0, the entered number cannot be correct.In general, any single error will result in a sum that is not 0 modulo 10.The advantage of the UPC scheme is that it will detect nearly all

errors involving the transposition of two adjacent digits as well as allerrors involving one digit. For doubters, let us say that the identifica-tion number given in Figure 0.4 is entered as 021000658798. Noticethat the last two digits preceding the check digit have been trans-posed. But by calculating the dot product, we obtain 94 mod 10 � 0,so we have detected an error. In fact, the only undetected transposi-tion errors of adjacent digits a and b are those where |a 2 b| 5 5. Toverify this, we observe that a transposition error of the form

a1a2 ? ? ? aiai11 ? ? ? a12 → a1a2 ? ? ? ai11ai ? ? ? a12

is undetected if and only if

(a1, a2, . . . , ai11, ai, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 0.

That is, the error is undetected if and only if

(a1, a2, . . . , ai11, ai, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 105 (a1, a2, . . . , ai, ai11, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10.

This equality simplifies to either

(3ai11 1 ai) mod 10 5 (3ai 1 ai11) mod 10

or

(ai11 1 3ai) mod 10 5 (ai 1 3ai11) mod 10

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depending on whether i is even or odd. Both cases reduce to 2(ai11 2 ai)mod 10 5 0. It follows that |ai11 2 ai| 5 5, if ai11 � ai.

In 2005 United States companies began to phase in the use of a 13thdigit to be in conformance with the 13-digit product indentificationnumbers used in Europe. The weighing vector for 13-digit numbers is(1, 3, 1, 3, . . . , 3, 1).

Identification numbers printed on bank checks (on the bottom leftbetween the two colons) consist of an eight-digit number a1a2 ? ? ? a8and a check digit a9, so that

(a1, a2, . . . , a9) ? (7, 3, 9, 7, 3, 9, 7, 3, 9) mod 10 5 0.

As is the case for the UPC scheme, this method detects all single-digit errors and all errors involving the transposition of adjacent digits aand b except when |a 2 b| 5 5. But it also detects most errors of theform ? ? ? abc ? ? ? → ? ? ? cba ? ? ?, whereas the UPC method detects noerrors of this form.

In Chapter 5, we will examine more sophisticated means of assign-ing check digits to numbers.

What about error correction? Suppose you have a number such as73245018 and you would like to be sure that even if a single mistakewere made in entering this number into a computer, the computerwould nevertheless be able to determine the correct number. (Think ofit. You could make a mistake in dialing a telephone number but still getthe correct phone to ring!) This is possible using two check digits. Oneof the check digits determines the magnitude of any single-digit error,while the other check digit locates the position of the error. With thesetwo pieces of information, you can fix the error. To illustrate the idea, letus say that we have the eight-digit identification number a1a2 ? ? ? a8. Weassign two check digits a9 and a10 so that

(a1 1 a2 1 ? ? ? 1 a9 1 a10) mod 11 5 0

and

(a1, a2, . . . , a9, a10) ? (1, 2, 3, . . . , 10) mod 11 5 0

are satisfied.Let’s do an example. Say our number before appending the two

check digits is 73245018. Then a9 and a10 are chosen to satisfy

(7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 1 a9 1 a10) mod 11 5 0 (1)

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and

(7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1 0 ? 6 (2)1 1 ? 7 1 8 ? 8 1 a9 ? 9 1 a10 ? 10) mod 11 5 0.

Since 7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 5 30 and 30 mod 11 5 8,Equation (1) reduces to

(8 1 a9 1 a10) mod 11 5 0. (19)

Likewise, since (7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 10 ? 6 1 1 ? 7 1 8 ? 8) mod 11 5 10, Equation (2) reduces to

(10 1 9a9 1 10a10) mod 11 5 0. (29)

Since we are using mod 11, we may rewrite Equation (29) as

(21 2 2a9 2 a10) mod 11 5 0

and add this to Equation (19) to obtain 7 2 a9 5 0. Thus a9 5 7. Nowsubstituting a9 5 7 into Equation (19) or Equation (29), we obtain a10 5 7 as well. So, the number is encoded as 7324501877.

Now let us suppose that this number is erroneously entered into acomputer programmed with our encoding scheme as 7824501877 (anerror in position 2). Since the sum of the digits of the received numbermod 11 is 5, we know that some digit is 5 too large or 6 too small(assuming only one error has been made). But which one? Say theerror is in position i. Then the second dot product has the form a1 ? 1 1a2 ? 2 1 ? ? ? 1 (ai 1 5)i 1 ai11 ? (i 1 1) 1 ? ? ? 1 a10 ? 10 5(a1, a2, ? ? ? , a10) ? (1, 2, ? ? ? , 10) 1 5i. So, (7, 8, 2, 4, 5, 0, 1, 8, 7, 7) ?(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) mod 11 5 5i mod 11. Since the left-handside mod 11 is 10, we see that i 5 2. Our conclusion: The digit in posi-tion 2 is 5 too large. We have successfully corrected the error.

Mathematical InductionThere are two forms of proof by mathematical induction that we willuse. Both are equivalent to the Well Ordering Principle. The explicitformulation of the method of mathematical induction came in the 16thcentury. Francisco Maurolycus (1494–1575), a teacher of Galileo, usedit in 1575 to prove that 1 1 3 1 5 1 ? ? ? 1 (2n 2 1) 5 n2, and BlaisePascal (1623–1662) used it when he presented what we now callPascal’s triangle for the coefficients of the binomial expansion. Theterm mathematical induction was coined by Augustus De Morgan.

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Theorem 0.4 First Principle of Mathematical Induction

PROOF The proof is left as an exercise (Exercise 29).

So, to use induction to prove that a statement involving positive inte-gers is true for every positive integer, we must first verify that the state-ment is true for the integer 1. We then assume the statement is true forthe integer n and use this assumption to prove that the statement is truefor the integer n 1 1.

Our next example uses some facts about plane geometry. Recall thatgiven a straightedge and compass, we can construct a right angle.

EXAMPLE 6 We use induction to prove that given a straightedge, acompass, and a unit length, we can construct a line segment of length

for every positive integer n. The case when n 5 1 is given. Now weassume that we can construct a line segment of length . Then usethe straightedge and compass to construct a right triangle with height 1and base . The hypotenuse of the triangle has length . So,by induction, we can construct a line segment of length for everypositive integer n.

EXAMPLE 7 DEMOIVRE’S THEOREM We use induction to provethat for every positive integer n and every real number u, (cos u 1i sin u)n 5 cos nu 1 i sin nu, where i is the complex number .Obviously, the statement is true for n 5 1. Now assume it is true for n.We must prove that (cos u 1 i sin u)n11 5 cos(n 1 1)u 1 i sin(n 1 1)u.Observe that

(cos u 1 i sin u)n11 5 (cos u 1 i sin u)n(cos u 1 i sin u)5 (cos nu 1 i sin nu)(cos u 1 i sin u)5 cos nu cos u 1 i(sin nu cos u

1 sin u cos nu) 2 sin nu sin u.

Now, using trigonometric identities for cos(a 1 b) and sin(a 1 b), wesee that this last term is cos(n 1 1)u 1 i sin(n 1 1)u. So, by induction,the statement is true for all positive integers.

In many instances, the assumption that a statement is true for an in-teger n does not readily lend itself to a proof that the statement is true

" 21

"n"n 1 1"n

"n"n

Let S be a set of integers containing a. Suppose S has the property thatwhenever some integer n $ a belongs to S, then the integer n 1 1 alsobelongs to S. Then, S contains every integer greater than or equal to a.

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for the integer n 1 1. In such cases, the following equivalent form ofinduction may be more convenient. Some authors call this formulationthe strong form of induction.

Theorem 0.5 Second Principle of Mathematical Induction

PROOF The proof is left to the reader.

To use this form of induction, we first show that the statement is truefor the integer a. We then assume that the statement is true for all inte-gers that are greater than or equal to a and less than n, and use this as-sumption to prove that the statement is true for n.

EXAMPLE 8 We will use the Second Principle of MathematicalInduction with a 5 2 to prove the existence portion of the FundamentalTheorem of Arithmetic. Let S be the set of integers greater than 1 thatare primes or products of primes. Clearly, 2 [ S. Now we assume thatfor some integer n, S contains all integers k with 2 # k , n. We mustshow that n [ S. If n is a prime, then n [ S by definition. If n is not aprime, then n can be written in the form ab, where 1 , a , n and 1 ,b , n. Since we are assuming that both a and b belong to S, we knowthat each of them is a prime or a product of primes. Thus, n is also aproduct of primes. This completes the proof.

Notice that it is more natural to prove the Fundamental Theorem ofArithmetic with the Second Principle of Mathematical Induction thanwith the First Principle. Knowing that a particular integer factors as aproduct of primes does not tell you anything about factoring the nextlarger integer. (Does knowing that 5280 is a product of primes help youto factor 5281 as a product of primes?)

The following problem appeared in the “Brain Boggler” section ofthe January 1988 issue of the science magazine Discover.

EXAMPLE 9 The Quakertown Poker Club plays with blue chipsworth $5.00 and red chips worth $8.00. What is the largest bet thatcannot be made?

Let S be a set of integers containing a. Suppose S has the property thatn belongs to S whenever every integer less than n and greater than orequal to a belongs to S. Then, S contains every integer greater than orequal to a.

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To gain insight into this problem, we try various combinations ofblue and red chips and obtain 5, 8, 10, 13, 15, 16, 18, 20, 21, 23, 24, 25,26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40. It appears that theanswer is 27. But how can we be sure? Well, we need only prove thatevery integer greater than 27 can be written in the form a ? 5 1b ? 8, where a and b are nonnegative integers. This will solve the prob-lem, since a represents the number of blue chips and b the number of redchips needed to make a bet of a ? 5 1 b ? 8. For the purpose of contrast,we will give two proofs—one using the First Principle of MathematicalInduction and one using the Second Principle.

Let S be the set of all integers greater than or equal to 28 of the forma ? 5 1 b ? 8, where a and b are nonnegative. Obviously, 28 [ S. Nowassume that some integer n [ S, say, n 5 a ? 5 1 b ? 8. We must showthat n 1 1 [ S. First, note that since n $ 28, we cannot have botha and b less than 3. If a $ 3, then

n 1 1 5 (a ? 5 1 b ? 8) 1 (23 ? 5 1 2 ? 8)5 (a 2 3) ? 5 1 (b 1 2) ? 8.

(Regarding chips, this last equation says that we may increase a betfrom n to n 1 1 by removing three blue chips from the pot and addingtwo red chips.) If b $ 3, then

n 1 1 5 (a ? 5 1 b ? 8) 1 (5 ? 5 2 3 ? 8)5 (a 1 5) ? 5 1 (b 2 3) ? 8.

(The bet can be increased by 1 by removing three red chips and addingfive blue chips.) This completes the proof.

To prove the same statement by the Second Principle, we note thateach of the integers 28, 29, 30, 31, and 32 is in S. Now assume that forsome integer n . 32, S contains all integers k with 28 # k , n. Wemust show that n [ S. Since n 2 5 [ S, there are nonnegativeintegers a and b such that n 2 5 5 a ? 5 1 b ? 8. But then n 5(a 1 1) ? 5 1 b ? 8. Thus n is in S.

Equivalence RelationsIn mathematics, things that are considered different in one context maybe viewed as equivalent in another context. We have already seen onesuch example. Indeed, the sums 2 1 1 and 4 1 4 are certainly differentin ordinary arithmetic, but are the same under modulo 5 arithmetic.Congruent triangles that are situated differently in the plane are not thesame, but they are often considered to be the same in plane geometry.In physics, vectors of the same magnitude and direction can produce

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different effects—a 10-pound weight placed 2 feet from a fulcrum pro-duces a different effect than a 10-pound weight placed 1 foot from afulcrum. But in linear algebra, vectors of the same magnitude and di-rection are considered to be the same. What is needed to make thesedistinctions precise is an appropriate generalization of the notion ofequality; that is, we need a formal mechanism for specifying whether ornot two quantities are the same in a given setting. This mechanism is anequivalence relation.

Definition Equivalence Relation

An equivalence relation on a set S is a set R of ordered pairs ofelements of S such that

1. (a, a) [ R for all a [ S (reflexive property).2. (a, b) [ R implies (b, a) [ R (symmetric property).3. (a, b) [ R and (b, c) [ R imply (a, c) [ R (transitive property).

When R is an equivalence relation on a set S, it is customary to writeaRb instead of (a, b) [ R. Also, since an equivalence relation is just ageneralization of equality, a suggestive symbol such as <, ;, or , isusually used to denote the relation. Using this notation, the three condi-tions for an equivalence relation become a , a; a , b implies b , a; and a , b and b , c imply a , c. If , is an equivalence relationon a set S and a [ S, then the set [a] 5 {x [ S | x , a} is called theequivalence class of S containing a.

EXAMPLE 10 Let S be the set of all triangles in a plane. If a, b [ S,define a , b if a and b are similar—that is, if a and b have correspond-ing angles that are the same. Then, , is an equivalence relation on S.

EXAMPLE 11 Let S be the set of all polynomials with real coeffi-cients. If f, g [ S, define f , g if f 9 5 g9, where f 9 is the derivative of f.Then, , is an equivalence relation on S. Since two polynomials withequal derivatives differ by a constant, we see that for any f in S, [ f ] 5{ f 1 c | c is real}.

EXAMPLE 12 Let S be the set of integers and let n be a positive inte-ger. If a, b [ S, define a ; b if a mod n 5 b mod n (that is, if a 2 b isdivisible by n). Then, ; is an equivalence relation on S and [a] 5 {a 1kn | k [ S}. Since this particular relation is important in abstract alge-bra, we will take the trouble to verify that it is indeed an equivalencerelation. Certainly, a 2 a is divisible by n, so that a ; a for all a in S.Next, assume that a ; b, say, a 2 b 5 rn. Then, b 2 a 5 (2r)n, and

16 Integers and Equivalence Relations

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therefore b ; a. Finally, assume that a ; b and b ; c, say, a 2 b 5 rnand b 2 c 5 sn. Then, we have a 2 c 5 (a 2 b) 1 (b 2 c) 5 rn 1 sn 5(r 1 s)n, so that a ; c.

EXAMPLE 13 Let ; be as in Example 12 and let n 5 7. Then wehave 16 ; 2; 9 ; 25; and 24 ; 3. Also, [1] 5 {. . . , 220, 213, 26, 1,8, 15, . . .} and [4] 5 {. . . , 217, 210, 23, 4, 11, 18, . . .}.

EXAMPLE 14 Let S 5 {(a, b) | a, b are integers, b 2 0}. If (a, b), (c, d ) [ S, define (a, b) < (c, d ) if ad 5 bc. Then < is an equiv-alence relation on S. [The motivation for this example comes from frac-tions. In fact, the pairs (a, b) and (c, d) are equivalent if the fractions a/band c/d are equal.]

To verify that < is an equivalence relation on S, note that (a, b) < (a, b)requires that ab 5 ba, which is true. Next, we assume that (a, b) < (c, d),so that ad 5 bc. We have (c, d) < (a, b) provided that cb 5 da, which istrue from commutativity of multiplication. Finally, we assume that (a, b) <(c, d ) and (c, d) < (e, f ) and prove that (a, b) < (e, f ). This amounts tousing ad 5 bc and cf 5 de to show that af 5 be. Multiplying both sidesof ad 5 bc by f and replacing cf by de, we obtain adf 5 bcf 5 bde. Sinced 2 0, we can cancel d from the first and last terms.

Definition Partition

A partition of a set S is a collection of nonempty disjoint subsets of Swhose union is S. Figure 0.5 illustrates a partition of a set into foursubsets.

Figure 0.5 Partition of S into four subsets.

EXAMPLE 15 The sets {0}, {1, 2, 3, . . .}, and {. . . , 23, 22, 21}constitute a partition of the set of integers.

EXAMPLE 16 The set of nonnegative integers and the set of non-positive integers do not partition the integers, since both contain 0.

The next theorem reveals that equivalence relations and partitionsare intimately intertwined.

S

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Theorem 0.6 Equivalence Classes Partition

PROOF Let , be an equivalence relation on a set S. For any a [ S, thereflexive property shows that a [ [a]. So, [a] is nonempty and the unionof all equivalence classes is S. Now, suppose that [a] and [b] are distinctequivalence classes. We must show that [a] > [b] 5 0/. On the contrary,assume c [ [a] > [b]. We will show that [a] # [b]. To this end, let x [ [a].We then have c , a, c , b, and x , a. By the symmetric property, wealso have a , c. Thus, by transitivity, x , c, and by transitivity again,x , b. This proves [a] # [b]. Analogously, [b] # [a]. Thus, [a] 5 [b],in contradiction to our assumption that [a] and [b] are distinct equiva-lence classes.

To prove the converse, let P be a collection of nonempty disjointsubsets of S whose union is S. Define a , b if a and b belong to thesame subset in the collection. We leave it to the reader to show that , isan equivalence relation on S (Exercise 55).

Functions (Mappings)Although the concept of a function plays a central role in nearly everybranch of mathematics, the terminology and notation associated withfunctions vary quite a bit. In this section, we establish ours.

Definition Function (Mapping)

A function (or mapping) f from a set A to a set B is a rule that assignsto each element a of A exactly one element b of B. The set A is calledthe domain of f, and B is called the range of f. If f assigns b to a, thenb is called the image of a under f. The subset of B comprising all theimages of elements of A is called the image of A under f.

We use the shorthand f: A → B to mean that f is a mapping fromA to B. We will write f(a) 5 b or f: a → b to indicate that f carriesa to b.

There are often different ways to denote the same element of a set. Indefining a function in such cases one must verify that the functionvalues assigned to the elements depend not on the way the elementsare expressed but only on the elements themselves. For example, the

The equivalence classes of an equivalence relation on a set Sconstitute a partition of S. Conversely, for any partition P of S, thereis an equivalence relation on S whose equivalence classes are theelements of P.

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correspondence f from the rational numbers to the integers given byf(a/b) 5 a 1 b does not define a function since 1/2 5 2/4 but f (1/2) ?f (2/4). To verify that a correspondence is a function, you assume that x1 5 x2 and prove that f (x1) 5 (x2).

Definition Composition of Functions

Let f: A → B and c: B → C. The composition cf is the mapping fromA to C defined by (cf)(a) 5 c(f(a)) for all a in A. The compositionfunction cf can be visualized as in Figure 0.6.

Figure 0.6 Composition of functions f and c.

In calculus courses, the composition of f with g is written ( f 8 g)(x) andis defined by ( f 8 g)(x) 5 f (g(x)). When we compose functions, we omitthe “circle.”

There are several kinds of functions that occur often enough to begiven names.

Definition One-to-One Function

A function f from a set A is called one-to-one if for every a1, a2 [ A,f(a1) 5 f(a2) implies a1 5 a2.

The term one-to-one is suggestive, since the definition ensures thatone element of B can be the image of only one element of A. Alternatively,f is one-to-one if a1 � a2 implies f(a1) � f(a2). That is, different ele-ments of A map to different elements of B. See Figure 0.7.

Figure 0.7

a1 a1

a2 a2

(a1)φ

φ

φ

(a1) 5 (a2) (a2)

is one-to-one is not one-to-oneψ

ψ

ψ ψ

φ

a ( (a))φ (a)

ψψφ

ψφ

φ

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Definition Function from A onto B

A function f from a set A to a set B is said to be onto B if each elementof B is the image of at least one element of A. In symbols, f: A → B isonto if for each b in B there is at least one a in A such that f(a) 5 b.See Figure 0.8.

Figure 0.8

The next theorem summarizes the facts about functions we will need.

Theorem 0.7 Properties of Functions

PROOF We prove only part 1. The remaining parts are left as exercises(Exercise 51). Let a [ A. Then (g(ba))(a) 5 g((ba)(a)) 5 g(b(a(a))).On the other hand, ((gb)a)(a) 5 (gb)(a(a)) 5 g(b(a(a))). So, g(ba) 5(gb)a.

It is useful to note that if a is one-to-one and onto, the function a21

described in part 4 of Theorem 0.7 has the property that if a (s) 5 t,then a21(t) 5 s. That is, the image of t under a21 is the unique element sthat maps to t under a. In effect, a21 “undoes” what a does.

EXAMPLE 17 Let Z denote the set of integers, R the set of real num-bers, and N the set of nonnegative integers. The following table illus-trates the properties of one-to-one and onto.

Given functions a: A → B, b: B → C, and g: C → D, then

1. g(ba) 5 (gb)a (associativity).2. If a and b are one-to-one, then ba is one-to-one.3. If a and b are onto, then ba is onto.4. If a is one-to-one and onto, then there is a function a21 from B

onto A such that (a21a)(a) 5 a for all a in A and (aa21)(b) 5 bfor all b in B.

φ is onto is not ontoψ

ψφ

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0 | Preliminaries 21

Domain Range Rule One-to-one OntoZ Z x → x3 Yes NoR R x → x3 Yes YesZ N x → |x| No YesZ Z x → x2 No No

To verify that x → x3 is one-to-one in the first two cases, notice that ifx3 5 y3, we may take the cube roots of both sides of the equation to ob-tain x 5 y. Clearly, the mapping from Z to Z given by x → x3 is notonto, since 2 is the cube of no integer. However, x → x3 defines anonto function from R to R, since every real number is the cube of itscube root (that is, → b). The remaining verifications are left tothe reader.

Exercises

I was interviewed in the Israeli Radio for five minutes and I said that morethan 2000 years ago, Euclid proved that there are infinitely many primes.Immediately the host interrupted me and asked: “Are there still infinitelymany primes?”

NOGA ALON

1. For n 5 5, 8, 12, 20, and 25, find all positive integers less than nand relatively prime to n.

2. Determine gcd(24 ? 32 ? 5 ? 72, 2 ? 33 ? 7 ? 11) and lcm(23 ? 32 ? 5,2 ? 33 ? 7 ? 11).

3. Determine 51 mod 13, 342 mod 85, 62 mod 15, 10 mod 15, (82 ? 73)mod 7, (51 1 68) mod 7, (35 ? 24) mod 11, and (47 1 68) mod 11.

4. Find integers s and t such that 1 5 7 ? s 1 11 ? t. Show that s and tare not unique.

5. In Florida, the fourth and fifth digits from the end of a driver’s licensenumber give the year of birth. The last three digits for a male withbirth month m and birth date b are represented by 40(m 2 1) 1 b. Forfemales the digits are 40(m 2 1) 1 b 1 500. Determine the dates ofbirth of people who have last five digits 42218 and 53953.

6. For driver’s license numbers issued in New York prior toSeptember of 1992, the three digits preceding the last two of thenumber of a male with birth month m and birth date b are repre-sented by 63m 1 2b. For females the digits are 63m 1 2b 1 1.Determine the dates of birth and sex(es) corresponding to the num-bers 248 and 601.

3"b

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7. Show that if a and b are positive integers, then ab 5 lcm(a, b) ?gcd(a, b).

8. Suppose a and b are integers that divide the integer c. If a and b arerelatively prime, show that ab divides c. Show, by example, that ifa and b are not relatively prime, then ab need not divide c.

9. If a and b are integers and n is a positive integer, prove that a mod n 5b mod n if and only if n divides a 2 b.

10. Let a and b be integers and d 5 gcd(a, b). If a 5 da9 and b 5 db9,show that gcd(a9, b9) 5 1.

11. Let n be a fixed positive integer greater than 1. If a mod n 5 a9 and b mod n 5 b9, prove that (a 1 b) mod n 5 (a9 1 b9) mod n and (ab) mod n 5 (a9b9) mod n. (This exercise is referred to in Chapters6, 8, and 15.)

12. Let a and b be positive integers and let d 5 gcd(a, b) and m 5lcm(a, b). If t divides both a and b, prove that t divides d. If s is amultiple of both a and b, prove that s is a multiple of m.

13. Let n and a be positive integers and let d 5 gcd(a, n). Show that theequation ax mod n 5 1 has a solution if and only if d 5 1. (Thisexercise is referred to in Chapter 2.)

14. Show that 5n 1 3 and 7n 1 4 are relatively prime for all n.15. Prove that every prime greater than 3 can be written in the form

6n 1 1 or 6n 1 5.16. Determine 71000 mod 6 and 61001 mod 7.17. Let a, b, s, and t be integers. If a mod st 5 b mod st, show that

a mod s 5 b mod s and a mod t 5 b mod t. What condition on sand t is needed to make the converse true? (This exercise is referredto in Chapter 8.)

18. Determine 8402 mod 5.19. Show that gcd(a, bc) 5 1 if and only if gcd(a, b) 5 1 and

gcd(a, c) 5 1. (This exercise is referred to in Chapter 8.)20. Let p1, p2, . . . , pn be primes. Show that p1 p2 ? ? ? pn 1 1 is divisi-

ble by none of these primes.21. Prove that there are infinitely many primes. (Hint: Use Exercise 20.)22. For every positive integer n, prove that 1 1 2 1 ? ? ? 1 n 5

n(n 1 1)/2.23. For every positive integer n, prove that a set with exactly n elements

has exactly 2n subsets (counting the empty set and the entire set).24. For any positive integer n, prove that 2n 32n 2 1 is always divisible

by 17.25. Prove that there is some positive integer n such that n, n 1 1,

n 1 2, ? ? ? , n 1 200 are all composite.

22 Integers and Equivalence Relations

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0 | Preliminaries 23

26. (Generalized Euclid’s Lemma) If p is a prime and p dividesa1a2 ? ? ? an, prove that p divides ai for some i.

27. Use the Generalized Euclid’s Lemma (see Exercise 26) to establishthe uniqueness portion of the Fundamental Theorem of Arithmetic.

28. What is the largest bet that cannot be made with chips worth $7.00and $9.00? Verify that your answer is correct with both forms ofinduction.

29. Prove that the First Principle of Mathematical Induction is a conse-quence of the Well Ordering Principle.

30. The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . . In gen-eral, the Fibonacci numbers are defined by f1 5 1, f2 5 1, and forn $ 3, fn 5 fn21 1 fn22. Prove that the nth Fibonacci number fn sat-isfies fn , 2n.

31. In the cut “As” from Songs in the Key of Life, Stevie Wonder men-tions the equation 8 3 8 3 8 3 8 5 4. Find all integers n for whichthis statement is true, modulo n.

32. Prove that for every integer n, n3 mod 6 5 n mod 6.33. If it were 2:00 A.M. now, what time would it be 3736 hours from now?34. Determine the check digit for a money order with identification

number 7234541780.35. Suppose that in one of the noncheck positions of a money order

number, the digit 0 is substituted for the digit 9 or vice versa. Provethat this error will not be detected by the check digit. Prove that allother errors involving a single position are detected.

36. Suppose that a money order identification number and check digitof 21720421168 is erroneously copied as 27750421168. Will thecheck digit detect the error?

37. A transposition error involving distinct adjacent digits is one of theform . . . ab . . . → . . . ba . . . with a � b. Prove that the moneyorder check digit scheme will not detect such errors unless thecheck digit itself is transposed.

38. Determine the check digit for the Avis rental car with identificationnumber 540047. (See Example 6.)

39. Show that a substitution of a digit ai9 for the digit ai (ai9 � ai) ina noncheck position of a UPS number is detected if and onlyif |ai 2 ai9| � 7.

40. Determine which transposition errors involving adjacent digits aredetected by the UPS check digit.

41. Use the UPC scheme to determine the check digit for the number07312400508.

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42. Explain why the check digit for a money order for the number N isthe repeated decimal digit in the real number N 4 9.

43. The 10-digit International Standard Book Number (ISBN-10)a1a2a3a4a5a6a7a8 a9a10 has the property (a1, a2, . . . , a10) ? (10, 9, 8, 7,6, 5, 4, 3, 2, 1) mod 11 5 0. The digit a10 is the check digit. Whena10 is required to be 10 to make the dot product 0, the character X isused as the check digit. Verify the check digit for the ISBN-10 as-signed to this book.

44. Suppose that an ISBN-10 has a smudged entry where the questionmark appears in the number 0-716?-2841-9. Determine the missingdigit.

45. Suppose three consecutive digits abc of an ISBN-10 are scrambled asbca. Which such errors will go undetected?

46. The ISBN-10 0-669-03925-4 is the result of a transposition of twoadjacent digits not involving the first or last digit. Determine thecorrect ISBN-10.

47. Suppose the weighting vector for ISBN-10s was changed to (1, 2, 3,4, 5, 6, 7, 8, 9, 10). Explain how this would affect the check digit.

48. Use the two-check-digit error-correction method described in thischapter to append two check digits to the number 73445860.

49. Suppose that an eight-digit number has two check digits appendedusing the error-correction method described in this chapter and it isincorrectly transcribed as 4302511568. If exactly one digit is in-correct, determine the correct number.

50. The state of Utah appends a ninth digit a9 to an eight-digit driver’slicense number a1a2 . . . a8 so that (9a1 1 8a2 1 7a3 1 6a4 1 5a5 14a6 1 3a7 1 2a8 1 a9) mod 10 5 0. If you know that the licensenumber 149105267 has exactly one digit incorrect, explain why theerror cannot be in position 2, 4, 6, or 8.

51. Complete the proof of Theorem 0.7.52. Let S be the set of real numbers. If a, b [ S, define a , b if a 2 b

is an integer. Show that , is an equivalence relation on S. Describethe equivalence classes of S.

53. Let S be the set of integers. If a, b [ S, define aRb if ab $ 0. Is R anequivalence relation on S?

54. Let S be the set of integers. If a, b [ S, define aRb if a 1 b is even.Prove that R is an equivalence relation and determine the equivalenceclasses of S.

55. Complete the proof of Theorem 0.6 by showing that , is an equiva-lence relation on S.

24 Integers and Equivalence Relations

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0 | Preliminaries 25

56. Prove that none of the integers 11, 111, 1111, 11111, . . . is asquare of an integer.

57. (Cancellation Property) Suppose a, b and g are functions. If ag 5bg and g is one-to-one and onto, prove that a 5 b.

Computer Exercises

There is nothing more practical than a good theory.LEONID BREZHNEV

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software checks the validity of a Postal Service money ordernumber. Use it to verify that 39539881642 is valid. Now enter thesame number with one digit incorrect. Was the error detected? Enterthe number with the 9 in position 2 replaced with a 0. Was the errordetected? Explain why or why not. Enter the number with two dig-its transposed. Was the error detected? Explain why or why not.

2. This software checks the validity of a UPC number. Use it to verifythat 090146003386 is valid. Now enter the same number with onedigit incorrect. Was the error detected? Enter the number with twoconsecutive digits transposed. Was the error detected? Enter thenumber with the second 3 and the 8 transposed. Was the error de-tected? Explain why or why not. Enter the number with the 9 andthe 1 transposed. Was the error detected? Explain why or why not.

3. This software checks the validity of a UPS number. Use it to verifythat 8733456723 is valid. Now enter the same number with one digitincorrect. Was the error detected? Enter the number with two consecu-tive digits transposed. Was the error detected? Enter the number withthe 8 replaced by 1. Was the error detected? Explain why or why not.

4. This software checks the validity of an identification number on abank check. Use it to verify that 091902049 is valid. Now enter thesame number with one digit incorrect. Was the error detected?Enter the number with two consecutive digits transposed. Was theerror detected? Enter the number with the 2 and the 4 transposed.Was the error detected? Explain why or why not.

5. This software checks the validity of an ISBN-10. Use it to verify that0395872456 is valid. Now enter the same number with one digit in-correct. Was the error detected? Enter the number with two digitstransposed (they need not be consecutive). Was the error detected?

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6. This software determines the two check digits for the mod 11 dec-imal error-correction scheme discussed in this chapter. Run theprogram with the input 21355432, 20965744, 10033456. Thenenter these numbers with the two check digits appended with onedigit incorrect. Was the error corrected?

Suggested Readings

Linda Deneen, “Secret Encryption with Public Keys,” The UMAP Journal8 (1987): 9–29.

This well-written article describes several ways in which modulararithmetic can be used to code secret messages. They range from asimple scheme used by Julius Caesar to a highly sophisticated schemeinvented in 1978 and based on modular n arithmetic, where n has morethan 200 digits.

J. A. Gallian, “Assigning Driver’s License Numbers,” Mathematics Maga-zine 64 (1991): 13–22.

This article describes various methods used by the states to assign dri-ver’s license numbers. Several include check digits for error detection.This article can be downloaded at http://www.d.umn.edu/~jgallian/license.pdf

J. A. Gallian, “The Mathematics of Identification Numbers,” The CollegeMathematics Journal 22 (1991): 194–202.

This article is a comprehensive survey of check digit schemes that areassociated with identification numbers. This article can be downloadedat http://www.d.umn.edu/~jgallian/ident.pdf

J. A. Gallian and S. Winters, “Modular Arithmetic in the Marketplace,”The American Mathematical Monthly 95 (1988): 548–551.

This article provides a more detailed analysis of the check digitschemes presented in this chapter. In particular, the error detectionrates for the various schemes are given. This article can be downloadedat http://www.d.umn.edu/~jgallian/marketplace.pdf

26 Integers and Equivalence Relations

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27

P A R T 2

Groups

For online student resources, visit this textbook’s website athttp://college.hmco.com/PIC/gallian7e

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29

Introduction to Groups

Symmetry is a vast subject, significant in art and nature. Mathematics lies atit* root, and it would be hard to find a better one on which to demonstratethe working of the mathematical intellect.

HERMANN WEYL, Symmetry

1

Symmetries of a SquareSuppose we remove a square region from a plane, move it in some way,then put the square back into the space it originally occupied. Our goalin this chapter is to describe in some reasonable fashion all possibleways in which this can be done. More specifically, we want to describethe possible relationships between the starting position of the squareand its final position in terms of motions. However, we are interestedin the net effect of a motion, rather than in the motion itself. Thus, forexample, we consider a 908 rotation and a 4508 rotation as equal, sincethey have the same net effect on every point. With this simplifying con-vention, it is an easy matter to achieve our goal.

To begin, we can think of the square region as being transparent(glass, say), with the corners marked on one side with the colors blue,white, pink, and green. This makes it easy to distinguish between mo-tions that have different effects. With this marking scheme, we are nowin a position to describe, in simple fashion, all possible ways in whicha square object can be repositioned. See Figure 1.1. We now claim thatany motion—no matter how complicated—is equivalent to one of theseeight. To verify this claim, observe that the final position of the squareis completely determined by the location and orientation (that is, faceup or face down) of any particular corner. But, clearly, there are onlyfour locations and two orientations for a given corner, so there areexactly eight distinct final positions for the corner.

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30 Groups

Figure 1.1

Let’s investigate some consequences of the fact that every motion isequal to one of the eight listed in Figure 1.1. Suppose a square is repo-sitioned by a rotation of 908 followed by a flip about the horizontal axisof symmetry. In pictures,

Thus, we see that this pair of motions—taken together—is equal tothe single motion D. This observation suggests that we can composetwo motions to obtain a single motion. And indeed we can, since the

PHR90

G

BW

P W

BG

W B

GP

R0R0 = Rotation of 0° (no change in position)P W

BG

P W

BG

R90R90 = Rotation of 90° (counterclockwise)

P W

BG

W B

GP

R180 = Rotation of 180°P W

BG

B G

PW

R180

R270 = Rotation of 270°P W

BG

G P

WBR270

H = Flip about a horizontal axisP W

BG

G B

WPH

V = Flip about a vertical axisP W

BG

W P

GBV

D = Flip about the main diagonalP G

BWD

P W

BG

D� = Flip about the other diagonalP W

BG

B W

PGD�

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1 | Introduction to Groups 31

eight motions may be viewed as functions from the square region toitself, and as such we can combine them using function composition.

With this in mind, we may now write HR90 5 D. The eight motions R0,R90, R180, R270, H, V, D, and D9, together with the operation composition,form a mathematical system called the dihedral group of order 8 (theorder of a group is the number of elements it contains). It is denoted byD4. Rather than introduce the formal definition of a group here, let’slook at some properties of groups by way of the example D4.

To facilitate future computations, we construct an operation table orCayley table (so named in honor of the prolific English mathematicianArthur Cayley, who first introduced them in 1854) for D4 below. Thecircled entry represents the fact that D 5 HR90. (In general, ab denotesthe entry at the intersection of the row with a at the left and the columnwith b at the top.)

R0 R90 R180 R270 H V D D9

R0 R0 R90 R180 R270 H V D D9

R90 R90 R180 R270 R0 D9 D H VR180 R180 R270 R0 R90 V H D9 DR270 R270 R0 R90 R180 D D9 V HH H D� V D9 R0 R180 R90 R270

V V D9 H D R180 R0 R270 R90

D D V D9 H R270 R90 R0 R180D9 D9 H D V R90 R270 R180 R0

Notice how orderly this table looks! This is no accident. Perhaps themost important feature of this table is that it has been completely filledin without introducing any new motions. Of course, this is because, aswe have already pointed out, any sequence of motions turns out to bethe same as one of these eight. Algebraically, this says that if A and Bare in D4, then so is AB. This property is called closure, and it is one ofthe requirements for a mathematical system to be a group. Next, noticethat if A is any element of D4, then AR0 5 R0 A 5 A. Thus, combiningany element A on either side with R0 yields A back again. An elementR0 with this property is called an identity, and every group must haveone. Moreover, we see that for each element A in D4, there is exactlyone element B in D4 such that AB 5 BA 5 R0. In this case, B is said tobe the inverse of A and vice versa. For example, R90 and R270 areinverses of each other, and H is its own inverse. The term inverse is adescriptive one, for if A and B are inverses of each other, then B “un-does” whatever A “does,” in the sense that A and B taken together in ei-ther order produce R0, representing no change. Another striking feature

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32 Groups

of the table is that every element of D4 appears exactly once in eachrow and column. This feature is something that all groups must have,and, indeed, it is quite useful to keep this fact in mind when construct-ing the table in the first place.

Another property of D4 deserves special comment. Observe thatHD Z DH but R90R180 5 R180R90. Thus, in a group, ab may or may notbe the same as ba. If it happens that ab 5 ba for all choices of groupelements a and b, we say the group is commutative or—better yet—Abelian (in honor of the great Norwegian mathematician Niels Abel).Otherwise, we say the group is non-Abelian.

Thus far, we have illustrated, by way of D4, three of the four con-ditions that define a group—namely, closure, existence of an identity,and existence of inverses. The remaining condition required for a groupis associativity; that is, (ab)c 5 a(bc) for all a, b, c in the set. To be surethat D4 is indeed a group, we should check this equation for each of the83 5 512 possible choices of a, b, and c in D4. In practice, however,this is rarely done! Here, for example, we simply observe that the eightmotions are functions and the operation is function composition. Then,since function composition is associative, we do not have to check theequations.

The Dihedral GroupsThe analysis carried out above for a square can similarly be done foran equilateral triangle or regular pentagon or, indeed, any regular n-gon(n $ 3). The corresponding group is denoted by Dn and is called thedihedral group of order 2n.

The dihedral groups arise frequently in art and nature. Many of thedecorative designs used on floor coverings, pottery, and buildings haveone of the dihedral groups as a group of symmetry. Corporation logosare rich sources of dihedral symmetry [1]. Chrysler’s logo has D5 as asymmetry group, and that of Mercedes-Benz has D3. The ubiquitousfive-pointed star has symmetry group D5. The phylum Echinodermatacontains many sea animals (such as starfish, sea cucumbers, featherstars, and sand dollars) that exhibit patterns with D5 symmetry.

Chemists classify molecules according to their symmetry. Moreover,symmetry considerations are applied in orbital calculations, in determin-ing energy levels of atoms and molecules, and in the study of molecularvibrations. The symmetry group of a pyramidal molecule such as ammo-nia (NH3), depicted in Figure 1.2, has symmetry group D3.

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1 | Introduction to Groups 33

Figure 1.2 A pyramidal molecule with symmetry group D3.

Mineralogists determine the internal structures of crystals (that is,rigid bodies in which the particles are arranged in three-dimensionalrepeating patterns—table salt and table sugar are two examples) bystudying two-dimensional x-ray projections of the atomic makeup of the crystals. The symmetry present in the projections reveals theinternal symmetry of the crystals themselves. Commonly occurringsymmetry patterns are D4 and D6 (see Figure 1.3). Interestingly, it ismathematically impossible for a crystal to possess a Dn symmetry pat-tern with n 5 5 or n . 6.

Figure 1.3 X-ray diffraction photos revealing D4 symmetry patterns in crystals.

The dihedral group of order 2n is often called the group of sym-metries of a regular n-gon. A plane symmetry of a figure F in aplane is a function from the plane to itself that carries F onto F andpreserves distances; that is, for any points p and q in the plane, the distance from the image of p to the image of q is the same as the

N

H

HH

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34 Groups

distance from p to q. (The term symmetry is from the Greek wordsymmetros, meaning “of like measure.”) The symmetry group of aplane figure is the set of all symmetries of the figure. Symmetries inthree dimensions are defined analogously. Obviously, a rotation of aplane about a point in the plane is a symmetry of the plane, and a rota-tion about a line in three dimensions is a symmetry in three-dimensionalspace. Similarly, any translation of a plane or of three-dimensionalspace is a symmetry. A reflection across a line L is that function thatleaves every point of L fixed and takes any point q, not on L, to the pointq9 so that L is the perpendicular bisector of the line segment joiningq and q9 (see Figure 1.4). A reflection across a plane in three dimen-sions is defined analogously. Notice that the restriction of a 1808 rota-tion about a line L in three dimensions to a plane containing L is areflection across L in the plane. Thus, in the dihedral groups, the mo-tions that we described as flips about axes of symmetry in three dimen-sions (for example, H, V, D, D9) are reflections across lines in twodimensions. Just as a reflection across a line is a plane symmetry thatcannot be achieved by a physical motion of the plane in two dimen-sions, a reflection across a plane is a three-dimensional symmetry thatcannot be achieved by a physical motion of three-dimensional space.A cup, for instance, has reflective symmetry across the plane bisectingthe cup, but this symmetry cannot be duplicated with a physical mo-tion in three dimensions.

Figure 1.4

Many objects and figures have rotational symmetry but not reflectivesymmetry. A symmetry group consisting of the rotational symmetries of08, 3608/n, 2(3608)/n, . . . , (n 2 1)3608/n, and no other symmetries iscalled a cyclic rotation group of order n and is denoted by 7R360/n8. Cyclicrotation groups, along with dihedral groups, are favorites of artists, de-signers, and nature. Figure 1.5 illustrates with corporate logos the cyclicrotation groups of orders 2, 3, 4, 5, 6, 8, 16, and 20.

Further examples of the occurrence of dihedral groups and cyclicgroups in art and nature can be found in the references. A study of sym-metry in greater depth is given in Chapters 27 and 28.

q

q9

L

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1 | Introduction to Groups 35

Exercises

The only way to learn mathematics is to do mathematics.PAUL HALMOS, Hilbert Space Problem Book

1. With pictures and words, describe each symmetry in D3 (the set ofsymmetries of an equilateral triangle).

2. Write out a complete Cayley table for D3.3. Is D3 Abelian?4. Describe in pictures or words the elements of D5 (symmetries of a

regular pentagon).5. For n $ 3, describe the elements of Dn. (Hint: You will need to

consider two cases—n even and n odd.) How many elementsdoes Dn have?

6. In Dn, explain geometrically why a reflection followed by a reflec-tion must be a rotation.

7. In Dn, explain geometrically why a rotation followed by a rotationmust be a rotation.

8. In Dn, explain geometrically why a rotation and a reflection takentogether in either order must be a reflection.

9. Associate the number 11 with a rotation and the number 21 witha reflection. Describe an analogy between multiplying these twonumbers and multiplying elements of Dn.

Figure 1.5 Logos with cyclic rotation symmetry groups.

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36 Groups

10. If r1, r2, and r3 represent rotations from Dn and f1, f2, and f3 representreflections from Dn, determine whether r1r2 f1r3 f2 f3r3 is a rotationor a reflection.

11. Find elements A, B, and C in D4 such that AB 5 BC but A Z C.(Thus, “cross cancellation” is not valid.)

12. Explain what the following diagram proves about the group Dn.

13. Describe the symmetries of a nonsquare rectangle. Construct thecorresponding Cayley table.

14. Describe the symmetries of a parallelogram that is neither a rec-tangle nor a rhombus. Describe the symmetries of a rhombus thatis not a rectangle.

15. Describe the symmetries of a noncircular ellipse. Do the same fora hyperbola.

16. Consider an infinitely long strip of equally spaced H’s:

? ? ? H H H H ? ? ?

Describe the symmetries of this strip. Is the group of symmetriesof the strip Abelian?

17. For each of the snowflakes in the figure, find the symmetry groupand locate the axes of reflective symmetry (disregard imperfections).

Photographs of snowflakes from the Bentley and Humphrey atlas.

1 1

2

1

n

2

31

2

13

n2

n

n – 11

2n

F

FR360/ n

R360 /n

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1 | Introduction to Groups 37

18. Determine the symmetry group of the outer shell of the cross sec-tion of the human immunodeficiency virus (HIV) shown below.

19. Does an airplane propeller have a cyclic symmetry group or a di-hedral symmetry group?

20. Bottle caps that are pried off typically have 22 ridges around therim. Find the symmetry group of such a cap.

21. What group theoretic property do upper-case letters F, G, J, K, L,P, Q, R have that is not shared by the remaining upper-case lettersin the alphabet?

22. For each design below, determine the symmetry group (ignoreimperfections).

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23. What would the effect be if a six-bladed ceiling fan were designedso that the centerlines of two of the blades were at a 708 angle andall the other blades were set at a 588 angle?

Reference

1. B. B. Capitman, American Trademark Designs, New York: Dover, 1976.

Suggested Reading

Michael Field and Martin Golubitsky, Symmetry in Chaos, Oxford Uni-versity Press, 1992.

This book has many beautiful symmetric designs that arise inchaotic dynamic systems.

38 Groups

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3939

Niels Abel

He [Abel] has left mathematicianssomething to keep them busy for fivehundred years.

CHARLES HERMITE

NIELS HENRIK ABEL, one of the foremostmathematicians of the 19th century, wasborn in Norway on August 5, 1802. At theage of 16, he began reading the classic math-ematical works of Newton, Euler, Lagrange,and Gauss. When Abel was 18 years old, hisfather died, and the burden of supporting thefamily fell upon him. He took in privatepupils and did odd jobs, while continuing todo mathematical research. At the age of 19,Abel solved a problem that had vexed lead-ing mathematicians for hundreds of years.He proved that, unlike the situation for equa-tions of degree 4 or less, there is no finite(closed) formula for the solution of the gen-eral fifth-degree equation.

Although Abel died long before the ad-vent of the subjects that now make up ab-stract algebra, his solution to the quinticproblem laid the groundwork for many ofthese subjects. Just when his work was be-ginning to receive the attention it deserved,Abel contracted tuberculosis. He died onApril 6, 1829, at the age of 26.

In recognition of the fact that there is noNobel Prize for mathematics, in 2002 Norwayestablished the Abel Prize as the “Nobel Prizein mathematics” in honor of its native son. Atapproximately the $1,000,000 level, the AbelPrize is now seen as an award equivalent tothe Nobel Prize.

To find more information about Abel, visit:http://www-groups.dcs.st-and

.ac.uk/~history/

A 500-kroner bank note first issuedby Norway in 1948.

This stamp was issued in 1929to commemorate the 100thanniversary of Abel’s death.

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40

Groups

A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one.

PAUL R. HALMOS

2

Definition and Examples of GroupsThe term group was used by Galois around 1830 to describe sets ofone-to-one functions on finite sets that could be grouped together toform a set closed under composition. As is the case with most funda-mental concepts in mathematics, the modern definition of a group thatfollows is the result of a long evolutionary process. Although this defi-nition was given by both Heinrich Weber and Walter von Dyck in 1882,it did not gain universal acceptance until the 20th century.

Definition Binary Operation

Let G be a set. A binary operation on G is a function that assigns eachordered pair of elements of G an element of G.

A binary operation on a set G, then, is simply a method (or for-mula) by which the members of an ordered pair from G combine toyield a new member of G. This condition is called closure. The mostfamiliar binary operations are ordinary addition, subtraction, andmultiplication of integers. Division of integers is not a binary opera-tion on the integers because an integer divided by an integer need notbe an integer.

The binary operations addition modulo n and multiplication mod-ulo n on the set {0, 1, 2, . . . , n 2 1}, which we denote by Zn, play anextremely important role in abstract algebra. In certain situations wewill want to combine the elements of Zn by addition modulo n only;in other situations we will want to use both addition modulo n andmultiplication modulo n to combine the elements. It will be clear

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2 | Groups 41

from the context whether we are using addition only or addition andmultiplication. For example, when multiplying matrices with entriesfrom Zn, we will need both addition modulo n and multiplicationmodulo n.

Definition Group

Let G be a set together with a binary operation (usually called multipli-cation) that assigns to each ordered pair (a, b) of elements of G an ele-ment in G denoted by ab. We say G is a group under this operation ifthe following three properties are satisfied.

1. Associativity. The operation is associative; that is, (ab)c 5 a(bc) forall a, b, c in G.

2. Identity. There is an element e (called the identity) in G such thatae 5 ea 5 a for all a in G.

3. Inverses. For each element a in G, there is an element b in G(called an inverse of a) such that ab 5 ba 5 e.

In words, then, a group is a set together with an associative opera-tion such that there is an identity, every element has an inverse, and anypair of elements can be combined without going outside the set. Besure to verify closure when testing for a group (see Example 5). Noticethat if a is the inverse of b, then b is the inverse of a.

If a group has the property that ab 5 ba for every pair of elementsa and b, we say the group is Abelian. A group is non-Abelian if thereis some pair of elements a and b for which ab 2 ba. When encounter-ing a particular group for the first time, one should determine whetheror not it is Abelian.

Now that we have the formal definition of a group, our first job isto build a good stock of examples. These examples will be usedthroughout the text to illustrate the theorems. (The best way to graspthe meat of a theorem is to see what it says in specific cases.) As weprogress, the reader is bound to have hunches and conjectures thatcan be tested against the stock of examples. To develop a better un-derstanding of the following examples, the reader should supply themissing details.

EXAMPLE 1 The set of integers Z (so denoted because the Germanword for numbers is Zahlen), the set of rational numbers Q (for quo-tient), and the set of real numbers R are all groups under ordinary addi-tion. In each case, the identity is 0 and the inverse of a is 2a.

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42 Groups

EXAMPLE 2 The set of integers under ordinary multiplication is nota group. Since the number 1 is the identity, property 3 fails. For exam-ple, there is no integer b such that 5b 5 1.

EXAMPLE 3 The subset {1, 21, i, 2i} of the complex numbers is a group under complex multiplication. Note that 21 is its own inverse,whereas the inverse of i is 2i, and vice versa.

EXAMPLE 4 The set Q1 of positive rationals is a group under ordi-nary multiplication. The inverse of any a is 1/a 5 a21.

EXAMPLE 5 The set S of positive irrational numbers together with 1under multiplication satisfies the three properties given in the definitionof a group but is not a group. Indeed, ? 5 2, so S is not closedunder multiplication.

EXAMPLE 6 A rectangular array of the form is called a

2 3 2 matrix. The set of all 2 3 2 matrices with real entries is a groupunder componentwise addition. That is,

The identity is and the inverse of is

EXAMPLE 7 The set Zn 5 {0, 1, . . . , n 2 1} for n $ 1 is a group underaddition modulo n. For any j . 0 in Zn, the inverse of j is n 2 j. This group is usually referred to as the group of integers modulo n.

As we have seen, the real numbers, the 2 3 2 matrices with real en-tries, and the integers modulo n are all groups under the appropriate ad-dition. But what about multiplication? In each case, the existence ofsome elements that do not have inverses prevents the set from being agroup under the usual multiplication. However, we can form a group ineach case by simply throwing out the rascals. Examples 8, 9, and 11illustrate this.

EXAMPLE 8 The set R* of nonzero real numbers is a group underordinary multiplication. The identity is 1. The inverse of a is 1/a.

c2a 2b

2c 2dd .ca b

c ddc0 0

0 0d ,

ca1 b1

c1 d1d 1 ca2 b2

c2 d2d 5 ca1 1 a2

c1 1 c2

b1 1 b2

d1 1 d2d

ca b

c dd

"2"2

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2 | Groups 43

EXAMPLE 9† The determinant of the 2 3 2 matrix is the

number ad 2 bc. If A is a 2 3 2 matrix, det A denotes the determinantof A. The set

GL(2, R) 5

of 2 3 2 matrices with real entries and nonzero determinant is a non-Abelian group under the operation

.

The first step in verifying that this set is a group is to show that theproduct of two matrices with nonzero determinant also has nonzerodeterminant. This follows from the fact that for any pair of 2 3 2matrices A and B, det (AB) 5 (det A)(det B).

Associativity can be verified by direct (but cumbersome) calcula-

tions. The identity is ; the inverse of is

(explaining the requirement that ad 2 bc 2 0). This very importantnon-Abelian group is called the general linear group of 2 3 2 matricesover R.

EXAMPLE 10 The set of all 2 3 2 matrices with real number entriesis not a group under the operation defined in Example 9. Inverses donot exist when the determinant is 0.

Now that we have shown how to make subsets of the real numbersand subsets of the set of 2 3 2 matrices into multiplicative groups, wenext consider the integers under multiplication modulo n.

≥d

ad 2 bc

2b

ad 2 bc

2c

ad 2 bc

a

ad 2 bc

¥

ca b

c ddc1 0

0 1d

ca1 b1

c1 d1d ca2 b2

c2 d2d 5 ca1a2 1 b1c2

c1a2 1 d1c2

a1b2 1 b1d2

c1b2 1 d1d2d

e ca b

c dd ` a, b, c, d P R, ad 2 bc ? 0 f

ca b

c dd

†For simplicity, we have restricted our matrix examples to the 2 3 2 case. However,readers who have had linear algebra can readily generalize to n 3 n matrices.

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EXAMPLE 11 (L. Euler, 1761) By Exercise 13 in Chapter 0, aninteger a has a multiplicative inverse modulo n if and only if a and n arerelatively prime. So, for each n . 1, we define U(n) to be the set of allpositive integers less than n and relatively prime to n. Then U(n) is agroup under multiplication modulo n. (We leave it to the reader tocheck that this set is closed under this operation.)

For n 5 10, we have U(10) 5 {1, 3, 7, 9}. The Cayley table forU(10) is

(Recall that ab mod n is the unique integer r with the property a ? b 5nq 1 r, where 0 # r , n and a ? b is ordinary multiplication.) In thecase that n is a prime, U(n) 5 {1, 2, . . . , n 2 1}.

In his classic book Lehrbuch der Algebra, published in 1899, HeinrichWeber gave an extensive treatment of the groups U(n) and describedthem as the most important examples of finite Abelian groups.

EXAMPLE 12 The set {0, 1, 2, 3} is not a group under multiplica-tion modulo 4. Although 1 and 3 have inverses, the elements 0 and 2do not.

EXAMPLE 13 The set of integers under subtraction is not a group,since the operation is not associative.

With the examples given thus far as a guide, it is wise for the readerto pause here and think of his or her own examples. Study actively!Don’t just read along and be spoon-fed by the book.

EXAMPLE 14 For all integers n $ 1, the set of complex nth rootsof unity

(i.e., complex zeros of xn 2 1) is a group under multiplication. (SeeDeMoivre’s Theorem—Example 7 in Chapter 0.) Compare this groupwith the one in Example 3.

e cos k ? 360°

n1 i sin

k ? 360°n

` k 5 0, 1, 2, . . . , n 2 1 f

44 Groups

mod 10 1 3 7 9

1 1 3 7 93 3 9 1 77 7 1 9 39 9 7 3 1

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2 | Groups 45

The complex number a 1 bi can be represented geometrically as thepoint (a, b) in a plane coordinatized by a horizontal real axis and a ver-tical i or imaginary axis. The distance from the point a 1 bi to the ori-gin is and is often denoted by Ua 1 bi|. For any angle u, theline segment joining the complex number cos u 1 i sin u and the originforms an angle of u with the positive real axis. Thus, the six complexzeros of x6 5 1 are located at points around the circle of radius 1, 60°apart, as shown in Figure 2.1.

Figure 2.1

EXAMPLE 15 The set Rn 5 {(a1, a2, . . . , an) U a1, a2, . . . , an [ R}is a group under componentwise addition [i.e., (a1, a2, . . . , an) 1(b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn)].

EXAMPLE 16 For a fixed point (a, b) in R2, define Ta,b: R2 → R2

by (x, y) → (x 1 a, y 1 b). Then G 5 {Ta,b U a, b [ R} is a groupunder function composition. Straightforward calculations show thatTa,bTc,d 5 Ta1c,b1d. From this formula we may observe that G isclosed, T0,0 is the identity, the inverse of Ta,b is T2a,2b, and G is Abelian.Function composition is always associative. The elements of G arecalled translations.

EXAMPLE 17 The set of all 2 3 2 matrices with determinant 1 with en-tries from Q (rationals), R (reals), C (complex numbers), or Zp (p a prime)is a non-Abelian group under matrix multiplication. This group is calledthe special linear group of 2 3 2 matrices over Q, R, C, or Zp, respectively.

21

21

23

60°

–21

23–– i i

21

23+

23

i21

23+– i

Imaginary

Real–1 1

√√

√√

√

"a21b2

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46 Groups

If the entries are from F, where F is any of the above, we denote this groupby SL(2, F). For the group SL(2, F), the formula given in Example 9 for

the inverse of simplifies to When the matrix

entries are from Zp, we use modulo p arithmetic to compute determi-nants, matrix products, and inverses. To illustrate the case SL(2, Z5),

consider the element A 5 . Then det A 5 (3 ? 4 2 4 ? 4) mod 5 5

24 mod 5 5 1, and the inverse of A is . Note

that when the arithmetic is done modulo 5.

Example 9 is a special case of the following general construction.

EXAMPLE 18 Let F be any of Q, R, C, or Zp ( p a prime). The setGL(2, F) of all 2 3 2 matrices with nonzero determinants and entriesfrom F is a non-Abelian group under matrix multiplication. As inExample 17, when F is Zp, modulo p arithmetic is used to calculatedeterminants, the matrix products, and inverses. The formula given in

Example 9 for the inverse of remains valid for elements from

GL(2, Zp) provided we interpret division by ad 2 bc as multiplicationby the inverse of ad 2 bc modulo p. For example, in GL(2, Z7),

consider . Then the determinant (ad 2 bc) mod 7 is (12 2 30)

mod 7 5 218 mod 7 5 3 and the inverse of 3 is 5 [since (3 ? 5)

mod 7 5 1]. So, the inverse of is .

[The reader should check that in GL(2, Z7)].

EXAMPLE 19 The set {1, 2, . . . , n 2 1} is a group under multipli-cation modulo n if and only if n is prime.

EXAMPLE 20 The set of all symmetries of the infinite ornamentalpattern in which arrowheads are spaced uniformly a unit apart along

c4 5

6 3d c1 3

5 6d 5 c1 0

0 1d

c3 ? 5 2 ? 5

1 ? 5 4 ? 5d 5 c1 3

5 6dc4 5

6 3d

c4 5

6 3d

ca b

c dd

c3 4

4 4d c4 1

1 3d 5 c1 0

0 1d

c 4 24

24 3d 5 c4 1

1 3d

c3 4

4 4d

c d 2b

2c ad .ca b

c dd

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2 | Groups 47

a line is an Abelian group under composition. Let T denote a translationto the right by one unit, T 21 a translation to the left by one unit, and H a re-flection across the horizontal line of the figure. Then, every member of thegroup is of the form x1x2 ? ? ? xn, where each xi [{T, T21, H}. In this case, we say that T, T21, and H generate the group.

Table 2.1 summarizes many of the specific groups that we havepresented thus far.

As the examples above demonstrate, the notion of a group is a verybroad one indeed. The goal of the axiomatic approach is to find proper-ties general enough to permit many diverse examples having theseproperties and specific enough to allow one to deduce many interestingconsequences.

The goal of abstract algebra is to discover truths about algebraicsystems (that is, sets with one or more binary operations) that are inde-pendent of the specific nature of the operations. All one knows or needs to know is that these operations, whatever they may be, have

Table 2.1 Summary of Group Examples (F can be any of Q, R, C, or Zp; L is a reflection)

Form ofGroup Operation Identity Element Inverse Abelian

Z Addition 0 k 2k YesQ1 Multiplication 1 m/n, n/m Yes

m, n . 0Zn Addition mod n 0 k n 2 k YesR* Multiplication 1 x 1/x YesGL(2, F) Matrix

,No

multiplication

ad 2 bc 2 0U(n) Multiplication 1 k, Solution to Yes

mod n gcd(k, n) 5 1 kx mod n 5 1Rn Componentwise (0, 0, …, 0) (a1, a2, …, an) (2a1, 2a2, …, 2an) Yes

additionSL(2, F) Matrix No

multiplication

ad 2 bc 5 1Dn Composition R0 Ra, L R360 2 a, L No

c d

2c

2b

adca b

c ddc1 0

0 1d

ca b

c ddc1 0

0 1d

≥d

ad 2 bc

2b

ad 2 bc

2c

ad 2 bc

a

ad 2 bc

¥

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48 Groups

certain properties. We then seek to deduce consequences of theseproperties. This is why this branch of mathematics is called abstractalgebra. It must be remembered, however, that when a specific groupis being discussed, a specific operation must be given (at least implicitly).

Elementary Properties of GroupsNow that we have seen many diverse examples of groups, we wish todeduce some properties that they share. The definition itself raisessome fundamental questions. Every group has an identity. Could agroup have more than one? Every group element has an inverse. Couldan element have more than one? The examples suggest not. But exam-ples can only suggest. One cannot prove that every group has a uniqueidentity by looking at examples, because each example inherently hasproperties that may not be shared by all groups. We are forced torestrict ourselves to the properties that all groups have; that is, we mustview groups as abstract entities rather than argue by example. The nextthree theorems illustrate the abstract approach.

Theorem 2.1 Uniqueness of the Identity

PROOF Suppose both e and e9 are identities of G. Then,

1. ae 5 a for all a in G, and2. e9a 5 a for all a in G.

The choices of a 5 e9 in (1) and a 5 e in (2) yield e9e 5 e9 and e9e 5 e. Thus, e and e9 are both equal to e9e and so are equal to eachother.

Because of this theorem, we may unambiguously speak of “the iden-tity” of a group and denote it by “e” (because the German word foridentity is Einheit).

Theorem 2.2 Cancellation

In a group G, the right and left cancellation laws hold; that is, ba 5 ca implies b 5 c, and ab 5 ac implies b 5 c.

In a group G, there is only one identity element.

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2 | Groups 49

PROOF Suppose ba 5 ca. Let a9 be an inverse of a. Then, multi-plying on the right by a9 yields (ba)a9 5 (ca)a9. Associativity yields b(aa9) 5 c(aa9). Then, be 5 ce and, therefore, b 5 c as desired. Simi-larly, one can prove that ab 5 ac implies b 5 c by multiplying by a9 onthe left.

A consequence of the cancellation property is the fact that in aCayley table for a group, each group element occurs exactly once ineach row and column (see Exercise 23). Another consequence of thecancellation property is the uniqueness of inverses.

Theorem 2.3 Uniqueness of Inverses

PROOF Suppose b and c are both inverses of a. Then ab 5 e and ac 5 e, so that ab 5 ac. Canceling the a on both sides gives b 5 c, asdesired.

As was the case with the identity element, it is reasonable, in viewof Theorem 2.3, to speak of “the inverse” of an element g of a group;in fact, we may unambiguously denote it by g21. This notation is sug-gested by that used for ordinary real numbers under multiplication.Similarly, when n is a positive integer, the associative law allows us touse gn to denote the unambiguous product

gg ? ? ? g.

n factors

We define g0 5 e. When n is negative, we define gn 5 (g21)|n| [for ex-ample, g23 5 (g21)3]. Unlike for real numbers, in an abstract group wedo not permit noninteger exponents such as g1/2. With this notation, thefamiliar laws of exponents hold for groups; that is, for all integers m andn and any group element g, we have gmgn 5 gm1n and (gm)n 5 gmn.Although the way one manipulates the group expressions gmgn and(gm)n coincides with the laws of exponents for real numbers, the lawsof exponents fail to hold for expressions involving two group elements.Thus, for groups in general, (ab)n Z anbn (see Exercise 15).

Also, one must be careful with this notation when dealing with aspecific group whose binary operation is addition and is denoted by

For each element a in a group G, there is a unique element b in Gsuch that ab 5 ba 5 e.

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50 Groups

“1.” In this case, the definitions and group properties expressed inmultiplicative notation must be translated to additive notation. Forexample, the inverse of g is written as 2g. Likewise, for example, g3

Table 2.2

Multiplicative Group Additive Group

a ? b or ab Multiplication a 1 b Additione or 1 Identity or one 0 Zeroa21 Multiplicative inverse of a 2a Additive inverse of aan Power of a na Multiple of aab21 Quotient a 2 b Difference

means g 1 g 1 g and is usually written as 3g, whereas g23 means(2g) 1 (2g) 1 (2g) and is written as 23g. When additive notation is used, do not interpret “ng” as combining n and g under the groupoperation; n may not even be an element of the group! Table 2.2 showsthe common notation and corresponding terminology for groups un-der multiplication and groups under addition. As is the case for realnumbers, we use a 2 b as an abbreviation for a 1 (2b).

Because of the associative property, we may unambiguously writethe expression abc, for this can be reasonably interpreted as only (ab)cor a(bc), which are equal. In fact, by using induction and repeated ap-plication of the associative property, one can prove a general associa-tive property that essentially means that parentheses can be inserted ordeleted at will without affecting the value of a product involving anynumber of group elements. Thus,

a2(bcdb2) 5 a2b(cd )b2 5 (a2b)(cd )b2 5 a(abcdb)b,

and so on.Although groups do not have the property that (ab)n 5 anbn there is

a simple relationship between (ab)21 and a21 and b21.

Theorem 2.4 Socks-Shoes Property

PROOF Since (ab)(ab)21 5 e and (ab)(b21a21) 5 a(bb21)a21 5aea21 5 aa21 5 e, we have by Theorem 2.3 that (ab)21 5 b21a21.

For group elements a and b, (ab)21 5 b21a21.

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2 | Groups 51

Historical NoteWe conclude this chapter with a bit of history concerning the non-commutativity of matrix multiplication. In 1925, quantum theory wasreplete with annoying and puzzling ambiguities. It was WernerHeisenberg who recognized the cause. He observed that the product ofthe quantum-theoretical analogs of the classical Fourier series did notnecessarily commute. For all his boldness, this shook Heisenberg. Ashe later recalled [2, p. 94]:

In my paper the fact that XY was not equal to YX was very disagreeable to me. I feltthis was the only point of difficulty in the whole scheme, otherwise I would be per-fectly happy. But this difficulty had worried me and I was not able to solve it.

Heisenberg asked his teacher, Max Born, if his ideas were worth pub-lishing. Born was fascinated and deeply impressed by Heisenberg’s newapproach. Born wrote [1, p. 217]:

After having sent off Heisenberg’s paper to the Zeitschrift für Physik for publica-tion, I began to ponder over his symbolic multiplication, and was soon so involvedin it that I thought about it for the whole day and could hardly sleep at night. For Ifelt there was something fundamental behind it, the consummation of our endeav-ors of many years. And one morning, about the 10 July 1925, I suddenly saw light:Heisenberg’s symbolic multiplication was nothing but the matrix calculus, well-known to me since my student days from Rosanes’ lectures in Breslau.

Born and his student, Pascual Jordan, reformulated Heisenberg’s ideasin terms of matrices, but it was Heisenberg who was credited with theformulation. In his autobiography, Born laments [1, p. 219]:

Nowadays the textbooks speak without exception of Heisenberg’s matrices, Heisen-berg’s commutation law, and Dirac’s field quantization.

In fact, Heisenberg knew at that time very little of matrices and had to studythem.

Upon learning in 1933 that he was to receive the Nobel Prizewith Dirac and Schrödinger for this work, Heisenberg wrote to Born[1, p. 220]:

If I have not written to you for such a long time, and have not thanked you for yourcongratulations, it was partly because of my rather bad conscience with respect toyou. The fact that I am to receive the Nobel Prize alone, for work done in Göttingenin collaboration—you, Jordan, and I—this fact depresses me and I hardly knowwhat to write to you. I am, of course, glad that our common efforts are now appre-ciated, and I enjoy the recollection of the beautiful time of collaboration. I also be-lieve that all good physicists know how great was your and Jordan’s contribution tothe structure of quantum mechanics—and this remains unchanged by a wrong deci-sion from outside. Yet I myself can do nothing but thank you again for all the finecollaboration, and feel a little ashamed.

The story has a happy ending, however, because Born received theNobel Prize in 1954 for his fundamental work in quantum mechanics.

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52 Groups

Exercises

“For example,” is not proof.Jewish Proverb

1. Give two reasons why the set of odd integers under addition is nota group.

2. Referring to Example 13, verify the assertion that subtraction is notassociative.

3. Show that {1, 2, 3} under multiplication modulo 4 is not a groupbut that {1, 2, 3, 4} under multiplication modulo 5 is a group.

4. Show that the group GL(2, R) of Example 9 is non-Abelian by ex-hibiting a pair of matrices A and B in GL(2, R) such that AB 2 BA.

5. Find the inverse of the element in GL(2, Z11).

6. Give an example of group elements a and b with the property thata21ba 2 b.

7. Translate each of the following multiplicative expressions into itsadditive counterpart. Assume that the operation is commutative.a. a2b3

b. a22(b21c)2

c. (ab2)23c2 5 e8. Show that the set {5, 15, 25, 35} is a group under multiplication

modulo 40. What is the identity element of this group? Can you seeany relationship between this group and U(8)?

10. List the members of H 5 {x 2 | x [ D4} and K 5 {x [ D4 | x 2 5 e}.11. Prove that the set of all 2 3 2 matrices with entries from R and de-

terminant 11 is a group under matrix multiplication.12. For any integer n . 2, show that there are at least two elements in

U(n) that satisfy x2 5 1.

c2 6

3 5d

16509_ch02_p040-056 pp3 11/15/08 11:14 AM Page 52

Text not available due to copyright restrictions

2 | Groups 53

13. An abstract algebra teacher intended to give a typist a list of nine in-tegers that form a group under multiplication modulo 91. Instead,one of the nine integers was inadvertently left out, so that the list ap-peared as 1, 9, 16, 22, 53, 74, 79, 81. Which integer was left out?(This really happened!)

14. Let G be a group with the following property: Whenever a, b, andc belong to G and ab 5 ca, then b 5 c. Prove that G is Abelian.(“Cross cancellation” implies commutativity.)

15. (Law of Exponents for Abelian Groups) Let a and b be elements ofan Abelian group and let n be any integer. Show that (ab)n 5 anbn.Is this also true for non-Abelian groups?

16. (Socks-Shoes Property) Draw an analogy between the statement(ab)21 5 b21a21 and the act of putting on and taking off your socksand shoes. Find an example that shows that in a group, it is possibleto have (ab)22 2 b22a22. Find distinct nonidentity elements a andb from a non-Abelian group such that (ab)21 5 a21b21.

17. Prove that a group G is Abelian if and only if (ab)21 5 a21b21 forall a and b in G.

18. Prove that in a group, (a21)21 5 a for all a.

19. For any elements a and b from a group and any integer n, provethat (a21ba)n 5 a21bna.

20. If a1, a2, . . . , an belong to a group, what is the inverse of a1a2. . . an?

21. The integers 5 and 15 are among a collection of 12 integers thatform a group under multiplication modulo 56. List all 12.

22. Give an example of a group with 105 elements. Give two examplesof groups with 44 elements.

23. Prove that every group table is a Latin square†; that is, each ele-ment of the group appears exactly once in each row and each col-umn. (This exercise is referred to in this chapter.)

24. Construct a Cayley table for U(12).

25. Suppose the table below is a group table. Fill in the blank entries.

†Latin squares are useful in designing statistical experiments. There is also a close con-nection between Latin squares and finite geometries.

e a b c d

e e — — — —a — b — — eb — c d e —c — d — a b

d — — — — —

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54 Groups

26. Prove that if (ab)2 5 a2b2 in a group G, then ab 5 ba.27. Let a, b, and c be elements of a group. Solve the equation axb 5 c

for x. Solve a21xa 5 c for x.28. Prove that the set of all rational numbers of the form 3m6n, where

m and n are integers, is a group under multiplication.29. Let G be a finite group. Show that the number of elements x of G

such that x3 5 e is odd. Show that the number of elements x of Gsuch that x2 2 e is even.

30. Give an example of a group with elements a, b, c, d, and x suchthat axb 5 cxd but ab 2 cd. (Hence “middle cancellation” is notvalid in groups.)

31. Let R be any rotation in some dihedral group and F any reflectionin the same group. Prove that RFR 5 F.

32. Let R be any rotation in some dihedral group and F, any reflectionin the same group. Prove that FRF 5 R21 for all integers k.

33. Suppose that G is a group with the property that for every choiceof elements in G, axb 5 cxd implies ab 5 cd. Prove that G isAbelian. (“Middle cancellation” implies commutativity.)

34. In the dihedral group Dn, let R 5 R360/n and let F be any reflection.Write each of the following products in the form Ri or RiF, where 0 # i , n.a. In D4, FR22FR5

b. In D5, R23FR4FR22

c. In D6, FR5FR22F35. Prove that if G is a group with the property that the square of every

element is the identity, then G is Abelian. (This exercise is referredto in Chapter 26.)

36. Prove that the set of all 3 3 3 matrices with real entries of the form

is a group. (Multiplication is defined by

This group, sometimes called the Heisenberg group after theNobel Prize–winning physicist Werner Heisenberg, is intimately re-lated to the Heisenberg Uncertainty Principle of quantum physics.)

£1

a 1 a9

1

b9 1 ac9 1 b

c9 1 c

1

§.£ 1

a9

1

b9

c9 § 1

5 £ 1

a

1

b

c § 1

£1

a

1

b

c

1

§

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37. Prove the assertion made in Example 19 that the set {1, 2, . . . ,n 2 1} is a group under multiplication modulo n if and only if n isprime.

38. In a finite group, show that the number of nonidentity elementsthat satisfy the equation x5 5 e is a multiple of 4. If the stipulationthat the group be finite is omitted, what can you say about thenumber of nonidentity elements that satisfy the equation x5 5 e?

39. Let Show that G is a group under

matrix multiplication. Explain why each element of G has an inverseeven though the matrices have 0 determinant. (Compare with Exam-ple 10.)

Computer Exercises

Almost immediately after the war, Johnny [Von Neumann] and I also beganto discuss the possibilities of using computers heuristically to try to obtaininsights into questions of pure mathematics. By producing examples and byobserving the properties of special mathematical objects, one could hope toobtain clues as to the behavior of general statements which have beentested on examples.

S. M. ULAM, Adventures of a Mathematician

Software for the computer exercises in this chapter is available at the web-site:

http://www.d.umn.edu/~jgallian

1. This software prints the elements of U(n) and the inverse of each element.

2. This software determines the size of U(k). Run the program for k 5 9, 27, 81, 243, 25, 125, 49, 121. On the basis of this output, tryto guess a formula for the size of U( pn) as a function of the primep and the integer n. Run the program for k 5 18, 54, 162, 486, 50,250, 98, 242. Make a conjecture about the relationship between thesize of U(2pn) and the size of U( pn), where p is a prime greaterthan 2.

3. This software computes the inverse of any element in GL(2, Zp),where p is a prime.

4. This software determines the number of elements in GL(2, Zp) andSL(2, Zp). (The technical term for the number of elements in a groupis the order of the group.) Run the program for p 5 3, 5, 7, and 11.

G 5 e ca a

a ad 0 aPR, a ? 0 f .

2 | Groups 55

16509_ch02_p040-056 pp3 11/15/08 11:14 AM Page 55

Do you see a relationship between the orders of GL (2, Zp) andSL(2, Zp) and p 2 1? Does this relationship hold for p 5 2? Basedon these examples, does it appear that p always divides the orderof SL (2, Zp)? What about p 2 1? What about p 1 1? Guess aformula for the order of SL (2, Zp). Guess a formula for the orderof GL (2, Zp).

References

1. Max Born, My Life: Recollections of a Nobel Laureate, New York:Charles Scribner’s Sons, 1978.

2. J. Mehra and H. Rechenberg, The Historical Development of QuantumTheory, Vol. 3, New York: Springer-Verlag, 1982.

Suggested Readings

Marcia Ascher, Ethnomathematics, Pacific Grove, CA: Brooks/Cole,1991.

Chapter 3 of this book describes how the dihedral group of order 8 canbe used to encode the social structure of the kin system of family rela-tionships among a tribe of native people of Australia.

Arie Bialostocki, “An Application of Elementary Group Theory to CentralSolitaire,” The College Mathematics Journal, May 1998: 208–212.

The author uses properties of groups to analyze the peg board gamecentral solitaire (which also goes by the name peg solitaire).

J. E. White, “Introduction to Group Theory for Chemists,” Journal ofChemical Education 44 (1967): 128–135.

Students interested in the physical sciences may find this article worth-while. It begins with easy examples of groups and builds up to applica-tions of group theory concepts and terminology to chemistry.

56 Groups

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Finite Groups;Subgroups

In our own time, in the period 1960–1980, we have seen particle physicsemerge as the playground of group theory.

FREEMAN DYSON

57

3

Terminology and NotationAs we will soon discover, finite groups—that is, groups with finitelymany elements—have interesting arithmetic properties. To facilitatethe study of finite groups, it is convenient to introduce some terminol-ogy and notation.

Definition Order of a Group

The number of elements of a group (finite or infinite) is called itsorder. We will use |G| to denote the order of G.

Thus, the group Z of integers under addition has infinite order,whereas the group U(10) 5 {1, 3, 7, 9} under multiplication modulo10 has order 4.

Definition Order of an Element

The order of an element g in a group G is the smallest positive integern such that gn 5 e. (In additive notation, this would be ng 5 0.) If nosuch integer exists, we say that g has infinite order. The order of anelement g is denoted by |g|.

So, to find the order of a group element g, you need only compute thesequence of products g, g2, g3, . . . , until you reach the identity for the firsttime. The exponent of this product (or coefficient if the operation is addi-tion) is the order of g. If the identity never appears in the sequence, theng has infinite order.

EXAMPLE 1 Consider U(15) 5 {1, 2, 4, 7, 8, 11, 13, 14} undermultiplication modulo 15. This group has order 8. To find the order of

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58 Groups

the element 7, say, we compute the sequence 71 5 7, 72 5 4, 73 5 13,74 5 1, so |7| 5 4. To find the order of 11, we compute 111 5 11,112 5 1, so |11| 5 2. Similar computations show that |1| 5 1, |2| 5 4,|4| 5 2, |8| 5 4, |13| 5 4, |14| 5 2. [Here is a trick that makes thesecalculations easier. Rather than compute the sequence 131, 132, 133,134, we may observe that 13 5 22 mod 15, so that 132 5 (22)2 5 4,133 5 22 ? 4 5 28, 134 5 (22)(28) 5 1.]†

EXAMPLE 2 Consider Z10 under addition modulo 10. Since 1 ? 2 5 2,2 ? 2 5 4, 3 ? 2 5 6, 4 ? 2 5 8, 5 ? 2 5 0, we know that |2| 5 5. Similarcomputations show that |0| 5 1, |7| 5 10, |5| 5 2, |6| 5 5. (Here 2 ? 2 isan abbreviation for 2 1 2, 3 ? 2 is an abbreviation for 2 1 2 1 2, etc.)

EXAMPLE 3 Consider Z under ordinary addition. Here every nonzeroelement has infinite order, since the sequence a, 2a, 3a, . . . never includes0 when a � 0.

The perceptive reader may have noticed among our examples ofgroups in Chapter 2 that some are subsets of others with the samebinary operation. The group SL(2, R) in Example 17, for instance, isa subset of the group GL(2, R) in Example 9. Similarly, the group ofcomplex numbers {1, 21, i, 2i} under multiplication is a subset ofthe group described in Example 14 for n equal to any multiple of 4.This situation arises so often that we introduce a special term to de-scribe it.

Definition Subgroup

If a subset H of a group G is itself a group under the operation of G, wesay that H is a subgroup of G.

We use the notation H # G to mean that H is a subgroup of G. If wewant to indicate that H is a subgroup of G but is not equal to G itself,we write H , G. Such a subgroup is called a proper subgroup. Thesubgroup {e} is called the trivial subgroup of G; a subgroup that is not{e} is called a nontrivial subgroup of G.

Notice that Zn under addition modulo n is not a subgroup of Z underaddition, since addition modulo n is not the operation of Z.

Subgroup TestsWhen determining whether or not a subset H of a group G is a sub-

† The website www.google.com provides a convenient way to do modular arithmetic.For example, to compute 134 mod 15, just type 13ˆ4 mod 15 in the search box.

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3 | Finite Groups; Subgroups 59

group of G, one need not directly verify the group axioms. The nextthree results provide simple tests that suffice to show that a subset of agroup is a subgroup.

Theorem 3.1 One-Step Subgroup Test

PROOF Since the operation of H is the same as that of G, it is clearthat this operation is associative. Next, we show that e is in H. Since His nonempty, we may pick some x in H. Then, letting a 5 x and b 5 x inthe hypothesis, we have e 5 xx21 5 ab21 is in H. To verify that x21 isin H whenever x is in H, all we need to do is to choose a 5 e and b 5x in the statement of the theorem. Finally, the proof will be completewhen we show that H is closed; that is, if x, y belong to H, we mustshow that xy is in H also. Well, we have already shown that y21 is in Hwhenever y is; so, letting a 5 x and b 5 y21, we have xy 5 x(y21)21 5ab21 is in H.

Although we have dubbed Theorem 3.1 the “One-Step SubgroupTest,’’ there are actually four steps involved in applying the theorem.(After you gain some experience, the first three steps will be routine.)Notice the similarity between the last three steps listed below and thethree steps involved in the Principle of Mathematical Induction.

1. Identify the property P that distinguishes the elements of H; that is,identify a defining condition.

2. Prove that the identity has property P. (This verifies that H isnonempty.)

3. Assume that two elements a and b have property P.4. Use the assumption that a and b have property P to show that

ab21 has property P.

The procedure is illustrated in Examples 4 and 5.

EXAMPLE 4 Let G be an Abelian group with identity e. Then H 5{x [ G | x2 5 e} is a subgroup of G. Here, the defining property of His the condition x2 5 e. So, we first note that e2 5 e, so that H is non-empty. Now we assume that a and b belong to H. This means that a2 5 eand b2 5 e. Finally, we must show that (ab21)2 5 e. Since G isAbelian, (ab21)2 5 ab21ab21 5 a2(b21)2 5 a2(b2)21 5 ee21 5 e.Therefore, ab21 belongs to H and, by the One-Step Subgroup Test, His a subgroup of G.

Let G be a group and H a nonempty subset of G. If ab21 is in Hwhenever a and b are in H, then H is a subgroup of G. (In additivenotation, if a 2 b is in H whenever a and b are in H, then H is asubgroup of G.)

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60 Groups

In many instances, a subgroup will consist of all elements that havea particular form. Then the property P is that the elements have thatparticular form. This is illustrated in the following example.

EXAMPLE 5 Let G be an Abelian group under multiplication withidentity e. Then H 5 {x2 | x [ G} is a subgroup of G. (In words, H isthe set of all “squares.”) Since e2 5 e, the identity has the correct form.Next, we write two elements of H in the correct form, say, a2 and b2. Wemust show that a2(b2)21 also has the correct form; that is, a2(b2)21 is thesquare of some element. Since G is Abelian, we may write a2(b2)21 as(ab21)2, which is the correct form. Thus, H is a subgroup of G.

Beginning students often prefer to use the next theorem instead ofTheorem 3.1.

Theorem 3.2 Two-Step Subgroup Test

PROOF By Theorem 3.1, it suffices to show that a, b [ H implies ab21 [ H. So, we suppose that a, b [ H. Since H is closed under taking inverses, we also have b21 [ H. Thus, ab21 [ H by closure un-der multiplication.

When applying the “Two-Step Subgroup Test,” we proceed exactlyas in the case of the “One-Step Subgroup Test,” except we use the as-sumption that a and b have property P to prove that ab has property Pand that a21 has property P.

How do you prove that a subset of a group is not a subgroup? Hereare three possible ways, any one of which guarantees that the subset isnot a subgroup:

1. Show that the identity is not in the set.2. Exhibit an element of the set whose inverse is not in the set.3. Exhibit two elements of the set whose product is not in the set.

EXAMPLE 6 Let G be the group of nonzero real numbers undermultiplication, H 5 {x [ G | x 5 1 or x is irrational} and K 5{x [ G | x $ 1}. Then H is not a subgroup of G, since [ Hbut 5 2 o H. Also, K is not a subgroup, since 2 [ K but221 o K.

"2 ? "2"2

Let G be a group and let H be a nonempty subset of G. If ab is in Hwhenever a and b are in H (H is closed under the operation), and a21

is in H whenever a is in H (H is closed under taking inverses), then His a subgroup of G.

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When dealing with finite groups, it is easier to use the followingsubgroup test.

Theorem 3.3 Finite Subgroup Test

PROOF In view of Theorem 3.2, we need only prove that a21 [ Hwhenever a [ H. If a 5 e, then a21 5 a and we are done. If a � e,consider the sequence a, a2, . . . . By closure, all of these elementsbelong to H. Since H is finite, not all of these elements are distinct. Sayai 5 aj and i . j. Then, ai2j 5 e; and since a � e, i 2 j . 1. Thus,aai2j21 5 ai2j 5 e and, therefore, ai2j21 5 a21. But, i 2 j 2 1 $ 1implies ai2j21 [ H and we are done.

Examples of SubgroupsThe proofs of the next few theorems show how our subgroup testswork. We first introduce an important notation. For any element a froma group, we let �a� denote the set {an | n [ Z}. In particular, observethat the exponents of a include all negative integers as well as 0 and thepositive integers (a0 is defined to be the identity).

Theorem 3.4 �a� Is a Subgroup

PROOF Since a [ �a�, �a� is not empty. Let an, am [ �a�. Then,an(am)21 5 an2m [ �a�; so, by Theorem 3.1, �a� is a subgroup of G.

The subgroup �a� is called the cyclic subgroup of G generated by a. Inthe case that G 5 �a�, we say that G is cyclic and a is a generator of G.(A cyclic group may have many generators.) Notice that although thelist . . . , a22, a21, a0, a1, a2, . . . has infinitely many entries, the set {an | n [ Z} might have only finitely many elements. Also note that,since aiaj 5 ai1j 5 aj1i 5 ajai, every cyclic group is Abelian.

EXAMPLE 7 In U(10), �3� 5 {3, 9, 7, 1} 5 U(10), for 31 5 3,32 5 9, 33 5 7, 34 5 1, 35 5 34 ? 3 5 1 ? 3, 36 5 34 ? 32 5 9, . . . ; 321 5 7

Let G be a group, and let a be any element of G. Then, �a� is a sub-group of G.

Let H be a nonempty finite subset of a group G. If H is closed underthe operation of G, then H is a subgroup of G.

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62 Groups

Rn 5 e

Rn11 5 R R21 5 Rn21

R22 5 Rn22 Rn12 5 R2

(since 3 ? 7 5 1), 322 5 9, 323 5 3, 324 5 1, 325 5 324 ? 321 51 ? 7, 326 5 324 ? 322 5 1 ? 9 5 9, . . . .

EXAMPLE 8 In Z10, �2� 5 {2, 4, 6, 8, 0}. Remember, an means nawhen the operation is addition.

EXAMPLE 9 In Z, �21� 5 Z. Here each entry in the list . . . ,22(21), 21(21), 0(21), 1(21), 2(21), . . . represents a distinct groupelement.

EXAMPLE 10 In Dn, the dihedral group of order 2n, let R denote arotation of 360/n degrees. Then,

Rn 5 R360° 5 e, Rn11 5 R, Rn12 5 R2, . . . .

Similarly, R21 5 Rn21, R22 5 Rn22, . . . , so that �R� 5 {e, R, . . . ,Rn21}. We see, then, that the powers of R “cycle back” periodicallywith period n. Visually, raising R to successive positive powers is thesame as moving counterclockwise around the following circle onenode at a time, whereas raising R to successive negative powers is thesame as moving around the circle clockwise one node at a time.

In Chapter 4 we will show that |�a�| 5 |a|; that is, the order of thesubgroup generated by a is the order of a itself. (Actually, the definitionof |a| was chosen to ensure the validity of this equation.)

We next consider one of the most important subgroups.

Definition Center of a Group

The center, Z(G ), of a group G is the subset of elements in G thatcommute with every element of G. In symbols,

Z(G) 5 {a [ G | ax 5 xa for all x in G}.

[The notation Z(G) comes from the fact that the German word forcenter is Zentrum. The term was coined by J. A. de Seguier in 1904.]

Theorem 3.5 Center Is a Subgroup

The center of a group G is a subgroup of G.

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PROOF For variety, we shall use Theorem 3.2 to prove this result.Clearly, e [ Z(G), so Z(G) is nonempty. Now, suppose a, b [ Z(G).Then (ab)x 5 a(bx) 5 a(xb) 5 (ax)b 5 (xa)b 5 x(ab) for all x in G;and, therefore, ab [ Z(G).

Next, assume that a [ Z(G). Then we have ax 5 xa for all x in G.What we want is a21x 5 xa21 for all x in G. Informally, all we need doto obtain the second equation from the first one is simultaneously tobring the a’s across the equals sign:

ax 5 xa

becomes xa21 5 a21x. (Be careful here; groups need not be commuta-tive. The a on the left comes across as a21 on the left, and the a on theright comes across as a21 on the right.) Formally, the desired equationcan be obtained from the original one by multiplying it on the left andright by a21, like so:

a21(ax)a21 5 a21(xa)a21,(a21a)xa21 5 a21x(aa21),

exa21 5 a21xe,xa21 5 a21x.

This shows that a21 [ Z(G) whenever a is.

For practice, let’s determine the centers of the dihedral groups.

EXAMPLE 11 For n $ 3,

To verify this, first observe that since every rotation in Dn is a powerof R360/n, rotations commute with rotations. We now investigate when arotation commutes with a reflection. Let R be any rotation in Dn and let F be any reflection in Dn. Observe that since RF is a reflection we haveRF 5 (RF )21 5 F21 R21 5 FR21. Thus it follows that R and F commuteif and only if FR 5 RF 5 FR21. By cancellation, this holds if and only ifR 5 R21. But R 5 R21 only when R 5 R0 or R 5 R180, and R180 is in Dnonly when n is even. So, we have proved that Z(Dn) 5 {R0} when n isodd and Z(Dn) 5 {R0, R180} when n is even.

Although an element from a non-Abelian group does not necessarilycommute with every element of the group, there are always someelements with which it will commute. For example, every element a

when n is even,

when n is odd.Z(Dn) 5 e 5R0, R18065R06

3 | Finite Groups; Subgroups 63

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64 Groups

commutes with all powers of a. This observation prompts the next def-inition and theorem.

Definition Centralizer of a in G

Let a be a fixed element of a group G. The centralizer of a in G, C(a), isthe set of all elements in G that commute with a. In symbols, C(a) 5{g [ G | ga 5 ag}.

EXAMPLE 12 In D4, we have the following centralizers:

C(R0) 5 D4 5 C(R180),C(R90) 5 {R0, R90, R180, R270} 5 C(R270),

C(H) 5 {R0, H, R180, V} 5 C(V),C(D) 5 {R0, D, R180, D9} 5 C(D9).

Notice that each of the centralizers in Example 12 is actually a sub-group of D4. The next theorem shows that this was not a coincidence.

Theorem 3.6 C(a) Is a Subgroup

PROOF A proof similar to that of Theorem 3.5 is left to the reader tosupply (Exercise 25).

Notice that for every element a of a group G, Z(G) # C(a). Also,observe that G is Abelian if and only if C(a) 5 G for all a in G.

Exercises

The purpose of proof is to understand, not to verify.ARNOLD ROSS

1. For each group in the following list, find the order of the groupand the order of each element in the group. What relation do yousee between the orders of the elements of a group and the order ofthe group?

Z12, U(10), U(12), U(20), D4

2. Let Q be the group of rational numbers under addition and let Q*be the group of nonzero rational numbers under multiplication.In Q, list the elements in � �. In Q*, list the elements in � �.1

212

For each a in a group G, the centralizer of a is a subgroup of G.

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3. Let Q and Q* be as in Exercise 2. Find the order of each elementin Q and in Q*.

4. Prove that in any group, an element and its inverse have thesame order.

5. Without actually computing the orders, explain why the two ele-ments in each of the following pairs of elements from Z30 musthave the same order: {2, 28}, {8, 22}. Do the same for the follow-ing pairs of elements from U(15): {2, 8}, {7, 13}.

6. Suppose that a is a group element and a6 5 e. What are the possi-bilities for |a|? Provide reasons for your answer.

7. If a is a group element and a has infinite order, prove that am ? an

when m ? n.8. Let x belong to a group. If x2 2 e and x6 5 e, prove that x4 2 e and

x5 � e. What can we say about the order of x?9. Show that if a is an element of a group G, then |a| # |G|.

10. Show that U(14) 5 �3� 5 �5�. [Hence, U(14) is cyclic.] Is U(14) 5 �11�?

11. Show that U(20) 2 �k� for any k in U(20). [Hence, U(20) is notcyclic.]

12. Prove that an Abelian group with two elements of order 2 musthave a subgroup of order 4.

13. Find groups that contain elements a and b such that |a| 5 |b| 5 2anda. |ab| 5 3, b. |ab| 5 4, c. |ab| 5 5.Can you see any relationship among |a|, |b|, and |ab|?

14. Suppose that H is a proper subgroup of Z under addition and Hcontains 18, 30, and 40. Determine H.

15. Suppose that H is a proper subgroup of Z under addition and that Hcontains 12, 30 and 54. What are the possibilities for H?

16. Prove that the dihedral group of order 6 does not have a subgroupof order 4.

17. For each divisor k . 1 of n, let Uk(n) 5 {x [ U(n) | x mod k 5 1}.[For example, U3(21) 5 {1, 4, 10, 13, 16, 19} and U7(21) 5 {1, 8}.]List the elements of U4(20), U5(20), U5(30), and U10(30). Prove thatUk(n) is a subgroup of U(n). Let H 5 {x [ U(10) | x mod 3 5 1}. IsH a subgroup of U(10)? (This exercise is referred to in Chapter 8.)

18. If H and K are subgroups of G, show that H > K is a subgroup ofG. (Can you see that the same proof shows that the intersectionof any number of subgroups of G, finite or infinite, is again asubgroup of G?)

3 | Finite Groups; Subgroups 65

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66 Groups

19. Let G be a group. Show that Z(G) 5 >a[GC(a). [This means theintersection of all subgroups of the form C(a).]

20. Let G be a group, and let a [ G. Prove that C(a) 5 C(a21).21. For any group element a and any integer k, show that C(a) # C(ak).

Use this fact to complete the following statement: “In a group, if Ris an integer and x commutes with a, then . . . .” Is the converse true?

22. Complete the partial Cayley group table given below.

1 2 3 4 5 6 7 8

1 1 2 3 4 5 6 7 82 2 1 4 3 6 5 8 73 3 4 2 1 7 8 6 54 4 3 1 2 8 7 5 65 5 6 8 7 16 6 5 7 8 17 7 8 5 6 18 8 7 6 5 1

1 2 3 4 5 6 7 8

1 1 2 3 4 5 6 7 82 2 1 8 7 6 5 4 33 3 4 5 6 7 8 1 24 4 3 2 1 8 7 6 55 5 6 7 8 1 2 3 46 6 5 4 3 2 1 8 77 7 8 1 2 3 4 5 68 8 7 6 5 4 3 2 1

23. Suppose G is the group defined by the following Cayley table.

a. Find the centralizer of each member of G.b. Find Z(G).c. Find the order of each element of G. How are these orders arith-

metically related to the order of the group?

24. If a and b are distinct group elements, prove that either a2 2 b2 ora3 2 b3.

25. Prove Theorem 3.6.

26. If H is a subgroup of G, then by the centralizer C(H) of H we meanthe set {x [ G | xh 5 hx for all h [ H}. Prove that C(H) is a sub-group of G.

27. Must the centralizer of an element of a group be Abelian?

28. Must the center of a group be Abelian?

29. Let G be an Abelian group with identity e and let n be some fixed in-teger. Prove that the set of all elements of G that satisfy the equation

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xn 5 e is a subgroup of G. Give an example of a group G in whichthe set of all elements of G that satisfy the equation x2 5 e does notform a subgroup of G. (This exercise is referred to in Chapter 11.)

30. Suppose a belongs to a group and |a| 5 5. Prove that C(a) 5 C(a3).Find an element a from some group such that |a| 5 6 and C(a) �C(a3).

31. Determine all finite subgroups of R*, the group of nonzero realnumbers under multiplication.

32. Suppose n is an even positive integer and H is a subgroup of Zn.Prove that either every member of H is even or exactly half of themembers of H are even.

33. Suppose a group contains elements a and b such that |a| 5 4,|b| 5 2, and a3b 5 ba. Find |ab|.

34. Suppose a and b are group elements such that |a| 5 2, b ? e, andaba 5 b2. Determine |b|.

35. Let a be a group element of order n, and suppose that d is a posi-tive divisor of n. Prove that |ad | 5 n/d.

36. Consider the elements and from

SL(2, R). Find |A|, |B|, and |AB|. Does your answer surprise you?

37. Consider the element in SL(2, R). What is the order of

A? If we view as a member of SL(2, Zp) (p is a prime),

what is the order of A?38. For any positive integer n and any angle u, show that in the group

SL(2, R),

Use this formula to find the order of

(Geometrically, represents a rotation of the plane

u degrees.)39. Let G be the symmetry group of a circle. Show that G has elements

of every finite order as well as elements of infinite order.

ccos u 2 sin u

sin u cos ud

ccos 60° 2 sin 60°

sin 60° cos 60°d

and ccos "2° 2 sin "2°

sin "2° cos "2°d .

ccos u 2 sin u

sin u cos ud n

5 ccos nu 2 sin nu

sin nu cos nud .

A 5 c1 1

0 1d

A 5 c1 1

0 1d

B 5 c 0 1

21 21dA 5 c0 21

1 0d

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68 Groups

40. Let x belong to a group and |x| 5 6. Find |x2|, |x3|, |x4|, and |x5|. Lety belong to a group and |y| 5 9. Find |yi| for i 5 2, 3, . . . , 8. Dothese examples suggest any relationship between the order of thepower of an element and the order of the element?

41. D4 has seven cyclic subgroups. List them. Find a subgroup of D4 oforder 4 that is not cyclic.

42. U(15) has six cyclic subgroups. List them.43. Prove that a group of even order must have an element of order 2.44. Suppose G is a group that has exactly eight elements of order 3.

How many subgroups of order 3 does G have?45. Let H be a subgroup of a finite group G. Suppose that g belongs to

G and n is the smallest positive integer such that gn [ H. Prove thatn divides |g|.

46. Compute the orders of the following groups.a. U(3), U(4), U(12)b. U(5), U(7), U(35)c. U(4), U(5), U(20)d. U(3), U(5), U(15)On the basis of your answers, make a conjecture about the relation-ship among |U(r)|, |U(s)|, and |U(rs)|.

47. Let R* be the group of nonzero real numbers under multiplicationand let H 5 {x [ R* | x2 is rational}. Prove that H is a subgroup ofR*. Can the exponent 2 be replaced by any positive integer and stillhave H be a subgroup?

48. Compute |U(4)|, |U(10)|, and |U(40)|. Do these groups provide acounterexample to your answer to Exercise 46? If so, revise yourconjecture.

49. Find a cyclic subgroup of order 4 in U(40).50. Find a noncyclic subgroup of order 4 in U(40).

51. Let G 5 under addition. Let H 5

. Prove that H is a subgroup of G.

What if 0 is replaced by 1?52. Let H 5 {A [ GL(2, R)| det A is an integer power of 2}. Show that

H is a subgroup of GL(2, R).53. Let H be a subgroup of R under addition. Let K 5 {2a | a [ H}.

Prove that K is a subgroup of R* under multiplication.

e ca b

c ddP G | a 1 b 1 c 1 d 5 0f

e ca b

c d d 0 a, b, c, d [ Z f

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54. Let G be a group of functions from R to R*, where the operationof G is multiplication of functions. Let H 5 { f [ G | f(2) 5 1}.Prove that H is a subgroup of G. Can 2 be replaced by any realnumber?

55. Let G 5 GL(2, R) and inte-

under the operation of matrix multiplication. Prove or

disprove that H is a subgroup of GL(2, R).56. Let H 5 {a 1 bi | a, b [ R, ab $ 0}. Prove or disprove that H is a

subgroup of C under addition.57. Let H 5 {a 1 bi | a, b [ R, a2 1 b2 5 1}. Prove or disprove that

H is a subgroup of C* under multiplication. Describe the elementsof H geometrically.

58. The smallest subgroup containing a collection of elements S is thesubgroup H with the property that if K is any subgroup containingS then K also contains H. (So, the smallest subgroup containing S iscontained in every subgroup that contains S.) The notation for thissubgroup is �S�. In the group Z, finda. �8, 14�b. �8, 13�c. �6, 15�d. �m, n�e. �12, 18, 45�.In each part, find an integer k such that the subgroup is �k�.

59. Let G 5 GL(2, R).

a. Find C .

b. Find C .

c. Find Z(G).60. Let G be a finite group with more than one element. Show that G

has an element of prime order.61. Let a belong to a group and |a| 5 m. If n is relatively prime to m,

show that a can be written as the nth power of some element in thegroup.

62. Let G be a finite Abelian group and let a and b belong to G. Provethat the set Ka, bL 5 {aib j | i, j [ Z} is a subgroup of G. What canyou say about |Ka, bL| in terms of |a| and |b|?

a c0 1

1 0d b

a c1 1

1 0d b

gers fH 5 e ca 0

0 b d 0 a and b are nonzero

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70 Groups

Computer Exercises

A Programmer’s Lament

I really hate this damned machine;

I wish that they would sell it

It never does quite what I want

but only what I tell it.

DENNIE L. VAN TASSEL, The Compleat Computer

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software determines the cyclic subgroups of U(n) generatedby each k in U(n) (n , 100). Run the program for n 5 12, 15, and30. Compare the order of the subgroups with the order of the groupitself. What arithmetic relationship do these integers have?

2. The program lists the elements of Zn that generate all of Zn—that is,those elements k, 0 # k # n 2 1, for which Zn 5 �k�. How does thisset compare with U(n)? Make a conjecture.

3. This software does the following: For each pair of elements a and bfrom U(n) (n , 100), it prints |a|, |b|, and |ab| on the same line. Runthe program for several values of n. Is there an arithmetic relation-ship between |ab| and |a| and |b|?

4. This exercise repeats Exercise 3 for Zn using a 1 b in place of ab.5. This software computes the order of elements in GL(2, Zp). Enter

several choices for matrices A and B. The software returns |A|, |B|,|AB|, |BA|, |A21BA|, and |B21AB|. Do you see any relationship be-tween |A|, |B| and |AB|? Do you see any relationship between |AB|and |BA|? Make a conjecture about this relationship. Test your con-jecture for several other choices for A and B. Do you see any rela-tionship between |B| and |A21BA|? Do you see any relationshipbetween |A| and |B21AB|? Make a conjecture about this relation-ship. Test your conjecture for several other choices for A and B.

Suggested Readings

Ruth Berger, “Hidden Group Structure,” Mathematics Magazine 78(2005): 45–48.

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In this note, the author investigates groups obtained from U(n) by mul-tiplying each element by some k in U(n). Such groups have identitiesthat are not obvious.

J. Gallian and M. Reid, “Abelian Forcing Sets,” American MathematicalMonthly 100 (1993): 580–582.

A set S is called Abelian forcing if the only groups that satisfy (ab)n 5anbn for all a and b in the group and all n in S are the Abelian ones.This paper characterizes the Abelian forcing sets.

Gina Kolata, “Perfect Shuffles and Their Relation to Math,” Science 216(1982): 505–506.

This is a delightful nontechnical article that discusses how group the-ory and computers were used to solve a difficult problem about shuf-fling a deck of cards. Serious work on the problem was begun by anundergraduate student as part of a programming course.

Suggested Software

Allen Hibbard and Kenneth Levasseur, Exploring Abstract Algebra withMathematica, New York: Springer-Verlag, 1999.

This book, intended as a supplement for a course in abstract algebra,consists of 14 group labs, 13 ring labs, and documentation for theAbstract Algebra software on which the labs are based. The software usesthe Mathematica language, and only a basic familiarity with the programis required. The software can be freely downloaded at http://www.central.edu/eaam/ and can be used independently of the book. This arti-cle can be downloaded at http://www.d.umn.edu/~jgallian/forcing.pdf

3 | Finite Groups; Subgroups 71

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Cyclic Groups

72

4

Properties of Cyclic GroupsRecall from Chapter 3 that a group G is called cyclic if there is an ele-ment a in G such that G 5 {an | n [ Z}. Such an element a is called agenerator of G. In view of the notation introduced in the precedingchapter, we may indicate that G is a cyclic group generated by a bywriting G 5 �a�.

In this chapter, we examine cyclic groups in detail and determinetheir important characteristics. We begin with a few examples.

EXAMPLE 1 The set of integers Z under ordinary addition is cyclic.Both 1 and 21 are generators. (Recall that, when the operation is addi-tion, 1n is interpreted as

1 1 1 1 ? ? ? 1 1

n terms

when n is positive and as

(21) 1 (21) 1 ? ? ? 1 (21)

|n| terms

when n is negative.)

EXAMPLE 2 The set Zn 5 {0, 1, . . . , n 2 1} for n $ 1 is a cyclic group under addition modulo n. Again, 1 and 21 5 n 2 1 are generators.

The notion of a “group,” viewed only 30 years ago as the epitome ofsophistication, is today one of the mathematical concepts most widely used in physics, chemistry, biochemistry, and mathematics itself.

ALEXEY SOSINSKY, 1991

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4 | Cyclic Groups 73

Unlike Z, which has only two generators, Zn may have many genera-tors (depending on which n we are given).

EXAMPLE 3 Z8 5 k1l 5 k3l 5 k5l 5 k7l. To verify, for instance, thatZ8 5 k3l, we note that k3l 5 {3, 3 1 3, 3 1 3 1 3, . . .} is the set {3, 6,1, 4, 7, 2, 5, 0} 5 Z8. Thus, 3 is a generator of Z8. On the other hand, 2is not a generator, since k2l 5 {0, 2, 4, 6} 2 Z8.

EXAMPLE 4 (See Example 11 in Chapter 2.)U(10) 5 {1, 3, 7, 9} 5 {30, 31, 33, 32} 5 �3�. Also, {1, 3, 7, 9} 5{70, 73, 71, 72} 5 �7�. So both 3 and 7 are generators for U(10).

Quite often in mathematics, a “nonexample” is as helpful in under-standing a concept as an example. With regard to cyclic groups, U(8)serves this purpose; that is, U(8) is not a cyclic group. How can we ver-ify this? Well, note that U(8) 5 {1, 3, 5, 7}. But

�1� 5 {1}�3� 5 {3, 1}�5� 5 {5, 1}�7� 5 {7, 1}

so U(8) 2 �a� for any a in U(8).With these examples under our belts, we are now ready to tackle

cyclic groups in an abstract way and state their key properties.

Theorem 4.1 Criterion for ai 5 a j

Let G be a group, and let a belong to G. If a has infinite order, thenif and only if . If a has finite order, say, n, then �a� 5

{e, a, a2, . . . , an–1} and ai 5 aj if and only if n divides i – j.i 5 jai 5 aj

PROOF If a has infinite order, there is no nonzero n such that an is theidentity. Since ai 5 aj implies ai2j 5 e, we must have i 2 j 5 0, and thefirst statement of the theorem is proved.

Now assume that |a| 5 n. We will prove that �a� 5 {e, a, . . . , an21}.Certainly, the elements e, a, . . . , an21 are in �a�.

Now, suppose that ak is an arbitrary member of �a�. By the divisionalgorithm, there exist integers q and r such that

k 5 qn 1 r with 0 # r , n.

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74 Groups

Then ak 5 aqn1r 5 aqnar 5 (an)qar 5 ear 5 ar, so that ak [ {e, a,a2, . . . , an21}. This proves that �a� 5 {e, a, a2, . . . , an21}.

Next, we assume that ai 5 a j and prove that n divides i 2 j. Webegin by observing that ai 5 aj implies ai2j 5 e. Again, by the divisionalgorithm, there are integers q and r such that

i 2 j 5 qn 1 r with 0 # r , n.

Then ai2j 5 aqn1r, and therefore e 5 ai2j 5 aqn1r 5 (an)qar 5 eqar 5ear 5 ar. Since n is the least positive integer such that an is the identity,we must have r 5 0, so that n divides i 2 j.

Conversely, if i 2 j 5 nq, then ai2j 5 anq 5 eq 5 e, so that ai 5 aj.

Theorem 4.1 reveals the reason for the dual use of the notation andterminology for the order of an element and the order of a group.

Corollary 1 |a| 5 |�a�|

One special case of Theorem 4.1 occurs so often that it deservessingling out.

Corollary 2 ak 5 e Implies That |a| Divides k

PROOF Since ak 5 e 5 a0, we know by Theorem 4.1 that n divides k 2 0.

Theorem 4.1 and its corollaries for the case |a| 5 6 are illustrated inFigure 4.1.

What is important about Theorem 4.1 in the finite case is that it saysthat multiplication in �a� is essentially done by addition modulo n. Thatis, if (i 1 j) mod n 5 k, then aia j 5 ak. Thus, no matter what group Gis, or how the element a is chosen, multiplication in �a� works the sameas addition in Zn whenever |a| 5 n. Similarly, if a has infinite order,

Let G be a group and let a be an element of order n in G. If ak 5 e,then n divides k.

For any group element a, |a| 5 |�a�|.

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4 | Cyclic Groups 75

Figure 4.1

then multiplication in �a� works the same as addition in Z, since aia j 5ai1j and no modular arithmetic is done.

For these reasons, the cyclic groups Zn and Z serve as prototypes forall cyclic groups, and algebraists say that there is essentially only onecyclic group of each order. What is meant by this is that, althoughthere may be many different sets of the form {an | n [ Z}, there isessentially only one way to operate on these sets. Algebraists do notreally care what the elements of a set are; they care only about thealgebraic properties of the set—that is, the ways in which the elementsof a set can be combined. We will return to this theme in the chapteron isomorphisms (Chapter 6).

The next theorem provides a simple method for computing |ak|knowing only |a|, and its first corollary provides a simple way to tellwhen �ai� 5 �aj�.

Theorem 4.2 �ak� 5 �agcd(n,k)�

PROOF To simplify the notation, let d 5 gcd(n,k) and let k 5 dr.Since ak 5 (ad)r, we have by closure that �ak� # �ad�. By Theorem 0.2(the gcd theorem), there are integers s and t such that d 5 ns 1 kt. So,ad 5 ans1kt 5 ansakt 5 (an)s(ak)t 5 e(ak)t 5 (ak)t [ �ak�. This proves�ad� # �ak�. So, we have verified that �ak� 5 �agcd(n,k)�.

We prove the second part of the theorem by showing first that |ad| 5n/d for any divisor d of n. Clearly, (ad)n/d 5 an 5 e, so that |ad| # n/d. Onthe other hand, if i is a positive integer less than n/d, then (ad)i 2 e by de-finition of |a|. We now apply this fact with d 5 gcd(n,k) to obtain |ak| 5|�ak�| 5 |�agcd(n,k)�| 5 |agcd(n,k)| 5 n/gcd(n,k).

The advantage of Theorem 4.2 is that it allows us to replace one generator of a cyclic subgroup with a more convenient one. For example,

Let a be an element of order n in a group and let k be a positiveinteger. Then �ak� 5 �agcd(n,k)� and |ak| 5 n/gcd(n,k).

... a–6 = a0 = a6 ...

... a –5 = a = a7...

... a–4 = a 2 = a8 ...

... a–3 = a3 = a9...

... a–2 = a4 = a 10...

... a–1 = a 5 = a 11...

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76 Groups

if |a| 5 30, we have �a26� 5 �a2�, �a23� 5 �a�, �a22� 5 �a2�, �a21� 5 �a3�.From this we can easily see that |a23| 5 30 and |a22| 5 15. Moreover, ifone wants to list the elements of, say, �a21�, it is easier to list the elementsof �a3� instead. (Try it doing it both ways!).

Theorem 4.2 establishes an important relationship between the orderof an element in a finite cyclic group and the order of the group.

Corollary 1 Orders of Elements in Finite Cyclic Groups

Corollary 2 Criterion for �ai� � �aj� and |ai | � |aj |

PROOF Theorem 4.2 shows that �ai� 5 �agcd(n,i)� and �a j� 5 �agcd(n,j)�,so that the proof reduces to proving that �agcd(n,i)� 5 �agcd(n, j)� if andonly if gcd(n, i) 5 gcd(n, j). Certainly, gcd(n, i) 5 gcd(n, j) impliesthat �agcd(n, i)� 5 �agcd(n, j)�. On the other hand, �agcd(n,i)� 5 �agcd(n, j)�implies that |agcd(n,i)|5 |agcd(n, j)|, so that by the second conclusion ofTheorem 4.2, we have n/gcd(n, i) 5 n/gcd(n, j), and therefore gcd(n, i) 5gcd(n, j).

The second part of the corollary follows from the first part andCorollary 1 of Theorem 4.1.

The next two corollaries are important special cases of the precedingcorollary.

Corollary 3 Generators of Finite Cyclic Groups

Corollary 4 Generators of Zn

The value of Corollary 3 is that once one generator of a cyclic group hasbeen found, all generators of the cyclic group can easily be determined.

An integer k in Zn is a generator of Zn if and only if gcd(n, k) 5 1.

Let |a| 5 n. Then �a� 5 �aj� if and only if gcd(n, j) 5 1 and |a| 5 |�aj�| if and only if gcd(n, j) 5 1.

Let |a| 5 n. Then �ai� 5 �aj� if and only if gcd(n, i) 5 gcd(n, j) and |ai| 5 |aj| if and only if gcd(n, i) 5 gcd(n, j) .

In a finite cyclic group, the order of an element divides the order of the group.

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4 | Cyclic Groups 77

For example, consider the subgroup of all rotations in D6. Clearly, onegenerator is R60. And, since |R60| 5 6, we see by Corollary 3 that the onlyother generator is (R60)5 5 R300. Of course, we could have readily deducedthis information without the aid of Corollary 3 by direct calculations. So,to illustrate the real power of Corollary 3, let us use it to find all genera-tors of the cyclic group U(50). First, note that direct computations showthat |U(50)| 5 20 and that 3 is one of its generators. Thus, in view ofCorollary 3, the complete list of generators for U(50) is

3 mod 50 5 3, 311 mod 50 5 47,33 mod 50 5 27, 313 mod 50 5 23,37 mod 50 5 37, 317 mod 50 5 13,39 mod 50 5 33, 319 mod 50 5 17.

Admittedly, we had to do some arithmetic here, but it certainly entailedmuch less work than finding all the generators by simply determiningthe order of each element of U(50) one by one.

The reader should keep in mind that Theorem 4.2 and its corollariesapply only to elements of finite order.

Classification of Subgroupsof Cyclic Groups

The next theorem tells us how many subgroups a finite cyclic group hasand how to find them.

Theorem 4.3 Fundamental Theorem of Cyclic Groups

Before we prove this theorem, let’s see what it means. Understand-ing what a theorem means is a prerequisite to understanding its proof.Suppose G 5 �a� and G has order 30. The first and second parts of thetheorem say that if H is any subgroup of G, then H has the form �a30/k� forsome k that is a divisor of 30. The third part of the theorem says that Ghas one subgroup of each of the orders 1, 2, 3, 5, 6, 10, 15, and 30—andno others. The proof will also show how to find these subgroups.

PROOF Let G 5 �a� and suppose that H is a subgroup of G. We mustshow that H is cyclic. If it consists of the identity alone, then clearly H iscyclic. So we may assume that H 2 {e}. We now claim that H contains

Every subgroup of a cyclic group is cyclic. Moreover, if |�a�| 5 n,then the order of any subgroup of �a� is a divisor of n; and, for eachpositive divisor k of n, the group �a� has exactly one subgroup oforder k—namely, �an/ k�.

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78 Groups

an element of the form at, where t is positive. Since G 5 �a�, everyelement of H has the form at; and when at belongs to H with t , 0, thena2t belongs to H also and 2t is positive. Thus, our claim is verified. Nowlet m be the least positive integer such that am [ H. By closure, �am� # H.We next claim that H 5 �am�. To prove this claim, it suffices to let b be anarbitrary member of H and show that b is in �am�. Since b [ G 5 �a�, wehave b 5 ak for some k. Now, apply the division algorithm to k and m toobtain integers q and r such that k 5 mq 1 r where 0 # r , m. Then ak 5amq1r 5 amqar, so that ar 5 a2mqak. Since ak 5 b [ H and a2mq 5(am)2q is in H also, ar [ H. But, m is the least positive integer such thatam [ H, and 0 # r , m, so r must be 0. Therefore, b 5 ak 5 amq 5(am)q [ �am�. This proves the assertion of the theorem that every sub-group of a cyclic group is cyclic.

To prove the next portion of the theorem, suppose that |�a�| 5 n andH is any subgroup of �a�. We have already shown that H 5 �am�, wherem is the least positive integer such that am [ H. Using e 5 b 5 an as inthe preceding paragraph, we have n 5 mq.

Finally, let k be any positive divisor of n. We will show that �an/k� isthe one and only subgroup of �a� of order k. From Theorem 4.2, we seethat �an/k� has order n/gcd(n, n/k) 5 n/(n/k) 5 k. Now let H be anysubgroup of �a� of order k. We have already shown above that H 5 �am�,where m is a divisor of n. Then m 5 gcd(n, m) and k 5 |am| 5 |agcd(n,m)| 5n/gcd (n, m) 5 n/m. Thus, m 5 n/k and H 5 �an/k�.

Returning for a moment to our discussion of the cyclic group �a�,where a has order 30, we may conclude from Theorem 4.3 that the sub-groups of �a� are precisely those of the form �am�, where m is a divisorof 30. Moreover, if k is a divisor of 30, the subgroup of order k is�a30/k�. So the list of subgroups of �a� is:

�a� 5 {e, a, a2, . . . , a29} order 30,�a2� 5 {e, a2, a4, . . . , a28} order 15,�a3� 5 {e, a3, a6, . . . , a27} order 10,�a5� 5 {e, a5, a10, a15, a20, a25} order 6,�a6� 5 {e, a6, a12, a18, a24} order 5,

�a10� 5 {e, a10, a20} order 3,�a15� 5 {e, a15} order 2,�a30� 5 {e} order 1.

In general, if �a� has order n and k divides n, then �an/k� is the uniquesubgroup of order k.

Taking the group in Theorem 4.3 to be Zn and a to be 1, we obtainthe following important special case.

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4 | Cyclic Groups 79

Corollary Subgroups of Zn

EXAMPLE 5 The list of subgroups of Z30 is

�1� 5 {0, 1, 2, . . . , 29} order 30,�2� 5 {0, 2, 4, . . . , 28} order 15,�3� 5 {0, 3, 6, . . . , 27} order 10,�5� 5 {0, 5, 10, 15, 20, 25} order 6,�6� 5 {0, 6, 12, 18, 24} order 5,

�10� 5 {0, 10, 20} order 3,�15� 5 {0, 15} order 2,�30� 5 {0} order 1.

By combining Theorems 4.2 and 4.3, we can easily count the num-ber of elements of each order in a finite cyclic group. For convenience,we introduce an important number-theoretic function called the Eulerphi function. Let f(1) 5 1, and for any integer n . 1, let f(n) denotethe number of positive integers less than n and relatively prime to n.Notice that by definition of the group U(n), |U(n)| 5 f(n). The first 12values of f(n) are given in Table 4.1.

For each positive divisor k of n, the set �n/k� is the unique subgroupof Zn of order k; moreover, these are the only subgroups of Zn.

Table 4.1 Values of f(n)

n 1 2 3 4 5 6 7 8 9 10 11 12

f(n) 1 1 2 2 4 2 6 4 6 4 10 4

Theorem 4.4 Number of Elements of Each Order in a Cyclic Group

PROOF By Theorem 4.3, the group has exactly one subgroup of order d—call it �a�. Then every element of order d also generates the sub-group �a� and, by Corollary 3 of Theorem 4.2, an element ak generates�a� if and only if gcd(k, d) 5 1. The number of such elements is preciselyf(d).

If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is f(d).

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80 Groups

Notice that for a finite cyclic group of order n, the number of elementsof order d for any divisor d of n depends only on d. Thus, Z8, Z640, andZ80000 each have f(8) 5 4 elements of order 8.

Although there is no formula for the number of elements of eachorder for arbitrary finite groups, we still can say something importantin this regard.

Corollary Number of Elements of Order d in a Finite Group

PROOF If a finite group has no elements of order d, the statement istrue, since f(d) divides 0. Now suppose that a [ G and |a| 5 d. ByTheorem 4.4, we know that �a� has f(d) elements of order d. If allelements of order d in G are in �a�, we are done. So, suppose that thereis an element b in G of order d that is not in �a�. Then, �b� also has f(d)elements of order d. This means that we have found 2f(d) elements oforder d in G provided that �a� and �b� have no elements of order d incommon. If there is an element c of order d that belongs to both �a� and�b�, then we have �a� 5 �c� 5 �b�, so that b [ �a�, which is a contradic-tion. Continuing in this fashion, we see that the number of elements oforder d in a finite group is a multiple of f(d).

On its face, the value of Theorem 4.4 and its corollary seem limited for large values of n because it is tedious to determine the number ofpositive integers less than or equal to n and relatively prime to nby examining them one by one. However, the following properties of the

function make computing simple: For any prime p, 5(see Exercise 71) and for relatively prime m and n,

5 Thus, (40) 5

The relationships among the various subgroups of a group can beillustrated with a subgroup lattice of the group. This is a diagram that in-cludes all the subgroups of the group and connects a subgroup H at onelevel to a subgroup K at a higher level with a sequence of line segmentsif and only if H is a proper subgroup of K. Although there are manyways to draw such a diagram, the connections between the subgroupsmust be the same. Typically one attempts to present the diagram in aneye-pleasing fashion. The lattice diagram for Z30 is shown in Figure 4.2.Notice that �10� is a subgroup of both �2� and �5�, but �6� is not a sub-group of �10�.

f(52)f(3) 5 (25 2 5) ? 2 5 40.f(8)f(5) 5 4 ? 4 5 16; f(75) 5ff(m)f(n).

f(mn)pn 2 pn21f(pn)f(n)f

In a finite group, the number of elements of order d is divisible by f(d).

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4 | Cyclic Groups 81

Figure 4.2 Subgroup lattice of Z30.

The precision of Theorem 4.3 can be appreciated by comparing theease with which we are able to identify the subgroups of Z30 with that ofdoing the same for, say, U(30) or D30. And these groups have relativelysimple structures among noncyclic groups.

We will prove in Chapter 7 that a certain portion of Theorem 4.3extends to arbitrary finite groups; namely, the order of a subgroup di-vides the order of the group itself. We will also see, however, that a finitegroup need not have exactly one subgroup corresponding to each divisorof the order of the group. For some divisors, there may be none at all,whereas for other divisors, there may be many. Indeed, D4, the dihedralgroup of order 8, has five subgroups of order 2 and three of order 4.

One final remark about the importance of cyclic groups is appropri-ate. Although cyclic groups constitute a very narrow class of finitegroups, we will see in Chapter 11 that they play the role of buildingblocks for all finite Abelian groups in much the same way that primesare the building blocks for the integers and that chemical elements arethe building blocks for the chemical compounds.

Exercises

It is not unreasonable to use the hypothesis.ARNOLD ROSS

1. Find all generators of Z6, Z8, and Z20.2. Suppose that �a�, �b�, and �c� are cyclic groups of orders 6, 8, and

20, respectively. Find all generators of �a�, �b�, and �c�.3. List the elements of the subgroups and in . Let a be a

group element of order 30. List the elements of the subgroups and .�a10�

�a20�Z30�10��20�

<10>

<0>

<6> <15>

<3>

<5><2>

<1>

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82 Groups

4. List the elements of the subgroups �3� and �15� in Z18. Let a be agroup element of order 18. List the elements of the subgroups �a3�and �a15�.

5. List the elements of the subgroups �3� and �7� in U(20).6. What do Exercises 3, 4, and 5 have in common? Try to make a gen-

eralization that includes these three cases.7. Find an example of a noncyclic group, all of whose proper sub-

groups are cyclic.8. Let a be an element of a group and let |a| 5 15. Compute the or-

ders of the following elements of G.a. a3, a6, a9, a12

b. a5, a10

c. a2, a4, a8, a14

9. How many subgroups does Z20 have? List a generator for each ofthese subgroups. Suppose that G 5 �a� and |a| 5 20. How manysubgroups does G have? List a generator for each of these sub-groups.

10. In Z24 list all generators for the subgroup of order 8. Let G 5 �a�and let |a| 5 24. List all generators for the subgroup of order 8.

11. Let G be a group and let a [ G. Prove that �a21� 5 �a�.12. In Z find all generators of the subgroup . If a has infinite order,

find all generators of the subgroup .13. In Z24 find a generator for �21� > �10�. Suppose that |a| 5 24. Find

a generator for �a21� > �a10�. In general, what is a generator for thesubgroup �am� > �an�?

14. Suppose that a cyclic group G has exactly three subgroups: Gitself, {e}, and a subgroup of order 7. What is |G|? What can yousay if 7 is replaced with p where p is a prime?

15. Let G be an Abelian group and let H 5 {g [ G| |g| divides 12}.Prove that H is a subgroup of G. Is there anything special about 12here? Would your proof be valid if 12 were replaced by some otherpositive integer? State the general result.

16. Find a collection of distinct subgroups �a1�, �a2�, . . . , �an� of Z240

with the property that �a1� , �a2� , ? ? ? , �an� with n as large as possible.

17. Complete the following statement: |a| 5 |a2| if and only if |a| . . . .18. If a cyclic group has an element of infinite order, how many ele-

ments of finite order does it have?19. List the cyclic subgroups of U(30).

�a3�

�3�

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4 | Cyclic Groups 83

20. Suppose that G is an Abelian group of order 35 and every elementof G satisfies the equation x35 5 e. Prove that G is cyclic. Doesyour argument work if 35 is replaced with 33?

21. Let G be a group and let a be an element of G.a. If a12 5 e, what can we say about the order of a?b. If am 5 e, what can we say about the order of a?c. Suppose that |G| 5 24 and that G is cyclic. If a8 2 e and a12 2 e,

show that �a� 5 G.22. Prove that a group of order 3 must be cyclic.23. Let Z denote the group of integers under addition. Is every sub-

group of Z cyclic? Why? Describe all the subgroups of Z. Let a bea group element with infinite order. Describe all subgroups of .

24. For any element a in any group G, prove that �a� is a subgroup ofC(a) (the centralizer of a).

25. If d is a positive integer, d 2 2, and d divides n, show that the num-ber of elements of order d in Dn is f(d). How many elements oforder 2 does Dn have?

26. Find all generators of Z. Let a be a group element that has infiniteorder. Find all generators of .

27. Prove that C*, the group of nonzero complex numbers under multi-plication, has a cyclic subgroup of order n for every positive integer n.

28. Let a be a group element that has infinite order. Prove that �ai� 5�aj� if and only if i 5 �j.

29. List all the elements of order 8 in Z8000000. How do you know yourlist is complete? Let a be a group element such that |a| .List all elements of order 8 in . How do you know your list iscomplete?

30. Suppose a and b belong to a group, a has odd order, and aba21 5b21. Show that b2 5 e.

31. Let G be a finite group. Show that there exists a fixed positive integern such that an 5 e for all a in G. (Note that n is independent of a.)

32. Determine the subgroup lattice for Z12.33. Determine the subgroup lattice for , where p and q are distinct

primes.34. Determine the subgroup lattice for Z8.35. Determine the subgroup lattice for , where p is a prime and n is

some positive integer.36. Prove that a finite group is the union of proper subgroups if and

only if the group is not cyclic.37. Show that the group of positive rational numbers under multiplica-

tion is not cyclic.

Zpn

Zp2q

�a�

5 8000000

�a�

�a�

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84 Groups

38. Consider the set {4, 8, 12, 16}. Show that this set is a group undermultiplication modulo 20 by constructing its Cayley table. Whatis the identity element? Is the group cyclic? If so, find all of itsgenerators.

39. Give an example of a group that has exactly 6 subgroups (includingthe trivial subgroup and the group itself). Generalize to exactly nsubgroups for any positive integer n.

40. Let m and n be elements of the group Z. Find a generator for thegroup �m� > �n�.

41. Suppose that a and b are group elements that commute and haveorders m and n. If �a� > �b� 5 {e}, prove that the group contains anelement whose order is the least common multiple of m and n.Show that this need not be true if a and b do not commute.

42. Prove that an infinite group must have an infinite number of sub-groups.

43. Let p be a prime. If a group has more than p 2 1 elements of order p,why can’t the group be cyclic?

44. Suppose that G is a cyclic group and that 6 divides |G|. How manyelements of order 6 does G have? If 8 divides |G|, how many ele-ments of order 8 does G have? If a is one element of order 8, listthe other elements of order 8.

45. List all the elements of Z40 that have order 10. Let |x| 5 40. List allthe elements of �x� that have order 10.

46. Reformulate the corollary of Theorem 4.4 to include the case whenthe group has infinite order.

47. Determine the orders of the elements of D33 and how many thereare of each.

48. If G is a cyclic group and 15 divides the order of G, determine thenumber of solutions in G of the equation x15 5 e. If 20 dividesthe order of G, determine the number of solutions of x20 5 e.Generalize.

49. If G is an Abelian group and contains cyclic subgroups of orders 4and 5, what other sizes of cyclic subgroups must G contain?Generalize.

50. If G is an Abelian group and contains cyclic subgroups of orders 4and 6, what other sizes of cyclic subgroups must G contain?Generalize.

51. Prove that no group can have exactly two elements of order 2.52. Given the fact that U(49) is cyclic and has 42 elements, deduce the

number of generators that U(49) has without actually finding any ofthe generators.

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4 | Cyclic Groups 85

53. Let a and b be elements of a group. If |a| 5 10 and |b| 5 21, showthat �a� > �b� 5 {e}.

54. Let a and b belong to a group. If |a| and |b| are relatively prime,show that �a� > �b� 5 {e}.

55. Let a and b belong to a group. If |a| 5 24 and |b| 5 10, what arethe possibilities for |�a� > �b�|?

56. Prove that U(2n) (n $ 3) is not cyclic.57. Suppose that a group G has at least nine elements x such that x8 5

e. Can you conclude that G is not cyclic? What if G has at least fiveelements x such that x4 5 e? Generalize.

58. Prove that Zn has an even number of generators if n . 2. Whatdoes this tell you about f(n)?

59. If |a5| 5 12, what are the possibilities for |a|? If |a4| 5 12, whatare the possibilities for |a|?

60. Suppose that |x| 5 n. Find a necessary and sufficient condition onr and s such that �xr� # �xs�.

61. Suppose a is a group element such that and .Determine .

62. Let a be group element such that . For each part find a di-visor k of 48 such thata.b.c. .

63. Let p be a prime. Show that in a cyclic group of order pn 21, everyelement is a pth power (that is, every element can be written in theform ap for some a).

64. Prove that is a cyclic subgroup of

GL(2, R).65. Let a and b belong to a group. If |a| 5 12, |b| 5 22, and �a� > �b� 2

{e}, prove that a6 5 b11.66. Suppose that G is a finite group with the property that every non-

identity element has prime order (for example, D3 and D5). If Z(G)is not trivial, prove that every nonidentity element of G has thesame order.

67. Let G be the set of all polynomials of the form ax2 1 bx 1 c withcoefficients from the set {0, 1, 2}. We can make G a group underaddition by adding the polynomials in the usual way, except thatwe use modulo 3 to combine the coefficients. With this operation,prove that G is a group of order 27 that is not cyclic.

H 5 e c1 n

0 1d 0 n[Z f

�a18� 5 �ak��a14� 5 �ak��a21� 5 �ak�

|a| 5 48|a|

|a22| 5 20|a28| 5 10

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86 Groups

68. Let and be rational numbers. Prove that the group under addition is cyclic. Gen-

eralize to the case where you have , , . . . , rationals.69. Let a and b belong to some group. Suppose that and

and m and n are relatively prime. If for some inte-ger k, prove that mn divides k.

70. For every integer n greater than 2, prove that the group is not cyclic.

71. Prove that for any prime p and positive integer n, 5.

72. Give an example of an infinite group that has exactly two elementsof order 4.

Computer Exercises

The nerds are running the world now.JOE PISCOPO

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software determines if U(n) is cyclic. Run the program for n 5 8, 32, 64, and 128. Make a conjecture. Run the program for n 53, 9, 27, 81, 243, 5, 25, 125, 7, 49, 11, and 121. Make a conjecture.Run the program for n 5 12, 20, 28, 44, 52, 15, 21, 33, 39, 51, 57,69, 35, 55, 65, and 85. Make a conjecture.

2. For any pair of positive integers m and n, let Zm % Zn 5 {(a, b) |a [ Zm, b [ Zn}. For any pair of elements (a, b) and (c, d) in Zm %

Zn, define (a, b) 1 (c, d) 5 ((a 1 c) mod m, (b 1 d) mod n). [Forexample, in Z3 % Z4, we have (1, 2) 1 (2, 3) 5 (0, 1).] This soft-ware checks whether or not Zm % Zn is cyclic. Run the program forthe following choices of m and n: (2, 2), (2, 3), (2, 4), (2, 5), (3, 4),(3, 5), (3, 6), (3, 7), (3, 8), (3, 9), and (4, 6). On the basis of this out-put, guess how m and n must be related for Zm % Zn to be cyclic.

3. In this exercise, a, b [ U(n). Define �a, b� 5 {aib j | 0 # i , |a|,0 # j , |b|}. This software computes the orders of �a, b�, �a�, �b�,and �a� > �b�. Run the program for the following choices of a, b,and n: (21, 101, 550), (21, 49, 550), (7, 11, 100), (21, 31, 100), and

pn 2 pn21

f(pn)

U(n2 2 1)

ak 5 bk|b| 5 n|a| 5 m

rkr2r1

5n1r1 1 n2r2 |n1 and n2 are integers6 G 5r2r1

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4 | Cyclic Groups 87

(63, 77, 100). On the basis of your output, make a conjecture aboutarithmetic relationships among |�a, b�|, |�a�|, |�b�|, and |�a� > �b�|.

4. For each positive integer n, this software gives the order of U(n)and the order of each element in U(n). Do you see any relationshipbetween the order of U(n) and the order of its elements? Run theprogram for n 5 8, 16, 32, 64, and 128. Make a conjecture aboutthe number of elements of order 2 in U(2k) when k is at least 3.Make a conjecture about the number of elements of order 4 inU(2k) when k is at least 4. Make a conjecture about the number ofelements of order 8 in U(2k) when k is at least 5. Make a conjectureabout the maximum order of any element in U(2k) when k is at least3. Try to find a formula for an element of order 4 in U(2k) when k isat least 4.

5. For each positive integer n, this software lists the number of ele-ments of U(n) of each order. For each order d of some element ofU(n), this software lists f(d) and the number of elements of order d.(Recall that f(d) is the number of positive integers less than orequal to d and relatively prime to d). Do you see any relationshipbetween the number of elements of order d and f(d)? Run the pro-gram for n 5 3, 9, 27, 81, 5, 25, 125, 7, 49, and 343. Make a con-jecture about the number of elements of order d and f(d) when n isa power of an odd prime. Run the program for n 5 6, 18, 54, 162,10, 50, 250, 14, 98, and 686. Make a conjecture about the numberof elements of order d and f(d) when n is twice a power of an oddprime. Make a conjecture about the number of elements of variousorders in U( pk) and U(2pk) where p is an odd prime.

6. For each positive integer n, this software gives the order of U(n).Run the program for n 5 9, 27, 81, and 243. Try to guess a formulafor the order of U(3k) when k is at least 2. Run the program for n 518, 54, 162, and 486. How does the order of U(2 ? 3k) appear to be re-lated to the order of U(3k)? Run the program for n 5 25, 125, and625. Try to guess a formula for the order of U(5k) when k is at least 2.Run the program for n 5 50, 250, and 1250. How does the order ofU(2 ? 5k) appear to be related to the order of U(5k)? Run the programfor n 5 49 and 343. Try to guess a formula for the order of U(7k)when k is at least 2. Run the program for n 5 98 and 686. How doesthe order of U(2 ? 7k) appear to be related to the order of U(7k)?Based on your guesses for U(3k), U(5k), and U(7k), guess a formulafor the order of U( pk) when p is an odd prime and k is at least 2.What about the order of U(2pk) when p is an odd prime and k is atleast 2. Does your formula also work when k is 1?

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88 Groups

Suggested Reading

Deborah L. Massari, “The Probability of Regenerating a Cyclic Group,”Pi Mu Epsilon Journal 7 (1979): 3–6.

In this easy-to-read paper, it is shown that the probability of a ran-domly chosen element from a cyclic group being a generator of thegroup depends only on the set of prime divisors of the order of thegroup, and not on the order itself. This article, written by an under-graduate student, received first prize in a Pi Mu Epsilon Paper Contest.

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89

J. J. Sylvester

I really love my subject.J. J. SYLVESTER

†F. Cajori, Teaching and History of Mathematics in the U.S., Washington, 1890, 265–266.

JAMES JOSEPH SYLVESTER was the most influ-ential mathematician in America in the 19thcentury. Sylvester was born on September 3,1814, in London and showed his mathemati-cal genius early. At the age of 14, he studiedunder De Morgan and won several prizes forhis mathematics, and at the unusually youngage of 25, he was elected a Fellow of theRoyal Society.

After receiving B.A. and M.A. degreesfrom Trinity College in Dublin in 1841,Sylvester began a professional life that wasto include academics, law, and actuarial ca-reers. In 1876, at the age of 62, he was ap-pointed to a prestigious position at the newlyfounded Johns Hopkins University. Duringhis seven years at Johns Hopkins, Sylvesterpursued research in pure mathematics with tremendous vigor and enthusiasm. He also founded the American Journal ofMathematics, the first journal in Americadevoted to mathematical research. Sylvesterreturned to England in 1884 to a professor-ship at Oxford, a position he held until hisdeath on March 15, 1897.

Sylvester’s major contributions tomathematics were in the theory of equations,matrix theory, determinant theory, and in-variant theory (which he founded withCayley). His writings and lectures—floweryand eloquent, pervaded with poetic flights,emotional expressions, bizarre utterances,and paradoxes—reflected the personality ofthis sensitive, excitable, and enthusiastic

man. We quote three of his students.† E. W.Davis commented on Sylvester’s teachingmethods.

Sylvester’s methods! He had none. “Three lec-tures will be delivered on a New UniversalAlgebra,” he would say; then, “The coursemust be extended to twelve.” It did last all therest of that year. The following year the coursewas to be Substitutions-Theorie, by Netto. Weall got the text. He lectured about three times,following the text closely and stopping sharpat the end of the hour. Then he began to thinkabout matrices again. “I must give one lecturea week on those,” he said. He could not con-fine himself to the hour, nor to the one lecturea week. Two weeks were passed, and Nettowas forgotten entirely and never mentionedagain. Statements like the following were notinfrequent in his lectures: “I haven’t provedthis, but I am as sure as I can be of anythingthat it must be so. From this it will follow,etc.” At the next lecture it turned out that whathe was so sure of was false. Never mind, hekept on forever guessing and trying, andpresently a wonderful discovery followed,then another and another. Afterward he wouldgo back and work it all over again, and sur-prise us with all sorts of side lights. He thenmade another leap in the dark, more treasureswere discovered, and so on forever.

FPO

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90

Sylvester’s enthusiasm for teaching and hisinfluence on his students are captured in thefollowing passage written by Sylvester’s firststudent at Johns Hopkins, G. B. Halsted.

A short, broad man of tremendous vitality, . . .Sylvester’s capacious head was ever lost in the highest cloud-lands of pure mathematics.Often in the dead of night he would get hisfavorite pupil, that he might communicate the very last product of his creative thought.Everything he saw suggested to him some-thing new in the higher algebra. This transmu-tation of everything into new mathematics was a revelation to those who knew himintimately. They began to do it themselves.

Another characteristic of Sylvester, whichis very unusual among mathematicians, washis apparent inability to remember mathemat-ics! W. P. Durfee had the following to say.

Sylvester had one remarkable peculiarity. Heseldom remembered theorems, propositions,etc., but had always to deduce them when hewished to use them. In this he was the veryantithesis of Cayley, who was thoroughlyconversant with everything that had beendone in every branch of mathematics.

I remember once submitting to Sylvestersome investigations that I had been engagedon, and he immediately denied my first state-ment, saying that such a proposition had neverbeen heard of, let alone proved. To his aston-ishment, I showed him a paper of his own inwhich he had proved the proposition; in fact, Ibelieve the object of his paper had been thevery proof which was so strange to him.

For more information about Sylvester,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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4 | Supplementary Exercises for Chapters 1–4 91

Supplementary Exercises for Chapters 1–4

If you really want something in this life, you have to work for it—Now quiet,they’re about to announce the lottery numbers!

HOMER SIMPSON

True/False questions for Chapters 1–4 are available on the web at:

http://www.d.umn.edu/~jgallian/TF

1. Let G be a group and let H be a subgroup of G. For any fixed x inG, define xHx21 5 {xhx21 | h [ H}. Prove the following.a. xHx21 is a subgroup of G.b. If H is cyclic, then xHx21 is cyclic.c. If H is Abelian, then xHx21 is Abelian.The group xHx21 is called a conjugate of H. (Note that conjuga-tion preserves structure.)

2. Let G be a group and let H be a subgroup of G. Define N(H) 5{x [ G | xHx21 5 H}. Prove that N(H) (called the normalizer ofH) is a subgroup of G.†

3. Let G be a group. For each a [ G, define cl(a) 5 {xax21 | x [ G}.Prove that these subsets of G partition G. [cl(a) is called theconjugacy class of a.]

4. The group defined by the following table is called the group ofquaternions. Use the table to determine each of the following:a. The centerb. cl(a)c. cl(b)d. All cyclic subgroups

e a a2 a3 b ba ba2 ba3

e e a a2 a3 b ba ba2 ba3

a a a2 a3 e ba3 b ba ba2

a2 a2 a3 e a ba2 ba3 b baa3 a3 e a a2 ba ba2 ba3 bb b ba ba2 ba3 a2 a3 e aba ba ba2 ba3 b a a2 a3 eba2 ba2 ba3 b ba e a a2 a3

ba3 ba3 b ba ba2 a3 e a a2

†This very important subgroup was first used by L. Sylow in 1872 to prove the exis-tence of certain kinds of subgroups in a group. His work is discussed in Chapter 24.

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92 Groups

5. (Conjugation preserves order.) Prove that, in any group, |xax21| 5|a|. (This exercise is referred to in Chapter 24.)

6. Prove that, in any group, |ab| 5 |ba|.7. If a, b, and c are elements of a group, give an example to show that

it need not be the case that |abc| 5 |cba|.8. Let a and b belong to a group G. Prove that there is an element x in

G such that xax 5 b if and only if ab 5 c2 for some element c in G.9. Prove that if a is the only element of order 2 in a group, then a lies

in the center of the group.10. Let G be the plane symmetry group of the infinite strip of equally

spaced H’s shown below.

Let x be the reflection about Axis 1 and let y be the reflection aboutAxis 2. Calculate |x|, |y|, and |xy|. Must the product of elements offinite order have finite order?

11. What are the orders of the elements of D15? How many elementshave each of these orders?

12. Prove that a group of order 4 is Abelian.13. Prove that a group of order 5 must be cyclic.14. Prove that an Abelian group of order 6 must be cyclic.15. Let G be an Abelian group and let n be a fixed positive integer. Let

Gn 5 {gn | g [ G}. Prove that Gn is a subgroup of G. Give an ex-ample showing that Gn need not be a subgroup of G when G isnon-Abelian. (This exercise is referred to in Chapter 11.)

16. Let , where a and b are rational numbers notboth 0. Prove that G is a group under ordinary multiplication.

17. (1969 Putnam Competition) Prove that no group is the union oftwo proper subgroups. Does the statement remain true if “two” isreplaced by “three”?

18. Prove that the subset of elements of finite order in an Abeliangroup forms a subgroup. (This subgroup is called the torsion sub-group.) Is the same thing true for non-Abelian groups?

19. Let p be a prime and let G be an Abelian group. Show that the setof all elements whose orders are powers of p is a subgroup of G.

20. Suppose that a and b are group elements. If and ,determine the possibilities for .|a|

bab 5 a4|b| 5 2

G 5 5a 1 b"2 6

H H HHHAxis 1 Axis 2

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4 | Supplementary Exercises for Chapters 1–4 93

21. Suppose that a finite group is generated by two elements a and b(that is, every element of the group can be expressed as some prod-uct of a’s and b’s). Given that a35 b2 5 e and ba2 5 ab, constructthe Cayley table for the group. We have already seen an exampleof a group that satisfies these conditions. Name it.

22. If a is an element of a group and , prove that when k is relatively prime to n.

23. Let x and y belong to a group G. If xy [ Z(G), prove that xy 5 yx.24. Suppose that H and K are nontrivial subgroups of Q under addi-

tion. Show that H > K is a nontrivial subgroup of Q. Is this true ifQ is replaced by R?

25. Let H be a subgroup of G and let g be an element of G. Prove thatN(gHg21) 5 gN(H)g21. See Exercise 2 for the notation.

26. Let H be a subgroup of a group G and let |g| 5 n. If gm belongs toH and m and n are relatively prime, prove that g belongs to H.

27. Find a group that contains elements a and b such that |a| 5 2,|b| 5 11, and |ab| 5 2.

28. Suppose that G is a group with exactly eight elements of order 10.How many cyclic subgroups of order 10 does G have?

29. (1989 Putnam Competition) Let S be a nonempty set with an asso-ciative operation that is left and right cancellative (xy 5 xz impliesy 5 z, and yx 5 zx implies y 5 z). Assume that for every a in S theset {an | n 5 1, 2, 3, . . .} is finite. Must S be a group?

30. Let H1, H2, H3, . . . be a sequence of subgroups of a group with theproperty that H1 # H2 # H3 . . . . Prove that the union of the se-quence is a subgroup.

31. Let R* be the group of nonzero real numbers under multiplicationand let H 5{g [ R*| some nonzero integer power of g is a rationalnumber}. Prove that H is a subgroup of R*.

32. Suppose that a and b belong to a group, a and b commute, and |a|and |b| are relatively prime. Prove that |ab| 5 |a||b|. Give an exam-ple showing that |ab| need not be |a||b| when a and b commute but|a| and |b| are not relatively prime. (Don’t use .)

33. Let H 5 {A [ GL(2, R) | det A is rational}. Prove or disprove thatH is a subgroup of GL(2, R). What if “rational” is replaced by “aninteger”?

34. Suppose that G is a group that has exactly one nontrivial propersubgroup. Prove that G is cyclic and |G| 5 p2, where p is prime.

35. Suppose that G is a group and G has exactly two nontrivial propersubgroups. Prove that G is cyclic and |G| 5 pq, where p and q aredistinct primes, or that G is cyclic and |G| 5 p3, where p is prime.

a [ �b�

C(a) 5 C(ak)|a| 5 n

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94 Groups

36. If |a2| 5 |b2|, prove or disprove that |a| 5 |b|.37. (1995 Putnam Competition) Let S be a set of real numbers that is

closed under multiplication. Let T and U be disjoint subsets of Swhose union is S. Given that the product of any three (not neces-sarily distinct) elements of T is in T and that the product of anythree elements of U is in U, show that at least one of the two sub-sets T and U is closed under multiplication.

38. If p is an odd prime, prove that there is no group that has exactly pelements of order p.

39. Give an example of a group G with infinitely many distinct sub-groups H1, H2, H3, . . . such that H1 , H2 , H3 . . . .

40. Suppose a and b are group elements and b 2 e. If a21ba 5 b2 and |a| 5 3, find |b|. What is |b|, if |a| 5 5? What can you say about|b| in the case where |a| 5 k?

41. Let a and b belong to a group G. Show that there is an element g inG such that g21 abg 5 ba.

42. Suppose G is a group and x3y3 5 y3x3 for every x and y in G. Let H 5 {x [ G| |x| is relatively prime to 3}. Prove that elements of Hcommute with each other and that H is a subgroup of G. Is yourargument valid if 3 is replaced by an arbitrary positive integer n?Explain why or why not.

43. Let G be a finite group and let S be a subset of G that containsmore than half of the elements of G. Show that every element of Gcan be expressed in the form s1s2 where s1 and s2 belong to S.

44. Let G be a group and let f be a function from G to some set. Showthat H 5 {g [ G| f (xg) 5 f (x) for all x [ G} is a subgroup of G.In the case that G is the group of real numbers under addition and f (x) 5 sin x, describe H.

45. Let G be a cyclic group of order n and let H be the subgroup oforder d. Show that H 5 {x [ G| |x| divides d}.

46. Let a be an element of maximum order from a finite Abelian groupG. Prove that for any element b in G, |b| divides |a|. Show byexample that this need not be true for finite non-Abelian groups.

47. Define an operation * on the set of integers by a * .Show that the set of integers under this operation is a cyclic group.

48. Let n be an integer greater than 1. Find a noncyclic subgroup ofof order 4 that contains the element .2n 2 1U(4n)

b 5 a 1 b 2 1

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95

Permutation Groups

Wigner’s discovery about the electron permutation group was just thebeginning. He and others found many similar applications and nowadaysgroup theoretical methods—especially those involving characters andrepresentations—pervade all branches of quantum mechanics.

GEORGE MACKEY, Proceedings of the

American Philosophical Society

5

Definition and NotationIn this chapter, we study certain groups of functions, called permutationgroups, from a set A to itself. In the early and mid-19th century, groupsof permutations were the only groups investigated by mathematicians.It was not until around 1850 that the notion of an abstract group was in-troduced by Cayley, and it took another quarter century before the ideafirmly took hold.

Definitions Permutation of A, Permutation Group of A

A permutation of a set A is a function from A to A that is both one-to-one and onto. A permutation group of a set A is a set of permuta-tions of A that forms a group under function composition.

Although groups of permutations of any nonempty set A of objectsexist, we will focus on the case where A is finite. Furthermore, it iscustomary, as well as convenient, to take A to be a set of the form{1, 2, 3, . . . , n} for some positive integer n. Unlike in calculus, wheremost functions are defined on infinite sets and are given by formulas,in algebra, permutations of finite sets are usually given by an explicitlisting of each element of the domain and its corresponding functionalvalue. For example, we define a permutation a of the set {1, 2, 3, 4} byspecifying

a(1) 5 2, a(2) 5 3, a(3) 5 1, a(4) 5 4.

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5 c1 2 3 4 5

4 2 1 3 5d£

1 2 3 4 5

2 4 3 5 1

§£1 2 3 4 5

5 4 1 2 3

§gs 5

96 Groups

A more convenient way to express this correspondence is to write a inarray form as

Here a( j) is placed directly below j for each j. Similarly, the permuta-tion b of the set {1, 2, 3, 4, 5, 6} given by

b(1) 5 5, b(2) 5 3, b(3) 5 1, b(4) 5 6, b(5) 5 2, b(6) 5 4

is expressed in array form as

Composition of permutations expressed in array notation is carriedout from right to left by going from top to bottom, then again from topto bottom. For example, let

and

then

g 5 c1 2 3 4 5

5 4 1 2 3d ;

s 5 c1 2 3 4 5

2 4 3 5 1d

b 5 c1 2 3 4 5 6

5 3 1 6 2 4d.

a 5 c1 2 3 4

2 3 1 4d .

On the right we have 4 under 1, since (gs)(1) 5 g(s(1)) 5 g(2) 5 4,so gs sends 1 to 4. The remainder of the bottom row gs is obtained ina similar fashion.

We are now ready to give some examples of permutation groups.

EXAMPLE 1 Symmetric Group S3

Let S3 denote the set of all one-to-one functions from {1, 2, 3} to itself. Then S3, under function com-position, is a group with six elements. The six elements are

, , ,a2 5 c1 2 3

3 1 2da 5 c1 2 3

2 3 1de 5 c1 2 3

1 2 3d

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5 | Permutation Groups 97

, .

Note that ba 5 5 a2b 2 ab, so that S3 is non-Abelian.

The relation ba 5 a2b can be used to compute other products in S3without resorting to the arrays. For example, ba2 5 (ba)a 5 (a2b)a 5a2(ba) 5 a2(a2b) 5 a4b 5 ab.

Example 1 can be generalized as follows.

EXAMPLE 2 Symmetric Group Sn

Let A 5 {1, 2, . . . , n}. The setof all permutations of A is called the symmetric group of degree n and isdenoted by Sn. Elements of Sn have the form

It is easy to compute the order of Sn. There are n choices of a(1). Oncea(1) has been determined, there are n 2 1 possibilities for a(2) [since a is one-to-one, we must have a(1) 2 a(2)]. After choosing a(2), thereare exactly n 2 2 possibilities for a(3). Continuing along in this fashion,we see that Sn has n(n 2 1) ? ? ? 3 ? 2 ? 1 5 n! elements. We leave it to thereader to prove that Sn is non-Abelian when n $ 3 (Exercise 41).

The symmetric groups are rich in subgroups. The group S4 has 30subgroups, and S5 has well over 100 subgroups.

EXAMPLE 3 Symmetries of a Square As a third example, weassociate each motion in D4 with the permutation of the locations of eachof the four corners of a square. For example, if we label the four cornerpositions as in the figure below and keep these labels fixed for reference,we may describe a 90° counterclockwise rotation by the permutation

r 5 c1 2 3 4

2 3 4 1d ,

3

4

2

1

a 5 c 1 2 c n

a(1) a(2) ca(n)d .

c1 2 3

3 2 1d

a2b 5 c1 2 3

3 2 1dab 5 c1 2 3

2 1 3db 5 c1 2 3

1 3 2d ,

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98 Groups

whereas a reflection across a horizontal axis yields

These two elements generate the entire group (that is, every element issome combination of the r’s and f’s).

When D4 is represented in this way, we see that it is a subgroupof S4.

Cycle NotationThere is another notation commonly used to specify permutations. It iscalled cycle notation and was first introduced by the great French math-ematician Cauchy in 1815. Cycle notation has theoretical advantages inthat certain important properties of the permutation can be readily de-termined when cycle notation is used.

As an illustration of cycle notation, let us consider the permutation

This assignment of values could be presented schematically as follows:

Although mathematically satisfactory, such diagrams are cumber-some. Instead, we leave out the arrows and simply write a 5 (1, 2)(3, 4, 6)(5). As a second example, consider

In cycle notation, b can be written (2, 3, 1, 5)(6, 4) or (4, 6)(3, 1, 5, 2),since both of these unambiguously specify the function b. An expres-sion of the form (a1, a2, . . . , am) is called a cycle of length m or an m-cycle.

b 5 c1 2 3 4 5 6

5 3 1 6 2 4d.

2

1

α α

α α

α α

6

3 5

4

a 5 c1 2 3 4 5 6

2 1 4 6 5 3d.

f 5 c1 2 3 4

2 1 4 3d .

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5 | Permutation Groups 99

A multiplication of cycles can be introduced by thinking of a cycleas a permutation that fixes any symbol not appearing in the cycle.Thus, the cycle (4, 6) can be thought of as representing the

permutation In this way, we can multiply cycles

by thinking of them as permutations given in array form. Consider thefollowing example from S8. Let a 5 (13)(27)(456)(8) and b 5(1237)(648)(5). (When the domain consists of single-digit integers, it iscommon practice to omit the commas between the digits.) What is the cycle form of ab? Of course, one could say that ab 5(13)(27)(456)(8)(1237)(648)(5), but it is usually more desirable to ex-press a permutation in a disjoint cycle form (that is, the various cycleshave no number in common). Well, keeping in mind that function com-position is done from right to left and that each cycle that does not con-tain a symbol fixes the symbol, we observe that: (5) fixes 1; (648) fixes 1;(1237) sends 1 to 2; (8) fixes 2; (456) fixes 2; (27) sends 2 to 7; and (13)fixes 7. So the net effect of ab is to send 1 to 7. Thus we begin ab 5 (17 ? ? ?) ? ? ? . Now, repeating the entire process beginning with 7,we have, cycle by cycle, right to left, 7 → 7 → 7 → 1 → 1 → 1 → 1 → 3,so that ab 5 (173 ? ? ?) ? ? ? . Ultimately, we have ab 5 (1732)(48)(56).The important thing to bear in mind when multiplying cycles is to “keepmoving” from one cycle to the next from right to left. (Warning: Some au-thors compose cycles from left to right. When reading another text, besure to determine which convention is being used.)

To be sure you understand how to switch from one notation to theother and how to multiply permutations, we will do one more exampleof each.

If array notations for a and b, respectively, are

and

then, in cycle notation, a 5 (12)(3)(45), b 5 (153)(24), and ab 5(12)(3)(45)(153)(24).

To put ab in disjoint cycle form, observe that (24) fixes 1; (153)sends 1 to 5; (45) sends 5 to 4; and (3) and (12) both fix 4. So, ab sends1 to 4. Continuing in this way we obtain ab 5 (14)(253).

One can convert ab back to array form without converting eachcycle of ab into array form by simply observing that (14) means 1 goesto 4 and 4 goes to 1; (253) means 2 → 5, 5 → 3, 3 → 2.

One final remark about cycle notation: Mathematicians prefer not towrite cycles that have only one entry. In this case, it is understood that any

c1 2 3 4 5

5 4 1 2 3dc1 2 3 4 5

2 1 3 5 4d

c1 2 3 4 5 6

1 2 3 6 5 4d .

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100 Groups

missing element is mapped to itself. With this convention, the permutationa above can be written as (12)(45). Similarly,

can be written a 5 (134). Of course, the identity permutation consistsonly of cycles with one entry, so we cannot omit all of these! In thiscase, one usually writes just one cycle. For example,

e

can be written as e 5 (5) or e 5 (1). Just remember that missingelements are mapped to themselves.

Properties of PermutationsWe are now ready to state several theorems about permutations andcycles. The proof of the first theorem is implicit in our discussion ofwriting permutations in cycle form.

Theorem 5.1 Products of Disjoint Cycles

PROOF Let a be a permutation on A 5 {1, 2, . . . , n}. To write a indisjoint cycle form, we start by choosing any member of A, say a1, and let

a2 5 a(a1), a3 5 a(a(a1)) 5 a2(a1),

and so on, until we arrive at a1 5 am(a1) for some m. We know that suchan m exists because the sequence a1, a(a1), a2(a1), ? ? ? must be finite;so there must eventually be a repetition, say a i(a1) 5 a j(a1) for somei and j with i , j. Then a1 5 am(a1), where m 5 j 2 i. We express thisrelationship among a1, a2, . . . , am as

a 5 (a1, a2, . . . , am) ? ? ? .

The three dots at the end indicate the possibility that we may not haveexhausted the set A in this process. In such a case, we merely chooseany element b1 of A not appearing in the first cycle and proceed to cre-ate a new cycle as before. That is, we let b2 5 a(b1), b3 5 a2(b1), and soon, until we reach b1 5 a k(b1) for some k. This new cycle will have no

Every permutation of a finite set can be written as a cycle or as aproduct of disjoint cycles.

5 c1 2 3 4 5

1 2 3 4 5d

a 5 c1 2 3 4 5

3 2 4 1 5d

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5 | Permutation Groups 101

elements in common with the previously constructed cycle. For, if so,then a i(a1) 5 a j(b1) for some i and j. But then a i2j(a1) 5 b1, and there-fore b1 5 at for some t. This contradicts the way b1 was chosen.Continuing this process until we run out of elements of A, our permuta-tion will appear as

a 5 (a1, a2, . . . , am)(b1, b2, . . . , bk) ? ? ? (c1, c2, . . . , cs).

In this way, we see that every permutation can be written as a productof disjoint cycles.

Theorem 5.2 Disjoint Cycles Commute

PROOF For definiteness, let us say that a and b are permutations ofthe set

S 5 {a1, a2, . . . , am, b1, b2, . . . , bn, c1, c2, . . . , ck}

where the c’s are the members of S left fixed by both a and b (theremay not be any c’s). To prove that ab 5 ba, we must show that (ab)(x) 5(ba)(x) for all x in S. If x is one of the a elements, say ai, then

(ab)(ai) 5 a(b(ai)) 5 a(ai) 5 ai11,

since b fixes all a elements. (We interpret ai11 as a1 if i 5 m.) For thesame reason,

(ba)(ai) 5 b(a (ai)) 5 b(ai11) 5 ai11.

Hence, the functions of ab and ba agree on the a elements. A similarargument shows that ab and ba agree on the b elements as well.Finally, suppose that x is a c element, say ci. Then, since both a and bfix c elements, we have

(ab)(ci) 5 a(b(ci)) 5 a(ci) 5 ci

and

(ba)(ci) 5 b(a(ci)) 5 b(ci) 5 ci.

This completes the proof.

In demonstrating how to multiply cycles, we showed that theproduct (13)(27)(456)(8)(1237)(648)(5) can be written in disjoint cycle

If the pair of cycles a 5 (a1, a2, . . . , am) and b 5 (b1,b2, . . . , bn) have no entries in common, then ab 5 ba.

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102 Groups

form as (1732)(48)(56). Is economy in expression the only advantageto writing a permutation in disjoint cycle form? No. The next theoremshows that the disjoint cycle form has the enormous advantage ofallowing us to “eyeball” the order of the permutation.

Theorem 5.3 Order of a Permutation (Ruffini—1799)

PROOF First, observe that a cycle of length n has order n. (Verify thisyourself.) Next, suppose that a and b are disjoint cycles of lengths mand n, and let k be the least common multiple of m and n. It follows fromTheorem 4.1 that both ak and bk are the identity permutation e and, sincea and b commute, (ab)k 5 akbk is also the identity. Thus, we know byCorollary 2 to Theorem 4.1 (ak 5 e implies that |a| divides k) that theorder of ab—let us call it t—must divide k. But then (ab)t 5 atb t 5 e,so that at 5 b2t. However, it is clear that if a and b have no commonsymbol, the same is true for a t and b2t, since raising a cycle to a powerdoes not introduce new symbols. But, if a t and b2t are equal and haveno common symbol, they must both be the identity, because every sym-bol in a t is fixed by b2t and vice versa (remember that a symbol not ap-pearing in a permutation is fixed by the permutation). It follows, then,that both m and n must divide t. This means that k, the least commonmultiple of m and n, divides t also. This shows that k 5 t.

Thus far, we have proved that the theorem is true in the caseswhere the permutation is a single cycle or a product of two disjointcycles. The general case involving more than two cycles can be han-dled in an analogous way.

Theorem 5.3 is an enomously powerful tool for calculating the or-ders of permuations. We demonstrate this in the next example.

EXAMPLE 4 To determine the orders of the 5040 elements of , weneed only consider the possible disjoint cycle structures of theelements of . For convenience, we denote an n-cycle by (n). Then, ar-ranging all possible disjoint cycle structures of elements of according to longest cycle lengths left to right, we have

S7

S7

S7

The order of a permutation of a finite set written in disjoint cycleform is the least common multiple of the lengths of the cycles.

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5 | Permutation Groups 103

(7)(6) (1)(5) (2)(5) (1) (1)(4) (3)(4) (2) (1)(4) (1) (1) (1)(3) (3) (1)(3) (2) (2)(3) (2) (1) (1)(3) (1) (1) (1) (1) (1)(2) (2) (2) (1)(2) (2) (1) (1) (1)(2) (1) (1) (1) (1) (1)(1) (1) (1) (1) (1) (1) (1).

Now, from Theorem 5.3 we see that the orders of the elements of are 7, 6, 10, 5, 12, 4, 3, 2, and 1. To do the same for the elements of would be nearly as simple.

As we will soon see, a particularly important kind of permutation isa cycle of length 2—that is, a permutation of the form (ab) wherea 2 b. Many authors call these permutations transpositions, since theeffect of (ab) is to interchange or transpose a and b.

Theorem 5.4 Product of 2-Cycles

PROOF First, note that the identity can be expressed as (12)(12), andso it is a product of 2-cycles. By Theorem 5.1, we know that every per-mutation can be written in the form

(a1a2 ? ? ? ak)(b1b2 ? ? ? bt) ? ? ? (c1c2 ? ? ? cs).

A direct computation shows that this is the same as

(a1ak)(a1ak21) ? ? ? (a1a2)(b1bt)(b1bt21) ? ? ? (b1b2) ? ? ? (c1cs)(c1cs21) ? ? ? (c1c2).

This completes the proof.

The decompositions in the following example demonstrate this technique.

Every permutation in Sn, n . 1, is a product of 2-cycles.

S10

10! 5 3,628,800S7

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104 Groups

EXAMPLE 5

(12345) 5 (15)(14)(13)(12)(1632)(457) 5 (12)(13)(16)(47)(45)

The decomposition of a permutation into a product of 2-cycles givenin the proof of Theorem 5.4 is not the only way a permutation can bewritten as a product of 2-cycles. Although the next example shows thateven the number of 2-cycles may vary from one decomposition to an-other, we will prove in Theorem 5.5 (first proved by Cauchy) that thereis one aspect of a decomposition that never varies.

EXAMPLE 6

(12345) 5 (54)(53)(52)(51)(12345) 5 (54)(52)(21)(25)(23)(13)

We isolate a special case of Theorem 5.5 as a lemma.

Lemma

PROOF Clearly, r 2 1, since a 2-cycle is not the identity. If r 5 2, weare done. So, we suppose that r . 2, and we proceed by induction.Since (ij) 5 ( ji), the product br21br can be expressed in one of the fol-lowing forms shown on the right:

e 5 (ab)(ab)(ab)(bc) 5 (ac)(ab)(ac)(cb ) 5 (bc)(ab)(ab)(cd) 5 (cd)(ab).

If the first case occurs, we may delete br21br from the original productto obtain e 5 b1b2 ? ? ? br22. In the other three cases, we replace theform of br21br on the right by its counterpart on the left to obtain a newproduct of r 2-cycles that is still the identity, but where the rightmostoccurrence of the integer a is in the second-from-the-rightmost 2-cycleof the product instead of the rightmost 2-cycle. We now repeat the proce-dure just described with br22br21, and, as before, we obtain a product of(r 2 2) 2-cycles equal to the identity or a new product of r 2-cycles,where the rightmost occurrence of a is in the third 2-cycle from the right.

If e 5 b1b2 ? ? ? br, where the b’s are 2-cycles, then r is even.

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5 | Permutation Groups 105

Continuing this process, we must obtain a product of (r 2 2) 2-cyclesequal to the identity, because otherwise we have a product equal to theidentity in which the only occurrence of the integer a is in the leftmost 2-cycle, and such a product does not fix a, whereas the identity does. Hence,by the Second Principle of Mathematical Induction, r 2 2 is even, and ris even as well.

Theorem 5.5 Always Even or Always Odd

PROOF Observe that b1b2 ? ? ? br 5 g1g2 ? ? ? gs implies

e 5 g1g2 ? ? ? gsbr21 ? ? ? b2

21b121

5 g1g2 ? ? ? gsbr ? ? ? b2b1,

since a 2-cycle is its own inverse. Thus, the lemma on page 104 guar-antees that s 1 r is even. It follows that r and s are both even or bothodd.

Definition Even and Odd Permutations

A permutation that can be expressed as a product of an even numberof 2-cycles is called an even permutation. A permutation that can be expressed as a product of an odd number of 2-cycles is called anodd permutation.

Theorems 5.4 and 5.5 together show that every permutation can beunambiguously classified as either even or odd. The significance of thisobservation is given in Theorem 5.6.

Theorem 5.6 Even Permutations Form a Group

PROOF This proof is left to the reader (Exercise 13).

The subgroup of even permutations in Sn arises so often that we giveit a special name and notation.

The set of even permutations in Sn forms a subgroup of Sn.

If a permutation a can be expressed as a product of an even (odd)number of 2-cycles, then every decomposition of a into a product of2-cycles must have an even (odd) number of 2-cycles. In symbols, if

a 5 b1b2 ? ? ? br and a 5 g1g2 ? ? ? gs,

where the b’s and the g’s are 2-cycles, then r and s are both even orboth odd.

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106 Groups

Definition Alternating Group of Degree n

The group of even permutations of n symbols is denoted by An and iscalled the alternating group of degree n.

The next result shows that exactly half of the elements of Sn(n . 1)are even permutations.

Theorem 5.7

PROOF For each odd permutation a, the permutation (12)a is evenand (12)a 2 (12)b when a 2 b. Thus, there are at least as many evenpermutations as there are odd ones. On the other hand, for eacheven permutation a, the permutation (12)a is odd and (12)a 2 (12)bwhen a 2 b. Thus, there are at least as many odd permutations as thereare even ones. It follows that there are equal numbers of even and oddpermutations. Since |Sn| 5 n!, we have |An| 5 n!/2.

The names for the symmetric group and the alternating group of degreen come from the study of polynomials over n variables. A symmetricpolynomial in the variables x1, x2, . . . , xn is one that is unchanged underany transposition of two of the variables. An alternating polynomial isone that changes signs under any transposition of two of the variables. Forexample, the polynomial x1x2x3 is unchanged by any transposition of twoof the three variables, whereas the polynomial (x12x2)(x12x3)(x22x3)changes signs when any two of the variables are transposed. Since everymember of the symmetric group is the product of transpositions, the sym-metric polynomials are those that are unchanged by members of the sym-metric group. Likewise, since any member of the alternating group is theproduct of an even number of transpositions, the alternating polynomialsare those that are unchanged by members of the alternating group andchange sign by the other permutations of Sn.

The alternating groups are among the most important examples ofgroups. The groups A4 and A5 will arise on several occasions in laterchapters. In particular, A5 has great historical significance.

A geometric interpretation of A4 is given in Example 7, and a multi-plication table for A4 is given as Table 5.1.

EXAMPLE 7 ROTATIONS OF A TETRAHEDRON The 12 rota-tions of a regular tetrahedron can be conveniently described with theelements of A4. The top row of Figure 5.1 illustrates the identity andthree 180° “edge” rotations about axes joining midpoints of two edges.

For n . 1, An has order n!/2.

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5 | Permutation Groups 107

A A

B

C C

D

1

2

3

4

(1)

A

BB

C

DD

1

2

3

4

(12)(34) (13)(24)

C C

D

A A

B

1

2

3

4C

DD

A

BB

1

3

24

(14)(23)

A

CC

DBB

1

3

24

(123)

A A

DD

CB

1

3

24

(134)

D D

AA

BC

1

3

24

(243)

AD

C C

BB

1

2

3

4

(142)

C

BB

DAA

1

3

24

(132)1

D

AA

C

BB

3

24

(234) (124)1

A

DD

B

CC

3

24

2

(143)1

B

CC

A

DD

3

4

Table 5.1 The Alternating Group A4 of Even Permutations of {1, 2, 3, 4}

(In this table, the permutations of A4 are designated as a1, a2, . . . , a12 and an entry k inside the table represents ak. For example, a3 a8 5 a6.)

a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12

(1) 5 a1 1 2 3 4 5 6 7 8 9 10 11 12(12)(34) 5 a2 2 1 4 3 6 5 8 7 10 9 12 11(13)(24) 5 a3 3 4 1 2 7 8 5 6 11 12 9 10(14)(23) 5 a4 4 3 2 1 8 7 6 5 12 11 10 9

(123) 5 a5 5 8 6 7 9 12 10 11 1 4 2 3(243) 5 a6 6 7 5 8 10 11 9 12 2 3 1 4(142) 5 a7 7 6 8 5 11 10 12 9 3 2 4 1(134) 5 a8 8 5 7 6 12 9 11 10 4 1 3 2(132) 5 a9 9 11 12 10 1 3 4 2 5 7 8 6(143) 5 a10 10 12 11 9 2 4 3 1 6 8 7 5(234) 5 a11 11 9 10 12 3 1 2 4 7 5 6 8(124) 5 a12 12 10 9 11 4 2 1 3 8 6 5 7

Figure 5.1 Rotations of a regular tetrahedron.

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108 Groups

The second row consists of 120° “face” rotations about axes joining a ver-tex to the center of the opposite face. The third row consists of 2120° (or240°) “face” rotations. Notice that the four rotations in the second row canbe obtained from those in the first row by left-multiplying the four in thefirst row by the rotation (123), whereas those in the third row can be ob-tained from those in the first row by left-multiplying the ones in the firstrow by (132).

Many molecules with chemical formulas of the form AB4, such asmethane (CH4) and carbon tetrachloride (CCl4), have A4 as their sym-metry group. Figure 5.2 shows the form of one such molecule.

Many games and puzzles can be analyzed using permutations.

Figure 5.2 A tetrahedral AB4 molecule.

EXAMPLE 8 (Loren Larson) A Sliding Disk PuzzleConsider the puzzle shown below (the space in the middle is empty).

By sliding disks from one position to another along the linesindicated without lifting or jumping, can we obtain the followingarrangement?

1

2

3

4

5

6

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5 | Permutation Groups 109

To answer this question, we view the positions as numbered in thefirst figure above and consider two basic operations. Let r denote thefollowing operation: Move the disk in position 1 to the center position,then move the disk in position 6 to position 1, the disk in position 5 toposition 6, the disk in position 4 to position 5, the disk in position 3 toposition 4, then the disk in the middle position to position 3. Let sdenote the operation: Move the disk in position 1 to the center position,then move the disk in position 2 to position 1, then move the disk in po-sition 3 to position 2, and finally move the disk in the center to position 3.In permutation notation, we have r 5 (13456) and s 5 (132). Thepermutation for the arrangement we seek is (16523). Clearly, if we canexpress (16523) as a string of r’s and s’s, we can achieve the desiredarangement. Rather than attempt to find an appropriate combination ofr’s and s’s by hand, it is easier to employ computer software that is de-signed for this kind of problem. One such software program is GAP (seeSuggested Software at the end of this chapter). With GAP, all we need todo is use the following commands:

gap. G :5 SymmetricGroup(6);gap. r :5 (1,3,4,5,6); s :5 (1, 3, 2);gap. K :5 Subgroup(G,[r,s]);gap. Factorization(K,(1,6,5,2,3));

The first three lines inform the computer that our group is thesubgroup of S6 generated by r 5 (13456) and s 5 (132). The fourthline requests that (16523) be expressed in terms of r and s. The re-sponse to the command

gap. Size (K);

tells us that the order of the subgroup generated by r and s is 360. Then,observing that r and s are even permutations and that |A6| 5 360, wededuce that r and s can achieve any arrangement that corresponds to aneven permutation.

3

5

2

4

6

1

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110 Groups

GAP can even compute the 43,252,003,274,489,856,000 (431 quin-tillion) permutations of the Rubik’s Cube! Labeling the faces of thecube as shown here,

the group of permutations of the cube is generated by the following ro-tations of the six layers:

top 5 (1,3,8,6)(2,5,7,4)(9,33,25,17)(10,34,26,18)(11,35,27,19)left 5 (9,11,16,14)(10,13,15,12)(1,17,41,40)(4,20,44,37)(6,22,46,35)front 5 (17,19,24,22)(18,21,23,20)(6,25,43,16)(7,28,42,13)(8,30,41,11)right 5 (25,27,32,30)(26,29,31,28)(3,38,43,19)(5,36,45,21)(8,33,48,24)rear 5 (33,35,40,38)(34,37,39,36)(3,9,46,32)(2,12,47,29)(1,14,48,27)bottom 5 (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)

(16,24,32,40)

A Check Digit Scheme Based on D5

In Chapter 0, we presented several schemes for appending a check digitto an identification number. Among these schemes, only the Interna-tional Standard Book Number method was capable of detecting allsingle-digit errors and all transposition errors involving adjacent digits.However, recall that this success was achieved by introducing the al-phabetical character X to handle the case where 10 was required tomake the dot product 0 modulo 11.

In contrast, in 1969, J. Verhoeff [2] devised a method utilizing thedihedral group of order 10 that detects all single-digit errors and alltransposition errors involving adjacent digits without the necessity ofavoiding certain numbers or introducing a new character. To describethis method, consider the permutation s 5 (01589427)(36) and the di-hedral group of order 10 as represented in Table 5.2. (Here we use 0 through 4 for the rotations, 5 through 9 for the reflections, and * for theoperation of D5.)

14

6 7

2 3

85top

9 10 11

14 15 16 22 23 2412 left 13 20 21 28

17 18 19

4144

46 47

42 43

4845bottom

front right25

30

26

312927

3236 rear33

38

34

393735

40

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5 | Permutation Groups 111

Verhoeff’s idea is to view the digits 0 through 9 as the elements of thegroup D5 and to replace ordinary addition with calculations done in D5.In particular, to any string of digits a1a2 . . . an21, we append the checkdigit an so that s(a1) p s2(a2) p ? ? ? p s n 22(an22) p s n 21(an21) p

s n (an) 5 0. [Here s2(x) 5 s(s(x)), s3(x) 5 s(s2 (x)), and so on.]Since s has the property that s i (a) 2 s i(b) if a 2 b, all single-digit er-rors are detected. Also, because

a p s(b) 2 b p s(a) if a 2 b, (1)

as can be checked on a case-by-case basis (see Exercise 49), it followsthat all transposition errors involving adjacent digits are detected [sinceEquation (1) implies that s i(a) p s i11(b) 2 si(b) p s i11(a) if a 2 b].

From 1990 until 2002, the German government used a minor modi-fication of Verhoeff’s check-digit scheme to append a check digit to theserial numbers on German banknotes. Table 5.3 gives the values of thefunctions s, s2, . . . , s10 needed for the computations. [The functionalvalue s i ( j) appears in the row labeled with s i and the column labeled j.]Since the serial numbers on the banknotes use 10 letters of the alphabet inaddition to the 10 decimal digits, it is necessary to assign numerical val-ues to the letters to compute the check digit. This assignment is shown inTable 5.4.

To any string of digits a1a2 . . . a10 corresponding to a banknote serialnumber, the check digit a11 is chosen such that s (a1) p s 2(a2) p ? ? ? p

s9(a9) p s10(a10) p a11 5 0 [instead of s(a1) p s2(a2) p ? ? ? p s10(a10) ps11(a11) 5 0 as in the Verhoeff scheme].

Table 5.2 Multiplication for D5

* 0 1 2 3 4 5 6 7 8 9

0 0 1 2 3 4 5 6 7 8 91 1 2 3 4 0 6 7 8 9 52 2 3 4 0 1 7 8 9 5 63 3 4 0 1 2 8 9 5 6 74 4 0 1 2 3 9 5 6 7 85 5 9 8 7 6 0 4 3 2 16 6 5 9 8 7 1 0 4 3 27 7 6 5 9 8 2 1 0 4 38 8 7 6 5 9 3 2 1 0 49 9 8 7 6 5 4 3 2 1 0

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112 Groups

To trace through a specific example, consider the banknote (featur-ing the mathematician Gauss) shown in Figure 5.3 with the numberAG8536827U7. To verify that 7 is the appropriate check digit, we ob-serve that s(0) p s2(2) p s3(8) p s 4 (5) p s 5 (3) p s 6 (6) p s7(8) ps 8 (2) p s9(7) p s10(7) p 7 5 1 p 0 p 2 p 2 p 6 p 6 p 5 p 2 p 0 p 1 p7 5 0, as it should be. [To illustrate how to use the multiplication tablefor D5, we compute 1 p 0 p 2 p 2 5 (1 p 0) p 2 p 2 5 1 p 2 p 2 5(1 p 2) p 2 5 3 p 2 5 0.]

Figure 5.3 German banknote with serial number AG8536827U and check digit 7.

One shortcoming of the German banknote scheme is that it does notdistinguish between a letter and its assigned numerical value. Thus, a

Table 5.3 Powers of s

0 1 2 3 4 5 6 7 8 9

s 1 5 7 6 2 8 3 0 9 4s2 5 8 0 3 7 9 6 1 4 2s3 8 9 1 6 0 4 3 5 2 7s4 9 4 5 3 1 2 6 8 7 0s5 4 2 8 6 5 7 3 9 0 1s6 2 7 9 3 8 0 6 4 1 5s7 7 0 4 6 9 1 3 2 5 8s8 0 1 2 3 4 5 6 7 8 9s9 1 5 7 6 2 8 3 0 9 4s10 5 8 0 3 7 9 6 1 4 2

Table 5.4 Letter Values

A D G K L N S U Y Z

0 1 2 3 4 5 6 7 8 9

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5 | Permutation Groups 113

substitution of 7 for U (or vice versa) and the transposition of 7 and Uare not detected by the check digit. Moreover, the banknote schemedoes not detect all transpositions of adjacent characters involving thecheck digit itself. For example, the transposition of D and 8 in posi-tions 10 and 11 is not detected. Both of these defects can be avoided byusing the Verhoeff method with D18, the dihedral group of order 36, toassign every letter and digit a distinct value together with an appropri-ate function s (see Gallian [1]). Using this method to append a checkcharacter, all single-position errors and all transposition errors involv-ing adjacent digits will be detected.

Exercises

1. Find the order of each of the following permutations.a. (14)b. (147)c. (14762)d.

2. Write each of the following permutations as a product of disjointcycles.a. (1235)(413)b. (13256)(23)(46512)c. (12)(13)(23)(142)

3. What is the order of each of the following permutations?a. (124)(357)b. (124)(3567)c. (124)(35)d. (124)(357869)e. (1235)(24567)f. (345)(245)

(a1a2 . . . ak)

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Text not available due to copyright restrictions

114 Groups

4. What is the order of each of the following permutations?

a.

b.

5. What is the order of the product of a pair of disjoint cycles oflengths 4 and 6?

6. Show that A8 contains an element of order 15.7. What are the possible orders for the elements of S6 and A6? What

about A7? (This exercise is referred to in Chapter 25.)8. What is the maximum order of any element in A10?9. Determine whether the following permutations are even or odd.

a. (135)b. (1356)c. (13567)d. (12)(134)(152)e. (1243)(3521)

10. Show that a function from a finite set S to itself is one-to-one if andonly if it is onto. Is this true when S is infinite? (This exercise is re-ferred to in Chapter 6.)

11. Let n be a positive integer. If n is odd, is an n-cycle an odd or aneven permutation? If n is even, is an n-cycle an odd or an even per-mutation?

12. If a is even, prove that a21 is even. If a is odd, prove that a21 is odd.13. Prove Theorem 5.6.14. In Sn, let a be an r-cycle, b an s-cycle, and g a t-cycle. Complete

the following statements: ab is even if and only if r 1 s is ______;abg is even if and only if r 1 s 1 t is ______.

15. Let a and b belong to Sn. Prove that ab is even if and only if aand b are both even or both odd.

16. Associate an even permutation with the number 11 and an oddpermutation with the number 21. Draw an analogy between theresult of multiplying two permutations and the result of multiply-ing their corresponding numbers 11 or 21.

17. Let

a 5 and b 5 .c1 2 3 4 5 6

6 1 2 4 3 5dc1 2 3 4 5 6

2 1 3 5 4 6d

c1 2 3 4 5 6 7

7 6 1 2 3 4 5d

c1 2 3 4 5 6

2 1 5 4 6 3d

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5 | Permutation Groups 115

Compute each of the following.a. a21

b. bac. ab

18. Let

a 5 and b 5 .

Write a, b, and ab asa. products of disjoint cycles,b. products of 2-cycles.

19. Show that if H is a subgroup of Sn, then either every member of His an even permutation or exactly half of the members are even.(This exercise is referred to in Chapter 25.)

20. Compute the order of each member of A4. What arithmetic rela-tionship do these orders have with the order of A4?

21. Give two reasons why the set of odd permutations in is not asubgroup.

22. Let a and b belong to Sn. Prove that a21b21ab is an evenpermutation.

23. Use Table 5.1 to compute the following.a. The centralizer of a3 5 (13)(24).b. The centralizer of a12 5 (124).

24. How many elements of order 5 are in S7?25. How many elements of order 4 does have? How many elements

of order 2 does have?26. Prove that (1234) is not the product of 3-cycles.27. Let b [ S7 and suppose b4 5 (2143567). Find b.28. Let b 5 (123)(145). Write b99 in disjoint cycle form.29. Find three elements s in S9 with the property that s3 5

(157)(283)(469).30. What cycle is (a1a2 ? ? ? an)

21?31. Let G be a group of permutations on a set X. Let a [ X and define

stab(a) 5 {a [ G|a(a) 5 a}. We call stab(a) the stabilizer of a inG (since it consists of all members of G that leave a fixed). Provethat stab(a) is a subgroup of G. (This subgroup was introduced byGalois in 1832.) This exercise is referred to in Chapter 7.

32. Let b 5 (1, 3, 5, 7, 9, 8, 6)(2, 4, 10). What is the smallest positiveinteger n for which bn 5 b25?

S6

S6

Sn

c1 2 3 4 5 6 7 8

1 3 8 7 6 5 2 4dc1 2 3 4 5 6 7 8

2 3 4 5 1 7 8 6d

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116 Groups

33. Let a 5 (1, 3, 5, 7, 9)(2, 4, 6)(8, 10). If am is a 5-cycle, what canyou say about m?

34. Let H 5 {b [ S5|b(1) 5 1 and b(3) 5 3}. Prove that H is a sub-group of S5. How many elements are in H? Is your argument validwhen 5 is replaced by any ? How many elements are in Hwhen 5 is replaced by any ?

35. How many elements of order 5 are there in A6?36. In S4, find a cyclic subgroup of order 4 and a noncyclic subgroup

of order 4.37. Suppose that b is a 10-cycle. For which integers i between 2 and

10 is bi also a 10-cycle?38. In S3, find elements a and b such that |a| 5 2, |b| 5 2, and |ab| 5 3.39. Find group elements a and b such that |a| 5 3, |b| 5 3, and

|ab| 5 5.40. Represent the symmetry group of an equilateral triangle as a group

of permutations of its vertices (see Example 3).41. Prove that Sn is non-Abelian for all n $ 3.42. Let a and b belong to Sn. Prove that bab21 and a are both even or

both odd.43. Show that A5 has 24 elements of order 5, 20 elements of order 3, and

15 elements of order 2. (This exercise is referred to in Chapter 25.)44. Find a cyclic subgroup of that has order 4.45. Find a noncyclic subgroup of that has order 4.46. Suppose that H is a subgroup of of odd order. Prove that H is a

subgroup of .47. Show that every element in An for n $ 3 can be expressed as a

3-cycle or a product of three cycles.48. Show that for n $ 3, Z(Sn) 5 {e}.49. Verify the statement made in the discussion of the Verhoeff check

digit scheme based on D5 that a * s(b) 2 b * s(a) for distinct a andb. Use this to prove that si(a) * si11(b) 2 si(b) * si11(a) for all i.Prove that this implies that all transposition errors involving adjacentdigits are detected.

50. Use the Verhoeff check-digit scheme based on D5 to append acheck digit to 45723.

51. Prove that every element of Sn (n . 1) can be written as a productof elements of the form (1k).

52. (Indiana College Mathematics Competition) A card-shuffling ma-chine always rearranges cards in the same way relative to the orderin which they were given to it. All of the hearts arranged in order

An

Sn

A8

A8

n $ 3n $ 3

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5 | Permutation Groups 117

from ace to king were put into the machine, and then the shuffledcards were put into the machine again to be shuffled. If the cardsemerged in the order 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7, in whatorder were the cards after the first shuffle?

53. Show that a permutation with odd order is an even permutation.54. Let G be a group. Prove or disprove that H 5 {g2 | g [ G} is a sub-

group of G. (Compare with Example 5 in Chapter 3.)55. Determine integers n for which e is a sub-

group of .56. Given that b and g are in with , and

, determine b and g.57. Why does the fact that the orders of the elements of A4 are 1, 2, and

3 imply that |Z(A4)| 5 1?58. Label the four locations of tires on an automobile with the labels

1, 2, 3, and 4, clockwise. Let a represent the operation of switchingthe tires in positions 1 and 3 and switching the tires in positions2 and 4. Let b represent the operation of rotating the tires in posi-tions 2, 3, and 4 clockwise and leaving the tire in position 1 as is.Let G be the group of all possible combinations of a and b. Howmany elements are in G?

59. Shown below are four tire rotation patterns recommended by theDunlop Tire Company. Explain how these patterns can be repre-sented as permutations in S4 and find the smallest subgroup of S4that contains these four patterns. Is the subgroup Abelian?

b(1) 5 4gb 5 (1243)bg 5 (1432)S4

An

6H 5 5a [ An |a2 5

FRONT

Modified X

Rear Wheel DriveVehicles

4 Wheel DriveVehicles

FRONT

Modified X

X Tires tothe Driven Axle

Front Wheel DriveVehicles

Alternate Pattern

FRONT

X

FRONT

Normal

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118 Groups

Computer Exercises

Science is what we understand well enough to explain to a computer. Art is everything else we do.

DONALD KNUTH, The Art of Computer Programming, 1969

Software for Computer Exercise 1 in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software determines whether the two permutations (1x) and(123 ? ? ? n) generate Sn for various choices of x and n (that is,whether every element of Sn can be expressed as some product ofthese permutations). For n 5 4, run the program for x 5 2, 3, and4. For n 5 5, run the program for x 5 2, 3, 4, and 5. For n 5 6, runthe program for x 5 2, 3, 4, 5, and 6. For n 5 8, run the programfor x 5 2, 3, 4, 5, 6, 7, and 8. Conjecture a necessary and sufficientcondition involving x and n for (1x) and (123 ? ? ? n) to generate Sn.

2. Use a software package (see Suggested Software on page 120) toexpress the following permutations in terms of the r and s given inExample 8. (For GAP, the prompt brk. means that the permuta-tion entered is not in the group. In this situation, use Control-D toreturn to the main prompt. Be advised that GAP composes permu-tations from left to right as opposed to our method of right to left.)a. (456)b. (23)c. (12)(34)d. (12)(34)(56)

3. Repeat Example 8 for the puzzle shown here.

1

2

3

4

5

6

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5 | Permutation Groups 119

References

1. J. A. Gallian, “The Mathematics of Identification Numbers,” The CollegeMathematics Journal 22 (1991): 194–202.

2. J. Verhoeff, Error Detecting Decimal Codes, Amsterdam: Math-ematischCentrum, 1969.

Suggested Readings

Douglas E. Ensely, “Invariants Under Actions to Amaze Your Friends,” Math-ematics Magazine, Dec. 1999: 383–387.

This article explains some card tricks that are based on permutationgroups.

Dmitry Fomin, “Getting It Together with ‘Polynominoes,’ ” Quantum,Nov./Dec. 1991: 20–23.

In this article, permutation groups are used to analyze various sorts ofcheckerboard tiling problems.

J. A. Gallian, “Error Detection Methods,” ACM Computing Surveys 28(1996): 504–517.

This article gives a comprehensive survey of error-detection methods thatuse check digits. This article can be downloaded at http://www.d.umn.edu/~jgallian/detection.pdf

I. N. Herstein and I. Kaplansky, Matters Mathematical, New York: Chelsea,1978.

Chapter 3 of this book discusses several interesting applications of permu-tations to games.

Douglas Hofstadter, “The Magic Cube’s Cubies Are Twiddled by Cubists andSolved by Cubemeisters,” Scientific American 244 (1981): 20–39.

This article, written by a Pulitzer Prize recipient, discusses the group the-ory involved in the solution of the Magic (Rubik’s) Cube. In particular,permutation groups, subgroups, conjugates (elements of the form xyx21),commutators (elements of the form xyx21y21), and the “always even oralways odd” theorem (Theorem 5.5) are prominently mentioned. At onepoint, Hofstadter says, “It is this kind of marvelously concrete illustrationof an abstract notion of group theory that makes the Magic Cube one ofthe most amazing things ever invented for teaching mathematical ideas.”

John O. Kiltinen, Oval Track & Other Permutation Puzzles & Just EnoughGroup Theory to Solve Them, Mathematical Association of America,Washington, D.C., 2003.

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120 Groups

This book and the software that comes with it present the user with an arrayof computerized puzzles, plus tools to vary them in thousands of ways. Thebook provides the background needed to use the puzzle software to its fullestpotential, and also gives the reader a gentle, not-too-technical introduction tothe theory of permutation groups that is a prerequisite to a full understandingof how to solve puzzles of this type. The website http://www-instruct.nmu.edu/math_cs/kiltinen/web/mathpuzzles/ provides resources that expandupon the book. It also has news about puzzle software—modules that addfunctionality and fun to puzzles.

Will Oakley, “Portrait of Three Puzzle Graces,” Quantum, Nov./Dec. 1991:83–86.

The author uses permutation groups to analyze solutions to the 15 puzzle,Rubik’s Cube, and Rubik’s Clock.

A. White and R. Wilson, “The Hunting Group,” Mathematical Gazette 79(1995): 5–16.

This article explains how permutation groups are used in bell ringing.

S. Winters, “Error-Detecting Schemes Using Dihedral Groups,” UMAPJournal 11, no. 4 (1990): 299–308.

This article discusses error-detection schemes based on Dn for n odd.Schemes for both one and two check digits are analyzed.

Suggested Software

GAP is free for downloading. Versions are available for Unix, Windows,and Macintosh at:

http://www.gap-system.org

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Augustin Cauchy

You see that little young man? Well! He will supplant all of us in so far as we are mathematicians.

Spoken by Lagrange

to Laplace About the

11-year-old Cauchy

AUGUSTIN LOUIS CAUCHY was born onAugust 21, 1789, in Paris. By the time he was 11, both Laplace and Lagrange hadrecognized Cauchy’s extraordinary talentfor mathematics. In school he won prizes forGreek, Latin, and the humanities. At the ageof 21, he was given a commission inNapoleon’s army as a civil engineer. For thenext few years, Cauchy attended to his engi-neering duties while carrying out brilliantmathematical research on the side.

In 1815, at the age of 26, Cauchy wasmade Professor of Mathematics at the ÉcolePolytechnique and was recognized as theleading mathematician in France. Cauchyand his contemporary Gauss were amongthe last mathematicians to know the wholeof mathematics as known at their time, andboth made important contributions to nearly

every branch, both pure and applied, as wellas to physics and astronomy.

Cauchy introduced a new level of rigorinto mathematical analysis. We owe ourcontemporary notions of limit and continu-ity to him. He gave the first proof of theFundamental Theorem of Calculus. Cauchywas the founder of complex function theoryand a pioneer in the theory of permutationgroups and determinants. His total writtenoutput of mathematics fills 24 large volumes.He wrote more than 500 research papersafter the age of 50. Cauchy died at the age of67 on May 23, 1857.

For more information about Cauchy,visit:

http://www–groups.dcs.st-and.ac.uk/~history/

121

This stamp was issued by France in Cauchy’s honor.

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122

Isomorphisms6

MotivationSuppose an American and a German are asked to count a handful of ob-jects. The American says, “One, two, three, four, five, . . . ,” whereas theGerman says “Eins, zwei, drei, vier, fünf, . . .” Are the two doing differentthings? No. They are both counting the objects, but they are using differ-ent terminology to do so. Similarly, when one person says: “Two plusthree is five” and another says: “Zwei und drei ist fünf,” the two are inagreement on the concept they are describing, but they are using differentterminology to describe the concept. An analogous situation often occurswith groups; the same group is described with different terminology. Wehave seen two examples of this so far. In Chapter 1, we described the sym-metries of a square in geometric terms (e.g., R90), whereas in Chapter 5 wedescribed the same group by way of permutations of the corners. In bothcases, the underlying group was the symmetries of a square. In Chapter 4,we observed that when we have a cyclic group of order n generated by a,the operation turns out to be essentially that of addition modulo n, sincearas 5 ak, where k 5 (r 1 s) mod n. For example, each of U(43) and U(49)is cyclic of order 42. So, each has the form �a�, where aras 5 a (r 1 s) mod 42.

Definition and ExamplesIn this chapter, we give a formal method for determining whether twogroups defined in different terms are really the same. When this is thecase, we say that there is an isomorphism between the two groups. Thisnotion was first introduced by Galois about 175 years ago. The termisomorphism is derived from the Greek words isos, meaning “same” or“equal,” and morphe, meaning “form.” R. Allenby has colorfully

The basis for poetry and scientific discovery is the ability to comprehend the unlike in the like and the like in the unlike.

JACOB BRONOWSKI

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6 | Isomorphisms 123

defined an algebraist as “a person who can’t tell the difference betweenisomorphic systems.”

Definition Group Isomorphism

An isomorphism f from a group G to a group is a one-to-one map-ping (or function) from G onto that preserves the group operation.That is,

f(ab) 5 f(a)f(b) for all a, b in G.

If there is an isomorphism from G onto , we say that G and areisomorphic and write G < .

This definition can be visualized as shown in Figure 6.1. The pairsof dashed arrows represent the group operations.

Figure 6.1

It is implicit in the definition of isomorphism that isomorphicgroups have the same order. It is also implicit in the definition ofisomorphism that the operation on the left side of the equal sign is thatof G, whereas the operation on the right side is that of . The fourcases involving ? and 1 are shown in Table 6.1.

G

a

b

ab

(a)

(b)

G Gφ

φ

φφ

φ

φφ (a) (b)

GGG

GG

Table 6.1

G Operation Operation Operation Preservation

? ? f(a ? b) 5 f(a) ? f(b)? 1 f(a ? b) 5 f(a) 1 f(b)1 ? f(a 1 b) 5 f(a) ? f(b)1 1 f(a 1 b) 5 f(a) 1 f(b)

G

There are four separate steps involved in proving that a group G isisomorphic to a group .

Step 1 “Mapping.” Define a candidate for the isomorphism; that is, de-fine a function f from G to .

Step 2 “1–1.” Prove that f is one-to-one; that is, assume that f(a) 5f(b) and prove that a 5 b.

G

G

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124 Groups

Step 3 “Onto.” Prove that f is onto; that is, for any element in ,find an element g in G such that f(g) 5 .

Step 4 “O.P.” Prove that f is operation-preserving; that is, show thatf(ab) 5 f(a)f(b) for all a and b in G.

None of these steps is unfamiliar to you. The only one that may appearnovel is the fourth one. It requires that one be able to obtain the sameresult by combining two elements and then mapping, or by mappingtwo elements and then combining them. Roughly speaking, this saysthat the two processes—operating and mapping—can be done in eitherorder without affecting the result. This same concept arises in calculuswhen we say

or

Before going any further, let’s consider some examples.

EXAMPLE 1 Let G be the real numbers under addition and let bethe positive real numbers under multiplication. Then G and are iso-morphic under the mapping f(x) 5 2x. Certainly, f is a function fromG to . To prove that it is one-to-one, suppose that 2x 5 2y. Then log2 2x 5log2 2y, and therefore x 5 y. For “onto,” we must find for any positivereal number y some real number x such that f(x) 5 y; that is, 2x 5 y.Well, solving for x gives log2 y. Finally,

f(x 1 y) 5 2x1y 5 2x ? 2y 5 f(x)f(y)

for all x and y in G, so that f is operation-preserving as well.

EXAMPLE 2 Any infinite cyclic group is isomorphic to Z. Indeed, ifa is a generator of the cyclic group, the mapping ak → k is anisomorphism. Any finite cyclic group �a� of order n is isomorphic to Zn under the mapping ak → k mod n. That these correspondences arefunctions and are one-to-one is the essence of Theorem 4.1. Obviously,the mappings are onto. That the mappings are operation-preservingfollows from Exercise 11 in Chapter 0 in the finite case and from thedefinitions in the infinite case.

G

GG

3b

a

( f 1 g) dx 5 3b

a

f dx 1 3b

a

g dx.

limxSa

(

f (x) ? g(x) ) 5 lim xSa

f (x) limxSa

g(x)

gGg

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6 | Isomorphisms 125

EXAMPLE 3 The mapping from R under addition to itself given byf(x) 5 x3 is not an isomorphism. Although f is one-to-one and onto, itis not operation-preserving, since it is not true that (x 1 y)3 5 x3 1 y3

for all x and y.

EXAMPLE 4 U(10) < Z4 and U(5) < Z4. To verify this, one needonly observe that both U(10) and U(5) are cyclic of order 4. Then ap-peal to Example 2.

EXAMPLE 5 U(10) ] U(12). This is a bit trickier to prove. First,note that x2 5 1 for all x in U(12). Now, suppose that f is an isomor-phism from U(10) onto U(12). Then,

f(9) 5 f(3 ? 3) 5 f(3)f(3) 5 1

and

f(1) 5 f(1 ? 1) 5 f(1)f(1) 5 1.

Thus, f(9) 5 f(1), but 9 2 1, which contradicts the assumption thatf is one-to-one.

EXAMPLE 6 There is no isomorphism from Q, the group of rationalnumbers under addition, to Q*, the group of nonzero rational numbersunder multiplication. If f were such a mapping, there would be a ra-tional number a such that f(a) 5 21. But then

21 5 f(a) 5 f( a 1 a) 5 f( a)f( a) 5 [f( a)]2.

However, no rational number squared is 21.

EXAMPLE 7 Let G 5 SL(2, R), the group of 2 3 2 real matriceswith determinant 1. Let M be any 2 3 2 real matrix with determinant 1.Then we can define a mapping from G to G itself by fM(A) 5 MAM21

for all A in G. To verify that fM is an isomorphism, we carry out thefour steps.

Step 1 fM is a function from G to G. Here, we must show that fM(A)is indeed an element of G whenever A is. This follows from propertiesof determinants:

det (MAM21) 5 (det M)(det A)(det M)21 5 1 ? 1 ? 121 5 1.

Thus, MAM21 is in G.

Step 2 fM is one-to-one. Suppose that fM(A) 5 fM(B). Then MAM21 5MBM21 and, by left and right cancellation, A 5 B.

12

12

12

12

12

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126 Groups

Step 3 fM is onto. Let B belong to G. We must find a matrix A in Gsuch that fM(A) 5 B. How shall we do this? If such a matrix A is to ex-ist, it must have the property that MAM21 5 B. But this tells us exactlywhat A must be! For we can solve for A to obtain A 5 M21BM andverify that fM(A) 5 MAM21 5 M(M21BM)M21 5 B.

Step 4 fM is operation-preserving. Let A and B belong to G. Then,

fM(AB) 5 M(AB)M21 5 MA(M21M)BM21

5 (MAM21)(MBM21) 5 fM(A)fM(B).

The mapping fM is called conjugation by M.

Cayley’s TheoremOur first theorem is a classic result of Cayley. An important generaliza-tion of it will be given in Chapter 25.

Theorem 6.1 Cayley’s Theorem (1854)

PROOF To prove this, let G be any group. We must find a group ofpermutations that we believe is isomorphic to G. Since G is all we haveto work with, we will have to use it to construct . For any g in G,define a function Tg from G to G by

Tg(x) 5 gx for all x in G.

(In words, Tg is just multiplication by g on the left.) We leave it as anexercise (Exercise 23) to prove that Tg is a permutation on the set ofelements of G. Now, let 5 {Tg | g [ G}. Then, is a group underthe operation of function composition. To verify this, we first observethat for any g and h in G we have TgTh(x) 5 Tg(Th(x)) 5 Tg(hx) 5 g(hx) 5(gh)x 5 Tgh(x), so that TgTh 5 Tgh. From this it follows that Te is theidentity and (Tg)

21 5 Tg21 (see Exercise 9). Since function compositionis associative, we have verified all the conditions for to be a group.

The isomorphism f between G and is now ready-made. For everyg in G, define f(g) 5 Tg. If Tg 5 Th, then Tg(e) 5 Th(e) or ge 5 he.Thus, g 5 h and f is one-to-one. By the way was constructed, wesee that f is onto. The only condition that remains to be checked is thatf is operation-preserving. To this end, let a and b belong to G. Then

f(ab) 5 Tab 5 TaTb 5 f(a)f(b).

G

GG

GG

G

G

Every group is isomorphic to a group of permutations.

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6 | Isomorphisms 127

The group constructed above is called the left regular representa-tion of G.

EXAMPLE 8 For concreteness, let us calculate the left regular repre-sentation for U(12) 5 {1, 5, 7, 11}. Writing the permutations ofU(12) in array form, we have (remember, Tx is just multiplication by x)

, ,

, .

It is instructive to compare the Cayley table for U(12) and its left regu-lar representation .U(12)

T11 5 c 1 5 7 11

11 7 5 1dT7 5 c1 5 7 11

7 11 1 5d

T5 5 c1 5 7 11

5 1 11 7dT1 5 c1 5 7 11

1 5 7 11d

U(12)

G

T1 T5 T7 T11

T1 T1 T5 T7 T11T5 T5 T1 T11 T7T7 T7 T11 T1 T5T11 T11 T7 T5 T1

U(12)U(12) 1 5 7 11

1 1 5 7 115 5 1 11 77 7 11 1 5

11 11 7 5 1

It should be abundantly clear from these tables that U(12) and are only notationally different.

Cayley’s Theorem is important for two contrasting reasons. One isthat it allows us to represent an abstract group in a concrete way. A sec-ond is that it shows that the present-day set of axioms we have adoptedfor a group is the correct abstraction of its much earlier predecessor—agroup of permutations. Indeed, Cayley’s Theorem tells us that abstractgroups are not different from permutation groups. Rather, it is theviewpoint that is different. It is this difference of viewpoint that hasstimulated the tremendous progress in group theory and many otherbranches of mathematics in the 20th century.

It is sometimes very difficult to prove or disprove, whichever thecase may be, that two particular groups are isomorphic. For example, itrequires somewhat sophisticated techniques to prove the surprising factthat the group of real numbers under addition is isomorphic to thegroup of complex numbers under addition. Likewise, it is not easy to prove the fact that the group of nonzero complex numbers undermultiplication is isomorphic to the group of complex numbers with ab-solute value of 1 under multiplication. In geometric terms, this saysthat, as groups, the punctured plane and the unit circle are isomorphic.(See reference 1.)

U(12)

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128 Groups

PROOF We will restrict ourselves to proving only properties 1, 2, and 4,but observe that property 5 follows from properties 1 and 2, property 6follows from property 2, and property 7 follows from property 5. Forconvenience, let us denote the identity in G by e and the identity in by . Then, since e 5 ee, we have

f(e) 5 f(ee) 5 f(e)f(e).

Also, because f(e) [ , we have f(e) 5 f(e), as well. Thus, by can-cellation, 5 f(e). This proves property 1.

For positive integers, property 2 follows from the definition of anisomorphism and mathematical induction. If n is negative, then 2n ispositive, and we have from property 1 and the observation about thepositive integer case that e 5 f(e) 5 f(gng2n) 5 f(gn)f(g2n) 5f(gn)(f(g))2n. Thus, multiplying both sides on the right by (f(g))n, wehave (f(g))n 5 f(gn). Property 1 takes care of the case n 5 0.

To prove property 4, let G 5 �a� and note that, by closure, �f(a)� #. Because f is onto, for any element b in , there is an element ak in

G such that f(ak) 5 b. Thus, b 5 (f(a))k and so b [ �f(a)�. Thisproves that 5 �f(a)�.

Now suppose that 5 �f(a)�. Clearly, �a� # G. For any element b in G, we have f(b) [ �f(a)�. So, for some integer k we have

GG

GG

eeG

eG

Properties of IsomorphismsOur next two theorems give a catalog of properties of isomorphismsand isomorphic groups.

Theorem 6.2 Properties of Isomorphisms Acting on Elements

Suppose that f is an isomorphism from a group G onto a group .Then

1. f carries the identity of G to the identity of .2. For every integer n and for every group element a in G, f(an) 5

[f(a)]n.3. For any elements a and b in G, a and b commute if and only if

f(a) and f(b) commute.4. G 5 �a� if and only if 5 �f(a)�.5. |a| 5 |f(a)| for all a in G (isomorphisms preserve orders).6. For a fixed integer k and a fixed group element b in G, the

equation xk 5 b has the same number of solutions in G as doesthe equation xk 5 f(b) in .

7. If G is finite, then G and have exactly the same number ofelements of every order.

GG

G

G

G

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6 | Isomorphisms 129

f(b) 5 (f(a))k 5 f(ak). Because f is one-to-one, b 5 ak. This provesthat �a� 5 G.

When the group operation is addition, property 2 of Theorem 6.2 isf(na) 5 nf(a); property 4 says that an isomorphism between twocyclic groups takes a generator to a generator.

Property 6 is quite useful for showing that two groups are not iso-morphic. Often b is picked to be the identity. For example, consider C*and R*. Because the equation x4 5 1 has four solutions in C* but onlytwo in R*, no matter how one attempts to define an isomorphism fromC* to R*, property 6 cannot hold.

Theorem 6.3 Properties of Isomorphisms Acting on Groups

Suppose that f is an isomorphism from a group G onto a group .Then

1. f21 is an isomorphism from onto G.2. G is Abelian if and only if is Abelian.3. G is cyclic if and only if is cyclic.4. If K is a subgroup of G, then f(K) 5 {f(k) | k [ K} is a

subgroup of .G

GG

G

G

PROOF Properties 1 and 4 are left as exercises (Exercises 21 and 22).Property 2 is a direct consequence of property 3 of Theorem 6.2.Property 3 follows from property 4 of Theorem 6.2 and property 1 ofTheorem 6.3.

Theorems 6.2 and 6.3 show that isomorphic groups have many prop-erties in common. Actually, the definition is precisely formulated sothat isomorphic groups have all group-theoretic properties in common.By this we mean that if two groups are isomorphic, then any propertythat can be expressed in the language of group theory is true for one ifand only if it is true for the other. This is why algebraists speak of iso-morphic groups as “equal” or “the same.” Admittedly, calling suchgroups equivalent, rather than the same, might be more appropriate, butwe bow to long-standing tradition.

AutomorphismsCertain kinds of isomorphisms are referred to so often that they havebeen given special names.

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130 Groups

Definition Automorphism

An isomorphism from a group G onto itself is called an automorphismof G.

The isomorphism in Example 7 is an automorphism of SL(2, R).Two more examples follow.

EXAMPLE 9 The function f from C to C given by f(a 1 bi) 5a 2 bi is an automorphism of the group of complex numbers underaddition. The restriction of f to C* is also an automorphism of thegroup of nonzero complex numbers under multiplication. (SeeExercise 25.)

EXAMPLE 10 Let R2 5 {(a, b) | a, b [ R}. Then f(a, b) 5 (b, a)is an automorphism of the group R2 under componentwise addition.Geometrically, f reflects each point in the plane across the line y 5 x.More generally, any reflection across a line passing through theorigin or any rotation of the plane about the origin is an automor-phism of R2.

The isomorphism in Example 7 is a particular instance of an auto-morphism that arises often enough to warrant a name and notation ofits own.

Definition Inner Automorphism Induced by a

Let G be a group, and let a [ G. The function fa defined by fa(x) 5axa21 for all x in G is called the inner automorphism of G induced by a.

We leave it for the reader to show that fa is actually an automor-phism of G. (Use Example 7 as a model.)

EXAMPLE 11 The action of the inner automorphism of D4 inducedby R90 is given in the following table.

x → R90 x R9021

R0 → R90R0R90–1 5 R0

R90 → R90R90R9021 5 R90

R180 → R90R180R9021 5 R180

R270 → R90R270R9021 5 R270

H → R90HR9021 5 V

V → R90VR9021 5 H

D → R90DR9021 5 D9

D9 → R90D9R9021 5 D

fR90

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6 | Isomorphisms 131

When G is a group, we use Aut(G) to denote the set of all auto-morphisms of G and Inn(G) to denote the set of all inner automor-phisms of G. The reason these sets are noteworthy is demonstrated bythe next theorem.

Theorem 6.4 Aut(G) and Inn(G) Are Groups†

PROOF The proof of Theorem 6.4 is left as an exercise (Exercise 15).

The determination of Inn(G) is routine. If G 5 {e, a, b, c. . . .}, thenInn(G) 5 {fe, fa, fb, fc, . . .}. This latter list may have duplications,however, since fa may be equal to fb even though a 2 b (see Exercise33). Thus, the only work involved in determining Inn(G) is decidingwhich distinct elements give the distinct automorphisms. On the otherhand, the determination of Aut(G) is, in general, quite involved.

EXAMPLE 12 Inn(D4

)

To determine Inn(D4), we first observe that the complete list of innerautomorphisms is fR0

, fR90, fR180

, fR270, fH, fV, fD, and fD9. Our job is

to determine the repetitions in this list. Since R180 [ Z(D4), we havefR180

(x) 5 R180 xR18021 5 x, so that fR180

5 fR0. Also, fR270

(x) 5R270 xR270

21 5 R90R180 xR18021R90

21 5 R90 xR9021 5 fR90

(x). Similarly,since H 5 R180V and D9 5 R180D, we have fH 5 fV and fD 5 fD9.This proves that the previous list can be pared down to fR0

, fR90, fH,

and fD. We leave it to the reader to show that these are distinct(Exercise 13).

EXAMPLE 13 Aut(Z10

)

To compute Aut(Z10), we try to discover enough information about anelement a of Aut(Z10) to determine how a must be defined. Because Z10is so simple, this is not difficult to do. To begin with, observe that oncewe know a(1), we know a(k) for any k, because

The set of automorphisms of a group and the set of innerautomorphisms of a group are both groups under the operation of function composition.

†The group Aut(G) was first studied by O. Hölder in 1893 and, independently, by E. H. Moore in 1894.

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132 Groups

a(k) 5 a(1 1 1 1 ? ? ? 1 1)k terms

5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 ka(1).k terms

So, we need only determine the choices for a(1) that make a anautomorphism of Z10. Since property 5 of Theorem 6.2 tells us that|a(1)| 5 10, there are four candidates for a(1):

a(1) 5 1; a(1) 5 3; a(1) 5 7; a(1) 5 9.

To distinguish among the four possibilities, we refine our notation bydenoting the mapping that sends 1 to 1 by a1, 1 to 3 by a3, 1 to 7 by a7,and 1 to 9 by a9. So the only possibilities for Aut(Z10) are a1, a3, a7, anda9. But are all these automorphisms? Clearly, a1 is the identity. Let uscheck . Since implies ,

is well defined. Moreover, because is a generator of , itfollows that a3 is onto (and, by Exercise 10 in Chapter 5, it is also one-to-one). Finally, since a3(a 1 b) 5 3(a 1 b) 5 3a 1 3b 5 a3(a) 1 a3(b),we see that a3 is operation-preserving as well. Thus, a3 [ Aut(Z10). Thesame argument shows that a7 and a9 are also automorphisms.

This gives us the elements of Aut(Z10) but not the structure. For in-stance, what is a3a3? Well, (a3a3)(1) 5 a3(3) 5 3 ? 3 5 9 5 a9(1), soa3a3 5 a9. Similar calculations show that a 3

3 5 a7 and a 34 5 a1, so

that |a3| 5 4. Thus, Aut(Z10) is cyclic. Actually, the following Cayleytables reveal that Aut(Z10) is isomorphic to U(10).

Z10a3(1) 5 3a3

3x mod 10 5 3y mod 10x mod 10 5 y mod 10a3

U(10) 1 3 7 9

1 1 3 7 93 3 9 1 77 7 1 9 39 9 7 3 1

Aut(Z10) a1 a3 a7 a9

a1 a1 a3 a7 a9a3 a3 a9 a1 a7a7 a7 a1 a9 a3a9 a9 a7 a3 a1

With Example 13 as a guide, we are now ready to tackle the groupAut(Zn). The result is particularly nice, since it relates the two kinds ofgroups we have most frequently encountered thus far—the cyclicgroups Zn and the U-groups U(n).

Theorem 6.5 Aut(Zn) < U(n)

For every positive integer n, Aut(Zn) is isomorphic to U(n).

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6 | Isomorphisms 133

PROOF As in Example 13, any automorphism a is determined by thevalue of a(1), and a(1) [ U(n). Now consider the correspondencefrom Aut(Zn) to U(n) given by T: a → a(1). The fact that a(k) 5 ka(1)(see Example 13) implies that T is a one-to-one mapping. For if a andb belong to Aut(Zn) and a(1) 5 b(1), then a(k) 5 ka(1) 5 kb(1) 5b(k) for all k in Zn, and therefore a 5 b.

To prove that T is onto, let r [ U(n) and consider the mapping afrom Zn to Zn defined by a(s) 5 sr (mod n) for all s in Zn. We leave it asan exercise to verify that a is an automorphism of Zn (see Exercise 17).Then, since T(a) 5 a(1) 5 r, T is onto U(n).

Finally, we establish the fact that T is operation-preserving. Let a,b [ Aut(Zn). We then have

T(ab) 5 (ab)(1) 5 a(b(1)) 5 a(1 1 1 1 ? ? ? 1 1)

b(1) terms

5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 a(1)b(1)

b(1) terms5 T(a)T(b).

This completes the proof.

Exercises

Being a mathematician is a bit like being a manic depressive: you spendyour life alternating between giddy elation and black despair.

STEVEN G. KRANTZ, A Primer of Mathematical Writing

1. Find an isomorphism from the group of integers under addition tothe group of even integers under addition.

2. Find Aut(Z).3. Let R1 be the group of positive real numbers under multiplication.

Show that the mapping f(x) 5 is an automorphism of R1.4. Show that U(8) is not isomorphic to U(10).5. Show that U(8) is isomorphic to U(12).6. Prove that the notion of group isomorphism is transitive. That is, if

G, H, and K are groups and G < H and H < K, then G < K.7. Prove that S4 is not isomorphic to D12.8. Show that the mapping is an isomorphism from R+

under multiplication to R under addition.9. In the notation of Theorem 6.1, prove that Te is the identity and

that (Tg)21 5 Tg21.

a S log10 a

"x

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134 Groups

10. Let G be a group. Prove that the mapping a(g) 5 g21 for all g in Gis an automorphism if and only if G is Abelian.

11. For inner automorphisms fg, fh , and fgh, prove that fgfh 5 fgh.12. Find two groups G and H such that G ] H, but Aut(G) < Aut(H).13. Prove the assertion in Example 12 that the inner automorphisms

fR0, fR90

, fH, and fD of D4 are distinct.14. Find Aut(Z6).15. If G is a group, prove that Aut(G) and Inn(G) are groups.16. Prove that the mapping from U(16) to itself given by x → x3 is an

automorphism. What about x → x5 and x → x7? Generalize.17. Let r [ U(n). Prove that the mapping a: Zn → Zn defined by a(s) 5

sr mod n for all s in Zn is an automorphism of Zn. (This exercise isreferred to in this chapter.)

18. The group a[ Z is isomorphic to what familiar group?

What if Z is replaced by R?19. If and g are isomorphisms from the cyclic group to some

group and , prove that .20. Suppose that : is an automorphism with .

Determine a formula for .21. Prove Property 1 of Theorem 6.3.22. Prove Property 4 of Theorem 6.3.23. Referring to Theorem 6.1, prove that Tg is indeed a permutation on

the set G.24. Prove or disprove that U(20) and U(24) are isomorphic.25. Show that the mapping f(a 1 bi) 5 a 2 bi is an automorphism of

the group of complex numbers under addition. Show that f pre-serves complex multiplication as well—that is, f(xy) 5 f(x)f(y)for all x and y in C. (This exercise is referred to in Chapter 15.)

26. Let

G 5 {a 1 b | a, b rational}

and

H 5 a, b rational .

Show that G and H are isomorphic under addition. Prove that Gand H are closed under multiplication. Does your isomorphismpreserve multiplication as well as addition? (G and H are examplesof rings—a topic we will take up in Part 3.)

f`ca 2b

b ade

"2

f(x)f(11) 5 13Z50 S Z50f

f 5 gf(a) 5 g(a)�a�f

f`e c1 a

0 1d

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6 | Isomorphisms 135

27. Prove that Z under addition is not isomorphic to Q under addition.28. Prove that the quaternion group (see Exercise 4, Supplementary Exer-

cises for Chapters 1–4) is not isomorphic to the dihedral group D4.29. Let C be the complex numbers and

M 5 a, b [ R .

Prove that C and M are isomorphic under addition and that C* andM*, the nonzero elements of M, are isomorphic under multiplication.

30. Let Rn 5 {(a1, a2, . . . , an) | ai [ R}. Show that the mapping f:(a1, a2, . . . , an) → (2a1, 2a2, . . . , 2an) is an automorphism ofthe group Rn under componentwise addition. This automorphismis called inversion. Describe the action of f geometrically.

31. Consider the following statement: The order of a subgroup dividesthe order of the group. Suppose you could prove this for finitepermutation groups. Would the statement then be true for all finitegroups? Explain.

32. Suppose that G is a finite Abelian group and G has no element oforder 2. Show that the mapping g → g2 is an automorphism of G.Show, by example, that if G is infinite the mapping need not be anautomorphism.

33. Let G be a group and let g [ G. If z [ Z(G), show that the innerautomorphism induced by g is the same as the inner automorphisminduced by zg (that is, that the mappings fg and fzg are equal).

34. If a and g are elements of a group, prove that is isomorphic to.

35. Suppose that g and h induce the same inner automorphism of agroup G. Prove that h21g [ Z(G).

36. Combine the results of Exercises 33 and 35 into a single “if andonly if” theorem.

37. Let a belong to a group G and let |a| be finite. Let fa be the auto-morphism of G given by fa(x) 5 axa21. Show that |fa| divides |a|.Exhibit an element a from a group for which 1 , |fa| , |a|.

38. Let G 5 {0, 62, 64, 66, . . .} and H 5 {0, 63, 66, 69, . . .}.Show that G and H are isomorphic groups under addition. Doesyour isomorphism preserve multiplication? Generalize to the casewhen and , where m and n are integers.

39. Suppose that is an automorphism of such that and . Determine and .

40. In Aut(Z9), let ai denote the automorphism that sends 1 to i where gcd(i, 9) 5 1. Write a5 and a8 as permutations of {0, 1, . . . , 8} in disjoint cycle form. [For example, a2 5 (0)(124875)(36).]

f(H)f(D)f(V) 5 Vf(R90) 5 R270D4f

H 5 �n�G 5 �m�

C(gag21)C(a)

f`ca 2b

b ade

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136 Groups

41. Write the permutation corresponding to R90 in the left regular rep-resentation of D4 in cycle form.

42. Show that every automorphism f of the rational numbers Q underaddition to itself has the form f(x) 5 xf(1).

43. Prove that Q1, the group of positive rational numbers under multi-plication, is isomorphic to a proper subgroup.

44. Prove that Q, the group of rational numbers under addition, is notisomorphic to a proper subgroup of itself.

45. Prove that every automorphism of R*, the group of nonzero realnumbers under multiplication, maps positive numbers to positivenumbers and negative numbers to negative numbers.

46. Let G be a finite group. Show that in the disjoint cycle form of theright regular representation of G each cycle haslength .

47. Give a group-theoretic proof that Q under addition is not isomor-phic to R+ under multiplication.

Reference

1. J. R. Clay, “The Punctured Plane Is Isomorphic to the UnitCircle,” Journal of Number Theory 1 (1964): 500–501.

Computer Exercise

There is only one satisfying way to boot a computer.J. H. GOLDFUSS

Software for the computer exercise in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software computes the order of Aut(Dn). Run the program for n 5 3, 5, 7, and 11. Make a conjecture about the order when n isprime. Run the program for n 5 4, 8, 16, and 32. Make a conjectureabout the order when n is a power of 2. Run the program when n 56, 10, 14, and 22. Make a conjecture about the order when n is twicea prime. Run the program for n 5 9, 15, 21, and 33. Make a conjec-ture about the order when n is 3 times a prime. Try to deduce a gen-eral formula for the order of Aut(Dn).

0g 0 Tg(x) 5 xg

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137

Arthur Cayley

Cayley is forging the weapons for futuregenerations of physicists.

PETER TAIT

ARTHUR CAYLEY was born on August 16,1821, in England. His genius showed itself atan early age. He published his first researchpaper while an undergraduate of 20, and inthe next year he published eight papers.While still in his early twenties, he originatedthe concept of n-dimensional geometry.

After graduating from Trinity College,Cambridge, Cayley stayed on for three yearsas a tutor. At the age of 25, he began a 14-year career as a lawyer. During this period,he published approximately 200 mathemati-cal papers, many of which are now classics.

In 1863, Cayley accepted the newly es-tablished Sadlerian professorship of mathe-matics at Cambridge University. He spentthe rest of his life in that position. One of hisnotable accomplishments was his role in thesuccessful effort to have women admitted toCambridge.

Among Cayley’s many innovations inmathematics were the notions of an abstractgroup and a group algebra, and the matrixconcept. He made major contributions togeometry and linear algebra. Cayley and hislifelong friend and collaborator J. J. Sylvesterwere the founders of the theory of invariants,which was later to play an important role inthe theory of relativity.

Cayley’s collected works comprise 13volumes, each about 600 pages in length.He died on January 26, 1895.

To find more information about Cayley,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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138

Cosets and Lagrange’sTheorem7

Properties of CosetsIn this chapter, we will prove the single most important theorem in finitegroup theory—Lagrange’s Theorem. But first, we introduce a new andpowerful tool for analyzing a group—the notion of a coset. This notionwas invented by Galois in 1830, although the term was coined by G. A. Miller in 1910.

Definition Coset of H in G

Let G be a group and let H be a subset of G. For any a [ G, the set{ah | h [ H} is denoted by aH. Analogously, Ha 5 {ha | h [ H} andaHa21 5 {aha21 | h [ H}. When H is a subgroup of G, the set aH is calledthe left coset of H in G containing a, whereas Ha is called the right cosetof H in G containing a. In this case, the element a is called the cosetrepresentative of aH (or Ha). We use |aH| to denote the number of ele-ments in the set aH, and |Ha| to denote the number of elements in Ha.

EXAMPLE 1 Let G 5 S3 and H 5 {(1), (13)}. Then the left cosets ofH in G are

(1)H 5 H,(12)H 5 {(12), (12)(13)} 5 {(12), (132)} 5 (132)H,(13)H 5 {(13), (1)} 5 H,(23)H 5 {(23), (23)(13)} 5 {(23), (123)} 5 (123)H.

It might be difficult, at this point, for students to see the extremeimportance of this result [Lagrange’s Theorem]. As we penetrate the subjectmore deeply they will become more and more aware of its basic character.

I. N. HERSTEIN, Topics in Algebra

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7 | Cosets and Lagrange’s Theorem 139

EXAMPLE 2 Let _ 5 {R0, R180} in D4, the dihedral group of order 8.Then,

R0_ 5 _,R90_ 5 {R90, R270} 5 R270_,

R180_ 5 {R180, R0} 5 _,V_ 5 {V, H} 5 H_,D_ 5 {D, D9} 5 D9_.

EXAMPLE 3 Let H 5 {0, 3, 6} in Z9 under addition. In the case thatthe group operation is addition, we use the notation a 1 H instead ofaH. Then the cosets of H in Z9 are

0 1 H 5 {0, 3, 6} 5 3 1 H 5 6 1 H,1 1 H 5 {1, 4, 7} 5 4 1 H 5 7 1 H,2 1 H 5 {2, 5, 8} 5 5 1 H 5 8 1 H.

The three preceding examples illustrate a few facts about cosets thatare worthy of our attention. First, cosets are usually not subgroups.Second, aH may be the same as bH, even though a is not the same as b.Third, since in Example 1 (12)H 5 {(12), (132)} whereas H(12) 5{(12), (123)}, aH need not be the same as Ha.

These examples and observations raise many questions. When doesaH 5 bH? Do aH and bH have any elements in common? When does aH 5 Ha? Which cosets are subgroups? Why are cosets important? Thenext lemma and theorem answer these questions. (Analogous resultshold for right cosets.)

Lemma Properties of Cosets

PROOF

1. a 5 ae [ aH.2. To verify property 2, we first suppose that aH 5 H. Then a 5

ae [ aH 5 H. Next, we assume that a [ H and show that aH # H

Let H be a subgroup of G, and let a and b belong to G. Then,

1. a [ aH,2. aH 5 H if and only if a [ H,3. aH 5 bH if and only if a [ bH4. aH 5 bH or aH > bH 5 [,5. aH 5 bH if and only if a21b [ H,6. |aH| 5 |bH|,7. aH 5 Ha if and only if H 5 aHa21,8. aH is a subgroup of G if and only if a [ H.

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140 Groups

and H # aH. The first inclusion follows directly from the closure ofH. To show that H # aH, let h [ H. Then, since a [ H and h [ H, weknow that a21h [ H. Thus, h 5 eh 5 (aa21)h 5 a(a21h) [ aH.

3. If aH 5 bH, then a 5 ae [ aH 5 bH. Conversely, if a [ bH we have a 5 bh where h [ H, and therefore aH 5 (bh)H 5 b(hH) 5 bH.

4. Property 4 follows directly from property 3, for if there is an ele-ment c in aH y bH, then cH 5 aH and cH 5 bH.

5. Observe that aH 5 bH if and only if H 5 a21bH. The result nowfollows from property 2.

6. To prove that |aH| 5 |bH|, it suffices to define a one-to-one map-ping from aH onto bH. Obviously, the correspondence ah → bhmaps aH onto bH. That it is one-to-one follows directly from thecancellation property.

7. Note that aH 5 Ha if and only if (aH)a21 5 (Ha)a21 5 H(aa–1) 5H—that is, if and only if aHa21 5 H.

8. If aH is a subgroup, then it contains the identity e. Thus, aH >eH 2 0/; and, by property 4, we have aH 5 eH 5 H. Thus, fromproperty 2, we have a [ H. Conversely, if a [ H, then, again byproperty 2, aH 5 H.

Although most mathematical theorems are written in symbolic form,one should also know what they say in words. In the preceding lemma,property 1 says simply that the left coset of H containing a does contain a.Property 2 says that the H “absorbs” an element if and only if the ele-ment belongs to H. Property 3 shows that a left coset of H is uniquelydetermined by any one of its elements. In particular, any element of aleft coset can be used to represent the coset. Property 4 says—and this isvery important—that two left cosets of H are either identical or disjoint.Property 5 shows how we may transfer a question about equality of leftcosets of H to a question about H itself and vice versa. Property 6 saysthat all left cosets of H have the same size. Property 7 is analogous toproperty 5 in that it shows how a question about the equality of the leftand right cosets of H containing a is equivalent to a question about theequality of two subgroups of G. The last property of the lemma says thatH itself is the only coset of H that is a subgroup of G.

Note that properties 1, 4, and 6 of the lemma guarantee that the leftcosets of a subgroup H of G partition G into blocks of equal size.Indeed, we may view the cosets of H as a partitioning of G into equiva-lence classes under the equivalence relation defined by a , b if aH 5 bH (see Theorem 0.6).

In practice, the subgroup H is often chosen so that the cosets parti-tion the group in some highly desirable fashion. For example, if G is

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7 | Cosets and Lagrange’s Theorem 141

3-space R3 and H is a plane through the origin, then the coset (a, b, c) 1H (addition is done componentwise) is the plane passing through thepoint (a, b, c) and parallel to H. Thus, the cosets of H constitute a par-tition of 3-space into planes parallel to H. If G 5 GL(2, R) and H 5 SL(2, R), then for any matrix A in G, the coset AH is the set of all2 3 2 matrices with the same determinant as A. Thus,

H is the set of all 2 3 2 matrices of determinant 2

and

H is the set of all 2 3 2 matrices of determinant 23.

Property 4 of the lemma is useful for actually finding the distinctcosets of a subgroup. We illustrate this in the next example.

EXAMPLE 4 To find the cosets of H 5 {1, 15} in G 5 U(32) 5 {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31} we begin with H 5 {1, 15}. We can find a second coset by choosing any element notin H, say 3, as a coset representative. This gives the coset 3H 5 {3, 13}.We find our next coset by choosing a representative not already appear-ing in the two previously chosen cosets, say 5. This gives us the coset 5H 5{5, 11}. We continue to form cosets by picking elements from U(32)that have not yet appeared in the previous cosets as representatives ofthe cosets until we have accounted for every element of U(32). We thenhave the complete list of all distinct cosets of H.

Lagrange’s Theorem and ConsequencesWe are now ready to prove a theorem that has been around for morethan 200 years—longer than group theory itself! (This theorem was notoriginally stated in group theoretic terms.) At this stage, it should comeas no surprise.

Theorem 7.1 Lagrange’s Theorem†: |H| Divides |G|

If G is a finite group and H is a subgroup of G, then |H| divides |G|.Moreover, the number of distinct left (right) cosets of H in G is |G| / |H|.

c1 2

2 1d

c2 0

0 1d

†Lagrange stated his version of this theorem in 1770, but the first complete proof wasgiven by Pietro Abbati some 30 years later.

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142 Groups

PROOF Let a1H, a2H, . . . , arH denote the distinct left cosets of H inG. Then, for each a in G, we have aH 5 aiH for some i. Also, by prop-erty 1 of the lemma, a [ aH. Thus, each member of G belongs to oneof the cosets aiH. In symbols,

G 5 a1H < ? ? ? < ar H.

Now, property 4 of the lemma shows that this union is disjoint, so that

|G| 5 |a1H| 1 |a2H| 1 ? ? ? 1 |ar H|.

Finally, since |aiH| 5 |H| for each i, we have |G| 5 r|H|.

We pause to emphasize that Lagrange’s Theorem is a subgroup can-didate criterion; that is, it provides a list of candidates for the orders ofthe subgroups of a group. Thus, a group of order 12 may have sub-groups of order 12, 6, 4, 3, 2, 1, but no others. Warning! The converseof Lagrange’s Theorem is false. For example, a group of order 12 neednot have a subgroup of order 6. We prove this in Example 5.

A special name and notation have been adopted for the number ofleft (or right) cosets of a subgroup in a group. The index of a subgroupH in G is the number of distinct left cosets of H in G. This number is denoted by |G:H|. As an immediate consequence of the proof ofLagrange’s Theorem, we have the following useful formula for thenumber of distinct left (or right) cosets of H in G.

Corollary 1 |G:H| 5 |G|/|H|

Corollary 2 |a| Divides |G|

PROOF Recall that the order of an element is the order of the subgroupgenerated by that element.

Corollary 3 Groups of Prime Order Are Cyclic

A group of prime order is cyclic.

In a finite group, the order of each element of the group divides theorder of the group.

If G is a finite group and H is a subgroup of G, then |G:H| 5 |G|/|H|.

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7 | Cosets and Lagrange’s Theorem 143

PROOF Suppose that G has prime order. Let a [ G and a 2 e. Then,|�a�| divides |G| and |�a�| 2 1. Thus, |�a�| 5 |G| and the corollaryfollows.

Corollary 4 a|G| 5 e

PROOF By Corollary 2, |G| 5 |a|k for some positive integer k. Thus,a|G| 5 a|a|k 5 ek 5 e.

Corollary 5 Fermat’s Little Theorem

PROOF By the division algorithm, a 5 pm 1 r, where 0 # r , p.Thus, a mod p 5 r, and it suffices to prove that rp mod p 5 r. If r 5 0,the result is trivial, so we may assume that r [ U(p). [Recall that U(p) 5 {1, 2, . . . , p 2 1} under multiplication modulo p.] Then, by thepreceding corollary, rp21 mod p 5 1 and, therefore, rp mod p 5 r.

Fermat’s Little Theorem has been used in conjunction with comput-ers to test for primality of certain numbers. One case concerned thenumber p 5 2257 2 1. If p is prime, then we know from Fermat’s LittleTheorem that 10 p mod p 5 10 mod p and, therefore, 10 p11 mod p 5100 mod p. Using multiple precision and a simple loop, a computerwas able to calculate 10 p11 mod p 5 102257 mod p in a few seconds.The result was not 100, and so p is not prime.

EXAMPLE 5 The Converse of Lagrange’s Theorem Is False†

The group A4 of order 12 has no subgroups of order 6. To verify this,recall that A4 has eight elements of order 3 (a5 through a12 in the nota-tion of Table 5.1) and suppose that H is a subgroup of order 6. Let a beany element of order 3 in A4. Since H has index 2 in A4, at most two ofthe cosets H, aH, and a2H are distinct. But equality of any pair of thesethree implies that aH 5 H, so that a [ H. (For example, if H 5 a2H,multiply on the left by a.) Thus, a subgroup of A4 of order 6 would haveto contain all eight elements of order 3, which is absurd.

For every integer a and every prime p, ap mod p 5 a mod p.

Let G be a finite group, and let a [ G. Then, a|G| 5 e.

†The first counterexample to the converse of Lagrange’s Theorem was given by PaoloRuffini in 1799.

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144 Groups

For any prime p . 2, we know that Z2p and Dp are nonisomorphicgroups of order 2p. This naturally raises the question of whether therecould be other possible groups of these orders. Remarkably, with justthe simple machinery available to us at this point, we can answer thisquestion.

Theorem 7.2 Classification of Groups of Order 2p

PROOF We assume that G does not have an element of order 2p andshow that G < Dp. We begin by first showing that G must have an element of order p. By our assumption and Lagrange’s Theorem, anynonidentity element of G must have order 2 or p. Thus, to verify our as-sertion, we may assume that every nonidentity element of G has order 2.In this case, we have for all a and b in the group ab 5 (ab)21 5 b21a21 5ba, so that G is Abelian. Then, for any nonidentity elements a, b [ Gwith a 2 b, the set {e, a, b, ab} is closed and therefore is a subgroup ofG of order 4. Since this contradicts Lagrange’s Theorem, we haveproved that G must have an element of order p; call it a.

Now let b be any element not in kal. Then bkal 2 kal and G 5 kal < bkal. We next claim that |b| 5 2. To see this, observe that since kal and bkal are the only two distinct cosets of kal in G, we must have b2kal 5 kal or b2kal 5 bkal. We may rule out b2kal 5 bkal, for then bkal 5kal. On the other hand, b2kal 5 kal implies that b2 [ kal and, therefore,|b2| 5 1 or |b2| 5 p. But |b2| 5 p and |b| 2 2p imply that |b| 5 p. Thenkbl 5 kb2l and therefore b [ kal, which is a contradiction. Thus, anyelement of G not in kal has order 2.

Next consider ab. Since ab o kal, our argument above shows that|ab| 5 2. Then ab 5 (ab)21 5 b21a21 5 ba21. Moreover, this relationcompletely determines the multiplication table for G. [For example,a3(ba4) 5 a2(ab)a4 5 a2(ba21)a4 5 a(ab)a3 5 a(ba21)a3 5 (ab)a2 5(ba21)a2 5 ba.] Since the multiplication table for all noncyclic groupsof order 2p is uniquely determined by the relation ab 5 ba21, allnoncyclic groups of order 2p must be isomorphic to each other. But ofcourse, Dp, the dihedral group of order 2p, is one such group.

As an immediate corollary, we have that S3, the symmetric group ofdegree 3, is isomorphic to D3.

Let G be a group of order 2p, where p is a prime greater than 2. ThenG is isomorphic to Z2p or Dp.

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7 | Cosets and Lagrange’s Theorem 145

An Application of Cosets to Permutation Groups

Lagrange’s Theorem and its corollaries dramatically demonstrate thefruitfulness of the coset concept. We next consider an application ofcosets to permutation groups.

Definition Stabilizer of a Point

Let G be a group of permutations of a set S. For each i in S, let stabG(i) 5{f [ G | f(i) 5 i}. We call stabG(i) the stabilizer of i in G.

The student should verify that stabG(i) is a subgroup of G. (SeeExercise 31 in Chapter 5.)

Definition Orbit of a Point

Let G be a group of permutations of a set S. For each s in S, let orbG(s) 5{f(s) | f [ G}. The set orbG(s) is a subset of S called the orbit of sunder G. We use |orbG(s)| to denote the number of elements in orbG(s).

Example 6 should clarify these two definitions.

EXAMPLE 6 Let

G 5 {(1), (132)(465)(78), (132)(465), (123)(456),

(123)(456)(78), (78)}.

Then,

orbG(1) 5 {1, 3, 2}, stabG(1) 5 {(1), (78)},orbG(2) 5 {2, 1, 3}, stabG(2) 5 {(1), (78)},orbG(4) 5 {4, 6, 5}, stabG(4) 5 {(1), (78)},orbG(7) 5 {7, 8}, stabG(7) 5 {(1), (132)(465), (123)(456)}.

EXAMPLE 7 We may view D4 as a group of permutations of asquare region. Figure 7.1(a) illustrates the orbit of the point p under D4,and Figure 7.1(b) illustrates the orbit of the point q under D4. Observethat stabD

4( p) 5 {R0, D}, whereas stabD

4(q) 5 {R0}.

Figure 7.1

The preceding two examples also illustrate the following theorem.

q

(b)

p

(a)

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146 Groups

Theorem 7.3 Orbit-Stabilizer Theorem

PROOF By Lagrange’s Theorem, |G|/|stabG(i)| is the number of dis-tinct left cosets of stabG(i) in G. Thus, it suffices to establish a one-to-one correspondence between the left cosets of stabG(i) and the elements in the orbit of i. To do this, we define a correspondence Tby mapping the coset fstabG(i) to f(i) under T. To show that T is a well-defined function, we must show that astabG(i) 5 bstabG(i) implies a(i) 5b(i). But astabG(i) 5 bstabG(i) implies a21b [ stabG(i), so that (a21b) (i) 5 i and, therefore, b(i) 5 a(i). Reversing the argument fromthe last step to the first step shows that T is also one-to-one. We concludethe proof by showing that T is onto orbG(i). Let j [ orbG(i). Then a(i) 5 jfor some a [ G and clearly T(astabG(i)) 5 a(i) 5 j, so that T is onto.

We leave as an exercise the proof of the important fact that the orbitsof the elements of a set S under a group partition S (Exercise 33).

The Rotation Group of a Cubeand a Soccer Ball

It cannot be overemphasized that Theorem 7.3 and Lagrange’s Theorem(Theorem 7.1) are counting theorems.† They enable us to determine thenumbers of elements in various sets. To see how Theorem 7.3 works, wewill determine the order of the rotation group of a cube and a soccer ball.That is, we wish to find the number of essentially different ways inwhich we can take a cube or a soccer ball in a certain location in space,physically rotate it, and then still occupy its original location.

EXAMPLE 8 Let G be the rotation group of a cube. Label the sixfaces of the cube 1 through 6. Since any rotation of the cube must carryeach face of the cube to exactly one other face of the cube and differentrotations induce different permutations of the faces, G can be viewed asa group of permutations on the set {1, 2, 3, 4, 5, 6}. Clearly, there issome rotation about a central horizontal or vertical axis that carries facenumber 1 to any other face, so that |orbG(1)| 5 6. Next, we considerstabG(1). Here, we are asking for all rotations of a cube that leave facenumber 1 where it is. Surely, there are only four such motions—rotations of 0°, 90°, 180°, and 270°—about the line perpendicular to

Let G be a finite group of permutations of a set S. Then, for any i from S, |G| 5 |orbG (i)| |stabG(i)|.

†People who don’t count won’t count (Anatole France).

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7 | Cosets and Lagrange’s Theorem 147

the face and passing through its center (see Figure 7.2). Thus, byTheorem 7.3, |G| 5 |orbG(1)| |stabG(1)| 5 6 ? 4 5 24.

Figure 7.2 Axis of rotation of a cube.

Now that we know how many rotations a cube has, it is simple to de-termine the actual structure of the rotation group of a cube. Recall thatS4 is the symmetric group of degree 4.

Theorem 7.4 The Rotation Group of a Cube

PROOF Since the group of rotations of a cube has the same order asS4, we need only prove that the group of rotations is isomorphic to asubgroup of S4. To this end, observe that a cube has four diagonals andthat the rotation group induces a group of permutations on the four di-agonals. But we must be careful not to assume that different rotationscorrespond to different permutations. To see that this is so, all we needdo is show that all 24 permutations of the diagonals arise from rota-tions. Labeling the consecutive diagonals 1, 2, 3, and 4, it is obviousthat there is a 90° rotation that yields the permutation a 5 (1234); an-other 90° rotation about an axis perpendicular to our first axis yieldsthe permutation b 5 (1423). See Figure 7.3. So, the group of permuta-tions induced by the rotations contains the eight-element subgroup {e, a, a2, a3, b2, b2a, b2a2, b2a3} (see Exercise 37) and ab, which hasorder 3. Clearly, then, the rotations yield all 24 permutations since theorder of the rotation group must be divisible by both 8 and 3.

EXAMPLE 9 A traditional soccer ball has 20 faces that are regularhexagons and 12 faces that are regular pentagons. (The technical termfor this solid is truncated icosahedron.) To determine the number of ro-tational symmetries of a soccer ball using Theorem 7.3, we may choose

The group of rotations of a cube is isomorphic to S4.

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148 Groups

Figure 7.3

our set S to be the 20 hexagons or the 12 pentagons. Let us say that S isthe set of 12 pentagons. Since any pentagon can be carried to any otherpentagon by some rotation, the orbit of any pentagon is S. Also, thereare five rotations that fix (stabilize) any particular pentagon. Thus, bythe Orbit-Stabilizer Theorem, there are 12 ? 5 5 60 rotational symme-tries. (In case you are interested, the rotation group of a soccer ball isisomorphic to A5.)

In 1985, chemists Robert Curl, Richard Smalley, and Harold Krotocaused tremendous excitement in the scientific community when theycreated a new form of carbon by using a laser beam to vaporize graphite.The structure of the new molecule is composed of 60 carbon atomsarranged in the shape of a soccer ball! Because the shape of the new mol-ecule reminded them of the dome structures built by the architect R. Buckminster Fuller, Curl, Smalley, and Kroto named their discovery“buckyballs.” Buckyballs are the roundest, most symmetrical large mol-ecules known. Group theory has been particularly useful in illuminatingthe properties of buckyballs, since the absorption spectrum of a moleculedepends on its symmetries and chemists classify various molecular states

2

2

3

1

3

1

4

4

= (1423)β

2

2

3

1

3

1

4

4

= (1234)α

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7 | Cosets and Lagrange’s Theorem 149

according to their symmetry properties. The buckyball discovery spurreda revolution in carbon chemistry. In 1996, Curl, Smalley, and Krotoreceived the Nobel Prize in chemistry for their discovery.

Exercises

I don’t know, Marge. Trying is the first step towards failure.HOMER SIMPSON

1. Let H 5 {(1), (12)(34), (13)(24), (14)(23)}. Find the left cosets ofH in A4 (see Table 5.1 on page 107).

2. Let H be as in Exercise 1. How many left cosets of H in S4 arethere? (Determine this without listing them.)

3. Let H 5 {0, 63, 66, 69, . . .}. Find all the left cosets of H in Z.4. Rewrite the condition a21b [ H given in property 5 of the lemma

on page 139 in additive notation. Assume that the group is Abelian.5. Let H be as in Exercise 3. Use Exercise 4 to decide whether or not

the following cosets of H are the same.a. 11 1 H and 17 1 Hb. 21 1 H and 5 1 Hc. 7 1 H and 23 1 H

6. Let n be a positive integer. Let H 5 {0, 6n, 62n, 63n, . . .}. Findall left cosets of H in Z. How many are there?

7. Find all of the left cosets of {1, 11} in U(30).8. Suppose that a has order 15. Find all of the left cosets of �a5� in �a�.9. Let |a| 5 30. How many left cosets of �a4� in �a� are there? List them.

10. Let a and b be nonidentity elements of different orders in a groupG of order 155. Prove that the only subgroup of G that contains a and b is G itself.

11. Let H be a subgroup of R*, the group of nonzero real numbers un-der multiplication. If R+ # H # R*, prove that H 5 R+ or H 5 R*.

12. Let C* be the group of nonzero complex numbers under multiplica-tion and let H 5 {a + bi [ C*| a2 + b2 5 1}. Give a geometric de-scription of the coset (3 + 4i)H. Give a geometric description of thecoset (c + di)H.

13. Let G be a group of order 60. What are the possible orders for thesubgroups of G?

14. Suppose that K is a proper subgroup of H and H is a proper sub-group of G. If |K| 5 42 and |G| 5 420, what are the possibleorders of H?

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150 Groups

15. Let G be a group with |G| 5 pq, where p and q are prime. Provethat every proper subgroup of G is cyclic.

16. Recall that, for any integer n greater than 1, f(n) denotes the num-ber of positive integers less than n and relatively prime to n. Provethat if a is any integer relatively prime to n, then af(n) mod n 5 1.

17. Compute 515 mod 7 and 713 mod 11.18. Use Corollary 2 of Lagrange’s Theorem (Theorem 7.1) to prove

that the order of U(n) is even when n . 2.19. Suppose G is a finite group of order n and m is relatively prime to n.

If g [ G and gm 5 e, prove that g 5 e.20. Suppose H and K are subgroups of a group G. If |H| 5 12 and

|K| 5 35, find |H > K|. Generalize.21. Suppose that H is a subgroup of S4 and that H contains (12) and

(234.) Prove that H 5 S4.22. Suppose that H and K are subgroups of G and there are elements

a and b in G such that aH 8 bK. Prove that H 8 K.23. Suppose that G is an Abelian group with an odd number of elements.

Show that the product of all of the elements of G is the identity.24. Suppose that G is a group with more than one element and G has

no proper, nontrivial subgroups. Prove that |G| is prime. (Do notassume at the outset that G is finite.)

25. Let |G| 5 15. If G has only one subgroup of order 3 and only oneof order 5, prove that G is cyclic. Generalize to |G| 5 pq, where pand q are prime.

26. Let G be a group of order 25. Prove that G is cyclic or g5 5 e for all g in G.

27. Let |G| 5 33. What are the possible orders for the elements of G?Show that G must have an element of order 3.

28. Let |G| 5 8. Show that G must have an element of order 2.29. Can a group of order 55 have exactly 20 elements of order 11?

Give a reason for your answer.30. Determine all finite subgroups of C*, the group of nonzero com-

plex numbers under multiplication.31. Let H and K be subgroups of a finite group G with H # K # G.

Prove that |G:H| 5 |G:K| |K:H|.32. Show that Q, the group of rational numbers under addition, has no

proper subgroup of finite index.

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7 | Cosets and Lagrange’s Theorem 151

33. Let G be a group of permutations of a set S. Prove that the orbits ofthe members of S constitute a partition of S. (This exercise is re-ferred to in this chapter and in Chapter 29.)

34. Prove that every subgroup of Dn of odd order is cyclic.35. Let G 5 {(1), (12)(34), (1234)(56), (13)(24), (1432)(56), (56)(13),

(14)(23), (24)(56)}.a. Find the stabilizer of 1 and the orbit of 1.b. Find the stabilizer of 3 and the orbit of 3.c. Find the stabilizer of 5 and the orbit of 5.

36. Let G be a group of order pn where p is prime. Prove that the centerof G cannot have order pn21.

37. Prove that the eight-element set in the proof of Theorem 7.4 is agroup.

38. Prove that a group of order 12 must have an element of order 2.39. Suppose that a group contains elements of orders 1 through 10.

What is the minimum possible order of the group?40. Let G be a finite Abelian group and let n be a positive integer that

is relatively prime to |G|. Show that the mapping a → an is an au-tomorphism of G.

41. Show that in a group G of odd order, the equation x2 5 a has aunique solution for all a in G.

42. Let G be a group of order pqr, where p, q, and r are distinct primes.If H and K are subgroups of G with |H| 5 pq and |K| 5 qr, provethat |H > K| 5 q.

43. Let G 5 GL(2, R) and H 5 SL(2, R). Let A [ G and suppose thatdet A 5 2. Prove that AH is the set of all 2 3 2 matrices in G thathave determinant 2.

44. Let G be the group of rotations of a plane about a point P in theplane. Thinking of G as a group of permutations of the plane, de-scribe the orbit of a point Q in the plane. (This is the motivation forthe name “orbit.”)

45. Let G be the rotation group of a cube. Label the faces of the cube 1 through 6, and let H be the subgroup of elements of G that carryface 1 to itself. If s is a rotation that carries face 2 to face 1, give aphysical description of the coset Hs.

46. The group D4 acts as a group of permutations of the square regionsshown on the following page. (The axes of symmetry are drawn forreference purposes.) For each square region, locate the points in

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152 Groups

the orbit of the indicated point under D4. In each case, determinethe stabilizer of the indicated point.

47. Let G 5 GL(2, R), the group of 2 3 2 matrices over R with nonzerodeterminant. Let H be the subgroup of matrices of determinant 61.If a, b [ G and aH 5 bH, what can be said about det (a) anddet (b)? Prove or disprove the converse.

48. Calculate the orders of the following (refer to Figure 27.5 for illus-trations):a. The group of rotations of a regular tetrahedron (a solid with

four congruent equilateral triangles as faces)b. The group of rotations of a regular octahedron (a solid with

eight congruent equilateral triangles as faces)c. The group of rotations of a regular dodecahedron (a solid with

12 congruent regular pentagons as faces)d. The group of rotations of a regular icosahedron (a solid with 20

congruent equilateral triangles as faces)49. If G is a finite group with fewer than 100 elements and G has sub-

groups of orders 10 and 25, what is the order of G?50. A soccer ball has 20 faces that are regular hexagons and 12 faces

that are regular pentagons. Use Theorem 7.3 to explain why a soc-cer ball cannot have a 60° rotational symmetry about a line throughthe centers of two opposite hexagonal faces.

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7 | Cosets and Lagrange’s Theorem 153

Computer Exercise

In the fields of observation chance favors only the prepared mind.LOUIS PASTEUR

Software for the computer exercise in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software determines when Zn is the only group of order n inthe case that n 5 pq where p and q are distinct primes. Run thesoftware for n 5 3 ? 5, 3 ? 7, 3 ? 11, 3 ? 13, 3 ? 17, 3 ? 31, 5 ? 7, 5 ? 11,5 ? 13, 5 ? 17, 5 ? 31, 7 ? 11, 7 ? 13, 7 ? 17, 7 ? 19, and 7 ? 43. Conjec-ture a necessary and sufficient condition about p and q for Zpq tobe the only group of order pq, where p and q are distinct primes.

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154

Joseph Lagrange

JOSEPH LOUIS LAGRANGE was born in Italy ofFrench ancestry on January 25, 1736. He be-came captivated by mathematics at an earlyage when he read an essay by Halley onNewton’s calculus. At the age of 19, he be-came a professor of mathematics at the RoyalArtillery School in Turin. Lagrange made sig-nificant contributions to many branches ofmathematics and physics, among them thetheory of numbers, the theory of equations,ordinary and partial differential equations, thecalculus of variations, analytic geometry,fluid dynamics, and celestial mechanics. Hismethods for solving third- and fourth-degreepolynomial equations by radicals laid thegroundwork for the group-theoretic approachto solving polynomials taken by Galois.Lagrange was a very careful writer with aclear and elegant style.

At the age of 40, Lagrange was appointedHead of the Berlin Academy, succeedingEuler. In offering this appointment, Frederickthe Great proclaimed that the “greatest kingin Europe” ought to have the “greatest mathe-matician in Europe” at his court. In 1787,Lagrange was invited to Paris by Louis XVIand became a good friend of the king and hiswife, Marie Antoinette. In 1793, Lagrangeheaded a commission, which includedLaplace and Lavoisier, to devise a new system

Lagrange is the Lofty Pyramid of theMathematical Sciences.

NAPOLEON BONAPARTE

This stamp was issued byFrance in Lagrange’s honorin 1958.

of weights and measures. Out of this camethe metric system. Late in his life he wasmade a count by Napoleon. Lagrange died onApril 10, 1813.

To find more information about Lagrange,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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External Direct Products

The universe is an enormous direct product of representations of symmetry groups.

STEVEN WEINBERG†

Definition and ExamplesIn this chapter, we show how to piece together groups to make largergroups. In Chapter 9, we will show that we can often start with onelarge group and decompose it into a product of smaller groups in muchthe same way as a composite positive integer can be broken down intoa product of primes. These methods will later be used to give us a sim-ple way to construct all finite Abelian groups.

Definition External Direct Product

Let G1, G2, . . . , Gn be a finite collection of groups. The external directproduct of G1, G2, . . . , Gn, written as G1 % G2 % ? ? ? % Gn, is the set ofall n-tuples for which the ith component is an element of Gi and theoperation is componentwise.

In symbols,

G1 % G2 % ? ? ? % Gn 5 {(g1, g2, . . . , gn) | gi [ Gi},

where (g1, g2, . . . , gn)(g19, g29, . . . , gn9) is defined to be (g1g19,g2g29, . . . , gngn9). It is understood that each product gigi9 is performedwith the operation of Gi. We leave it to the reader to show that theexternal direct product of groups is itself a group (Exercise 1).

This construction is not new to students who have had linear algebra orphysics. Indeed, R2 5 R % R and R3 5 R % R % R—the operation beingcomponentwise addition. Of course, there is also scalar multiplication, but

†Weinberg received the 1979 Nobel Prize in physics with Sheldon Glashow and AbdusSalam for their construction of a single theory incorporating weak and electromagneticinteractions.

8

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156 Groups

we ignore this for the time being, since we are interested only in the groupstructure at this point.

EXAMPLE 1

U(8) % U(10) 5 {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3),(3, 7), (3, 9), (5, 1), (5, 3), (5, 7), (5, 9),(7, 1),(7, 3), (7, 7), (7, 9)}.

The product (3, 7)(7, 9) 5 (5, 3), since the first components are com-bined by multiplication modulo 8, whereas the second components arecombined by multiplication modulo 10.

EXAMPLE 2

Z2 % Z3 5 {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}.

Clearly, this is an Abelian group of order 6. Is this group related to an-other Abelian group of order 6 that we know, namely, Z6? Consider thesubgroup of Z2 % Z3 generated by (1, 1). Since the operation in each com-ponent is addition, we have (1, 1) 5 (1, 1), 2(1, 1) 5 (0, 2), 3(1, 1) 5(1, 0), 4(1, 1) 5 (0, 1), 5(1, 1) 5 (1, 2), and 6(1, 1) 5 (0, 0). Hence Z2 % Z3 is cyclic. It follows that Z2 % Z3 is isomorphic to Z6.

In Theorem 7.2 we classified the groups of order 2p where p is anodd prime. Now that we have defined Z2 % Z2, it is easy to classify thegroups of order 4.

EXAMPLE 3 Classification of Groups of Order 4A group of order 4 is isomorphic to Z4 or Z2 % Z2. To verify this, let G 5{e, a, b, ab}. If G is not cyclic, then it follows from Lagrange’s Theoremthat |a | 5 |b | 5 |ab | 5 2. Then the mapping e S (0, 0), a S (1, 0),b S (0, 1), and ab S (1, 1) is an isomorphism from G onto Z2 % Z2.

We see from Examples 2 and 3 that in some cases is isomor-phic to and in some cases it is not. Theorem 8.2 provides a simplecharacterization for when the isomorphism holds.

Properties of External Direct ProductsOur first theorem gives a simple method for computing the order of anelement in a direct product in terms of the orders of the componentpieces.

Zmn

Zm % Zn

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8 | External Direct Products 157

Theorem 8.1 Order of an Element in a Direct Product

PROOF Denote the identity of Gi by ei. Let s 5 lcm(|g1|, |g2|, . . . , |gn|)and t 5|(g1, g2, . . . , gn)|. Because s is a multiple of each |gi| implies that (g1, g2, . . . , gn)

s 5 (gs1, gs

2, . . . , gsn) 5 (e1, e2, . . . , en), we know that t # s. On

the other hand, from (gt1, gt

2, . . . , gtn) 5 (g1, g2, . . . , gn)

t 5(e1, e2, . . . , en) wesee that t is a common multiple of |g1|, |g2|, . . . , |gn|. Thus, s # t.

The next two examples are applications of Theorem 8.1.

EXAMPLE 4 We determine the number of elements of order 5 in Z25 % Z5. By Theorem 8.1, we may count the number of elements (a, b) in Z25 % Z5 with the property that 5 5 |(a, b)| 5 lcm(|a|, |b|).Clearly this requires that either |a| 5 5 and |b| 5 1 or 5, or |b| 5 5 and|a| 5 1 or 5. We consider two mutually exclusive cases.

Case 1 |a| 5 5 and |b| 5 1 or 5. Here there are four choices for a(namely, 5, 10, 15, and 20) and five choices for b. This gives 20 ele-ments of order 5.

Case 2 |a| 5 1 and |b| 5 5. This time there is one choice for a and fourchoices for b, so we obtain four more elements of order 5.

Thus, Z25 % Z5 has 24 elements of order 5.

EXAMPLE 5 We determine the number of cyclic subgroups of order10 in Z100 % Z25. We begin by counting the number of elements (a, b) oforder 10.

Case 1 |a| 5 10 and |b| 5 1 or 5. Since Z100 has a unique cyclic sub-group of order 10 and any cyclic group of order 10 has four generators(Theorem 4.4), there are four choices for a. Similarly, there are fivechoices for b. This gives 20 possibilities for (a, b).

Case 2 |a| 5 2 and |b| 5 5. Since any finite cyclic group of even orderhas a unique subgroup of order 2 (Theorem 4.4), there is only onechoice for a. Obviously, there are four choices for b. So, this caseyields four more possibilities for (a, b).

The order of an element in a direct product of a finite number offinite groups is the least common multiple of the orders of thecomponents of the element. In symbols,

|(g1, g2, . . . , gn)| 5 lcm(|g1|, |g2|, . . . , |gn|).

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158 Groups

Thus, Z100 % Z25 has 24 elements of order 10. Because each cyclicsubgroup of order 10 has four elements of order 10 and no two of thecyclic subgroups can have an element of order 10 in common, theremust be 24/4 5 6 cyclic subgroups of order 10. (This method is analo-gous to determining the number of sheep in a flock by counting legsand dividing by 4.)

The direct product notation is convenient for specifying certain sub-groups of a direct product.

EXAMPLE 6 For each divisor r of m and s of n the group has a subgroup isomorphic to (see Exercise 17). To find a sub-group of say isomorphic to we observe that is asubgroup of of order 6 and is a subgroup of of order 4, so

is the desired subgroup.

The next theorem and its first corollary characterize those directproducts of cyclic groups that are themselves cyclic.

Theorem 8.2 Criterion for G % H to be Cyclic

PROOF Let |G| 5 m and |H| 5 n, so that |G % H| 5 mn. To prove thefirst half of the theorem, we assume G % H is cyclic and show that m and n are relatively prime. Suppose that gcd(m, n) 5 d and (g, h) is agenerator of G % H. Since (g, h)mn/d 5 ((gm)n/d, (hn)m/d) 5 (e, e), wehave mn 5 |(g, h)| # mn/d. Thus, d 5 1.

To prove the other half of the theorem, let G 5 �g� and H 5 �h� andsuppose gcd(m, n) 5 1. Then, |(g, h)| 5 lcm(m, n) 5 mn 5 |G % H|,so that (g, h) is a generator of G % H.

As a consequence of Theorem 8.2 and an induction argument, weobtain the following extension of Theorem 8.2.

Corollary 1 Criterion for G1 % G2 % ? ? ? % Gn to Be Cyclic

An external direct product G1 % G2 % ? ? ? % Gn of a finite number of finite cyclic groups is cyclic if and only if |Gi| and |Gj| are relativelyprime when i 2 j.

Let G and H be finite cyclic groups. Then G % H is cyclic if and onlyif |G| and |H| are relatively prime.

�5� % �3�Z12�3�Z30

�5�Z6 % Z4Z30 % Z12

Zr % Zs

Zm % Zn

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8 | External Direct Products 159

Corollary 2 Criterion for Zn1n2 . . . nk< Zn1

% Zn2% . . . % Znk

By using the results above in an iterative fashion, one can expressthe same group (up to isomorphism) in many different forms. For ex-ample, we have

Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5 < Z2 % Z30.

Similarly,

Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5

< Z2 % Z3 % Z2 % Z5 < Z6 % Z10.

Thus, Z2 % Z30 < Z6 % Z10. Note, however, that Z2 % Z30 ] Z60.

The Group of Units Modulo n As An External Direct Product

The U-groups provide a convenient way to illustrate the precedingideas. We first introduce some notation. If k is a divisor of n, let

Uk(n) 5 {x [ U(n)| x mod k 5 1}.

For example, U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92}. It can be readilyshown that Uk(n) is indeed a subgroup of U(n). (See Exercise 17 inChapter 3.)

Theorem 8.3 U(n) as an External Direct Product

PROOF An isomorphism from U(st) to U(s) % U(t) is x → (x mod s,x mod t); an isomorphism from Us(st) to U(t) is x → x mod t; an isomor-phism from Ut(st) to U(s) is x → x mod s. We leave the verification thatthese mappings are operation-preserving, one-to-one, and onto to thereader. (See Exercises 11, 17, and 19 in Chapter 0; see also [1].)

As a consequence of Theorem 8.3, we have the following result.

Suppose s and t are relatively prime. Then U(st) is isomorphic to theexternal direct product of U(s) and U(t). In short,

U(st) < U(s) % U(t).

Moreover, Us(st) is isomorphic to U(t) and Ut(st) is isomorphic to U(s).

Let m 5 n1n2 ? ? ? nk. Then Zm is isomorphic to Zn1% Zn2

% ? ? ? % Znkif and only if ni and nj are relatively prime when i 2 j.

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160 Groups

Corollary

To see how these results work, let’s apply them to U(105). We obtain

U(105) < U(7) % U(15)U(105) < U(21) % U(5)U(105) < U(3) % U(5) % U(7).

Moreover,

U(7) < U15(105) 5 {1, 16, 31, 46, 61, 76}U(15) < U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92}U(21) < U5(105) 5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101}U(5) < U21(105) 5 {1, 22, 43, 64}U(3) < U35(105) 5 {1, 71}.

Among all groups, surely the cyclic groups Zn have the simpleststructures and, at the same time, are the easiest groups with which tocompute. Direct products of groups of the form Zn are only slightlymore complicated in structure and computability. Because of this, al-gebraists endeavor to describe a finite Abelian group as such a directproduct. Indeed, we shall soon see that every finite Abelian group canbe so represented. With this goal in mind, let us reexamine the U-groups. Using the corollary to Theorem 8.3 and the facts (see [2, p. 93]), first proved by Carl Gauss in 1801, that

U(2) < {0}, U(4) < Z2, U(2n) < Z2 % Z2n22 for n $ 3,

and

U( pn) < Zpn2pn21 for p an odd prime,

we now can write any U-group as an external direct product of cyclicgroups. For example,

U(105) 5 U(3 ? 5 ? 7) < U(3) % U(5) % U(7)

< Z2 % Z4 % Z6

and

U(720) 5 U(16 ? 9 ? 5) < U(16) % U(9) % U(5)

< Z2 % Z4 % Z6 % Z4.

Let m 5 n1n2 ? ? ? nk, where gcd(ni, nj ) 5 1 for i 2 j. Then,

U(m) < U(n1) % U(n2) % ? ? ? % U(nk).

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8 | External Direct Products 161

What is the advantage of expressing a group in this form? Well, for onething, we immediately see that the orders of the elements U(720) canonly be 1, 2, 3, 4, 6, and 12. This follows from the observations that anelement from Z2 % Z4 % Z6 % Z4 has the form (a, b, c, d), where|a| 5 1 or 2, |b| 5 1, 2, or 4, |c| 5 1, 2, 3, or 6, and |d| 5 1, 2, or 4, and that |(a, b, c, d)| 5 lcm(|a|, |b|, |c|, |d|). For another thing, we can read-ily determine the number of elements of order 12, say, that U(720) has.Because U(720) is isomorphic to Z2 % Z4 % Z6 % Z4, it suffices to cal-culate the number of elements of order 12 in Z2 % Z4 % Z6 % Z4. Butthis is easy. By Theorem 8.1, an element (a, b, c, d) has order 12 if andonly if lcm(|a|, |b|, |c|, |d|) 5 12. Since |a| 5 1 or 2, it does not matterhow a is chosen. So, how can we have lcm(|b|, |c|, |d|) 5 12? One wayis to have |b| 5 4, |c| 5 3 or 6, and d arbitrary. By Theorem 4.4, thereare two choices for b, four choices for c, and four choices for d. So, inthis case, we have 2 ? 4 ? 4 5 32 choices. The only other way to havelcm(|b|, |c|, |d|) 5 12 is for |d| 5 4, |c| 5 3 or 6, and |b| 5 1 or 2 (weexclude |b| 5 4, since this was already accounted for). This gives 2 ? 4 ?2 5 16 new choices. Finally, since a can be either of the two elementsin Z2, we have a total of 2(32 1 16) 5 96 elements of order 12.

These calculations tell us more. Since Aut(Z720) is isomorphic toU(720), we also know that there are 96 automorphisms of Z720 oforder 12. Imagine trying to deduce this information directly fromU(720) or, worse yet, from Aut(Z720)! These results beautifully illus-trate the advantage of being able to represent a finite Abelian group asa direct product of cyclic groups. They also show the value of our the-orems about Aut(Zn) and U(n). After all, theorems are labor-saving devices. If you want to convince yourself of this, try to provedirectly from the definitions that Aut(Z720) has exactly 96 elements oforder 12.

ApplicationsWe conclude this chapter with five applications of the material pre-sented here—three to cryptography, the science of sending and deci-phering secret messages, one to genetics, and one to electric circuits.

Data Security

Because computers are built from two-state electronic components,it is natural to represent information as strings of 0s and 1s calledbinary strings. A binary string of length n can naturally be thought ofas an element of Z2 % Z2 % ? ? ? % Z2 (n copies) where the parenthe-ses and the commas have been deleted. Thus the binary string

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162 Groups

11000110 corresponds to the element (1, 1, 0, 0, 0, 1, 1, 0) in Z2 % Z2 %

Z2 % Z2 % Z2 % Z2 % Z2 % Z2. Similarly, two binary strings a1a2 ? ? ? anand b1b2 ? ? ? bn are added componentwise modulo 2 just as theircorresponding elements in Z2 % Z2 % ? ? ? % Z2 are. For example,

11000111 1 01110110 5 10110001

and

10011100 1 10011100 5 00000000.

The fact that the sum of two binary sequences a1a2 ? ? ? an 1 b1b2 ? ? ?bn 5 00 ? ? ? 0 if and only if the sequences are identical is the basis fora data security system used to protect internet transactions.

Suppose that you want to purchase a compact disc from www.Amazon.com. Need you be concerned that a hacker will interceptyour credit-card number during the transaction? As you might expect,your credit-card number is sent to Amazon in a way that protects thedata. We explain one way to send credit-card numbers over the Websecurely. When you place an order with Amazon the company sendsyour computer a randomly generated string of 0’s and 1’s called a key.This key has the same length as the binary string corresponding toyour credit-card number and the two strings are added (think of thisprocess as “locking” the data). The resulting sum is then transmittedto Amazon. Amazon in turn adds the same key to the received stringwhich then produces the original string corresponding to your credit-card number (adding the key a second time “unlocks” the data).

To illustrate the idea, say you want to send an eight-digit binary stringsuch as s 5 10101100 to Amazon (actual credit-card numbers have verylong strings) and Amazon sends your computer the key k 5 00111101. Your computer returns the string s 1 k 5 10101100 100111101 5 10010001 to Amazon, and Amazon adds k to this string toget 10010001 1 00111101 5 10101100, which is the string represent-ing your credit-card number. If someone intercepts the number s 1 k 5 10010001 during transmission it is no value without knowing k.

The method is secure because the key sent by Amazon is randomlygenerated and used only one time. You can tell when you are using an en-cryption scheme on a web transaction by looking to see if the web ad-dress begins with “https” rather than the customary “http.” You will alsosee a small padlock in the status bar at the bottom of the browser window.

Application to Public Key Cryptography

In the mid-1970s, Ronald Rivest, Adi Shamir, and Leonard Adlemandevised an ingenious method that permits each person who is toreceive a secret message to tell publicly how to scramble messages

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8 | External Direct Products 163

sent to him or her. And even though the method used to scramble themessage is known publicly, only the person for whom it is intendedwill be able to unscramble the message. The idea is based on the factthat there exist efficient methods for finding very large prime numbers(say about 100 digits long) and for multiplying large numbers, but noone knows an efficient algorithm for factoring large integers (sayabout 200 digits long). So, the person who is to receive the messagechooses a pair of large primes p and q and chooses an integer r with1 , r , m, where m 5 lcm( p 2 1, q 2 1), such that r is relatively primeto m (any such r will do). This person calculates n 5 pq and announcesthat a message M is to be sent to him or her publicly as Mr mod n.Although r, n, and Mr are available to everyone, only the person whoknows how to factor n as pq will be able to decipher the message.

To present a simple example that nevertheless illustrates the princi-pal features of the method, say we wish to send the message “YES.” Weconvert the message into a string of digits by replacing A by 01, B by02, . . . , Z by 26, and a blank by 00. So, the message YES becomes250519. To keep the numbers involved from becoming too unwieldy,we send the message in blocks of four digits and fill in with blankswhen needed. Thus, the message YES is represented by the two blocks2505 and 1900. The person to whom the message is to be sent haspicked two primes p and q, say p 5 37 and q 5 73 (in actual practice,p and q would have 100 or so digits), and a number r that has no primedivisors in common with lcm( p 2 1, q 2 1) 5 72, say r 5 5, and haspublished n 5 37 ? 73 5 2701 and r 5 5 in a public directory. We willsend the “scrambled” numbers (2505)5 mod 2701 and (1900)5 mod2701 rather than 2505 and 1900, and the receiver will unscramble them.We show the work involved for us and the receiver only for the block2505. The arithmetic involved in computing these numbers is simpli-fied as follows:

2505 mod 2701 5 2505(2505)2 mod 2701 5 602(2505)4 mod 2701 5 (602)(602) mod 2701 5 470.

So, (2505)5 mod 2701 5 (2505)(470) mod 2701 5 2415.†

†To determine 25052 mod 2701 with a calculator, enter 2505 3 2505 to obtain62750025, then divide 6275025 by 2701 to obtain 2323.2228. Finally, enter 6275025 2(2323 3 2701) to obtain 602. Provided that the numbers are not too large, the Googlesearch engine at http://www.google.com will do modular arithmetic. For example, en-tering 2505^2 mod 2701 in the search box yields 602. Be careful, however, because en-tering 2505^5 mod 2701 computes the wrong value since 25055 is too large. Instead, wecan use Google to compute smaller powers such as 25053 mod 2701 (which yields 852)and 25052 mod 2701 and then compute (852 3 602) mod 2701 5 2415.

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Thus, the number 2415 is sent to the receiver. Now the receiver musttake this number and convert it back to 2505. To do so, the receivertakes the two factors of 2701, p 5 37 and q 5 73, and calculates theleast common multiple of p 2 1 5 36 and q 2 1 5 72, which is 72.(This is where the knowledge of p and q is necessary.) Next, the re-ceiver must find s 5 r21 in U(72)—that is, solve the equation 5 ? s 5 1mod 72. This number is 29. (There is a simple algorithm for findingthis number.) Then the receiver takes the number received, 2415, andcalculates (2415)29 mod 2701. This calculation can be simplified as fol-lows:

2415 mod 2701 5 2415(2415)2 mod 2701 5 766 (2415)4 mod 2701 5 (766)2 mod 2701 5 639 (2415)8 mod 2701 5 (639)2 mod 2701 5 470

(2415)16 mod 2701 5 (470)2 mod 2701 5 2119

So, (2415)29 mod 2701 5 (2415)16(2415)8(2415)4(2415) mod 2701 5(2119)(470)(639)(2415) mod 2701 5 ((2119)(470) mod 2701 3(639)(2415) mod 2701) mod 2701 5 (1962)(914) mod 2701 5 2505. [Wecompute the product (2119)(470)(639)(2415) in two stages so that wemay use a hand calculator.]

Thus the receiver correctly determines the code for “YE.” On theother hand, without knowing how pq factors, one cannot find the modu-lus (in our case, 72) that is needed to determine the intended message.

The procedure just described is called the RSA public key encryptionscheme in honor of the three people (Rivest, Shamir, and Adleman) whodiscovered the method. It is widely used in conjunction with web serversand browsers, e-mail programs, remote login sessions, and electronic fi-nancial transactions. The algorithm is summarized below.

Receiver

1. Pick very large primes p and q and compute n 5 pq.2. Compute the least common multiple of p 2 1 and q 2 1; let us call

it m.3. Pick r relatively prime to m.4. Find s such that rs mod m 5 1.5. Publicly announce n and r.

Sender

1. Convert the message to a string of digits. (In practice, the ASCIIcode is used.)

2. Break up the message into uniform blocks of digits; call them M1,M2, . . . , Mk.

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8 | External Direct Products 165

3. Check to see that the greatest common divisor of each Mi and n is1. If not, n can be factored and our code is broken. (In practice, theprimes p and q are so large that they exceed all Mi, so this step maybe omitted.)

4. Calculate and send Ri 5 Mir mod n.

Receiver

1. For each received message Ri, calculate Ris mod n.

2. Convert the string of digits back to a string of characters.

Why does this method work? Well, we know that U(n) < U( p) %U(q) < Zp21 % Zq21. Thus an element of the form xm in U(n) corre-sponds under an isomorphism to one of the form (mx1, mx2) in Zp21 %

Zq21. Since m is the least common multiple of p 2 1 and q 2 1, wemay write m 5 u( p 2 1) and m 5 v(q 2 1) for some u and v. Then(mx1, mx2) 5 (u( p 2 1)x1, v(q 2 1)x2) 5 (0, 0) in Zp21 % Zq21, and itfollows that xm 5 1 for all x in U(n). So, because each message Mi isan element of U(n) and r was chosen so that rs 5 1 1 tm for some t,we have, modulo n,

Ris 5 (Mi

r)s 5 Mirs 5 Mi

11tm 5 (Mim)tMi 5 1tMi 5 Mi.

In 2002, Ronald Rivest, Adi Shamir, and Leonard Adleman receivedthe Association for Computing Machinery A. M. Turing Award whichis considered the “Nobel Prize of Computing” for their contribution topublic key cryptography.

The software for Computer Exercise 5 in this chapter implementsthe RSA scheme for small primes.

Digital Signatures

With so many financial transactions now taking place electronically, theproblem of authenticity is paramount. How is a stockbroker to know thatan electronic message she receives that tells her to sell one stock and buyanother actually came from her client? The technique used in public keycryptography allows for digital signatures as well. Let us say that personA wants to send a secret message to person B in such a way that only Bcan decode the message and B will know that only A could have sent it.Abstractly, let EA and DA denote the algorithms that A uses for encryp-tion and decryption, respectively, and let EB and DB denote the algo-rithms that B uses for encryption and decryption, respectively. Herewe assume that EA and EB are available to the public, whereas DA isknown only to A and DB is known only to B and that DBEB and EADAapplied to any message leaves the message unchanged. Then A sends

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a message M to B as EB (DA(M)) and B decodes the received messageby applying the function EADB to it to obtain

(EADB) (EB(DA(M)) 5 EA(DBEB)(DA(M)) 5 EA(DA(M)) 5 M.

Notice that only A can execute the first step [i.e., create DA(M)] andonly B can implement the last step (i.e., apply EADB to the receivedmessage).

Transactions using digital signatures became legally binding in theUnited States in October 2000.

Application to Genetics†

The genetic code can be conveniently modeled using elements of Z4 %

Z4 % ? ? ? % Z4 where we omit the parentheses and the commas andjust use strings of 0s, 1s, 2s, and 3s and add componentwise modulo 4.A DNA molecule is composed of two long strands in the form of adouble helix. Each strand is made up of strings of the four nitrogenbases adenine (A), thymine (T), guanine (G), and cytosine (C). Eachbase on one strand binds to a complementary base on the other strand.Adenine always is bound to thymine, and guanine always is bound tocytosine. To model this process, we identify A with 0, T with 2, G with 1,and C with 3. Thus, the DNA segment ACGTAACAGGA and its com-plement segment TGCATTGTCCT are denoted by 03120030110 and21302212332. Noting that in Z4, 0 1 2 5 2, 2 1 2 5 0, 1 1 2 5 3, and3 1 2 5 1, we see that adding 2 to elements of Z4 interchanges 0 and 2and 1 and 3. So, for any DNA segment a1a2 ? ? ? an represented by ele-ments of Z4 % Z4 % ? ? ? % Z4, we see that its complementary segmentis represented by a1a2 ? ? ? an 1 22 ? ? ? 2.

Application to Electric Circuits

Many homes have light fixtures that are operated by a pair of switches.They are wired so that when either switch is thrown the light changesits status (from on to off or vice versa). Suppose the wiring is done sothat the light is on when both switches are in the up position. We canconveniently think of the states of the two switches as being matchedwith the elements of Z2 % Z2 with the two switches in the up positioncorresponding to (0, 0) and the two switches in the down position cor-responding to (1, 1). Each time a switch is thrown, we add 1 to thecorresponding component in the group Z2 % Z2. We then see that thelights are on when the switches correspond to the elements of the sub-group �(1, 1)� and are off when the switches correspond to the elements

†This discussion is adapted from [3].

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8 | External Direct Products 167

in the coset (1, 0) 1 �(1, 1)�. A similar analysis applies in the case ofthree switches with the subgroup {(0, 0, 0), (1, 1, 0), (0, 1, 1), (1, 0, 1)}corresponding to the lights-on situation.

Exercises

What’s the most difficult aspect of your life as a mathematician, DianeMaclagan, an assistant professor at Rutgers, was asked. “Trying to provetheorems,” she said. And the most fun? “Trying to prove theorems.”

1. Prove that the external direct product of any finite number ofgroups is a group. (This exercise is referred to in this chapter.)

2. Show that Z2 % Z2 % Z2 has seven subgroups of order 2.3. Let G be a group with identity eG and let H be a group with iden-

tity eH. Prove that G is isomorphic to G % {eH} and that H is iso-morphic to {eG} % H.

4. Show that G % H is Abelian if and only if G and H are Abelian.State the general case.

5. Prove or disprove that Z % Z is a cyclic group.6. Prove, by comparing orders of elements, that Z8 % Z2 is not iso-

morphic to Z4 % Z4.7. Prove that G1 % G2 is isomorphic to G2 % G1. State the general

case.8. Is Z3 % Z9 isomorphic to Z27? Why?9. Is Z3 % Z5 isomorphic to Z15? Why?

10. How many elements of order 9 does Z3 % Z9 have? (Do not do thisexercise by brute force.)

11. How many elements of order 4 does Z4 % Z4 have? (Do not do thisby examining each element.) Explain why Z4 % Z4 has the samenumber of elements of order 4 as does Z8000000 % Z400000. General-ize to the case Z4m % Z4n.

12. The dihedral group Dn of order 2n (n $ 3) has a subgroup of n ro-tations and a subgroup of order 2. Explain why Dn cannot be iso-morphic to the external direct product of two such groups.

13. Prove that the group of complex numbers under addition is iso-morphic to R % R.

14. Suppose that G1 < G2 and H1 < H2. Prove that G1 % H1 < G2 %

H2. State the general case.15. If G % H is cyclic, prove that G and H are cyclic. State the general

case.

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168 Groups

16. In Z40 % Z30, find two subgroups of order 12.17. If r is a divisor of m and s is a divisor of n, find a subgroup of Zm

% Zn isomorphic to Zr % Zs.18. Find a subgroup of Z12 % Z18 isomorphic to Z9 % Z4.19. Let G and H be finite groups and (g, h) [ G % H. State a neces-

sary and sufficient condition for �(g, h)� 5 �g� % �h�.20. Determine the number of elements of order 15 and the number of

cyclic subgroups of order 15 in Z30 % Z20.21. What is the order of any nonidentity element of Z3 % Z3 % Z3?

Generalize.22. Let m . 2 be an even integer and let n . 2 be an odd integer. Find

a formula for the number of elements of order 2 in Dm % Dn.23. Let M be the group of all real 2 3 2 matrices under addition. Let

N 5 R % R % R % R under componentwise addition. Prove thatM and N are isomorphic. What is the corresponding theorem forthe group of m 3 n matrices under addition?

24. The group S3 % Z2 is isomorphic to one of the following groups:Z12, Z6 % Z2, A4, D6. Determine which one by elimination.

25. Let G be a group, and let H 5 {(g, g) | g [ G}. Show that H is asubgroup of G % G. (This subgroup is called the diagonal of G % G.) When G is the set of real numbers under addition, de-scribe G % G and H geometrically.

26. Find a subgroup of Z4 % Z2 that is not of the form H % K, where His a subgroup of Z4 and K is a subgroup of Z2.

27. Find all subgroups of order 3 in Z9 % Z3.28. Find all subgroups of order 4 in Z4 % Z4.29. What is the largest order of any element in Z30 % Z20?30. How many elements of order 2 are in Z2000000 % Z4000000? Generalize.31. Find a subgroup of Z800 % Z200 that is isomorphic to Z2 % Z4.32. Find a subgroup of Z12 % Z4 % Z15 that has order 9.33. Prove that R* % R* is not isomorphic to C*. (Compare this with

Exercise 13.)34. Let

(See Exercise 36 in Chapter 2 for the definition of multiplication.)Show that H is an Abelian group of order 9. Is H isomorphic to Z9or to Z3 % Z3?

H 5 • £1 a b

0 1 0

0 0 1

§ † a, b [ Z3¶ .

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8 | External Direct Products 169

35. Let G 5 {3m6n | m, n [ Z} under multiplication. Prove that G is isomor-phic to Z % Z. Does your proof remain valid if G 5 {3m9n | m, n [ Z}?

36. Let (a1, a2, . . . , an) [ G1 % G2 % ? ? ? % Gn. Give a necessary andsufficient condition for |(a1, a2, . . . , an)| 5 `.

37. Prove that D3 % D4 ] D12% Z2.38. Determine the number of cyclic subgroups of order 15 in Z90 % Z36.39. If a group has exactly 24 elements of order 6, how many cyclic

subgroups of order 6 does it have?40. For any Abelian group G and any positive integer n, let Gn 5 {gn |

g [ G} (see Exercise 15, Supplementary Exercises for Chapters1– 4). If H and K are Abelian, show that (H % K)n 5 Hn % Kn.

41. Express Aut(U(25)) in the form Zm % Zn.42. Determine Aut(Z2 % Z2).43. Suppose that n1, n2, . . . , nk are positive even integers. How many

elements of order 2 does Zn1% Zn2

% . . . % Znkhave ? How many are

there if we drop the requirement that n1, n2, . . . , nk must be even?44. Is Z10 % Z12 % Z6 ^ Z60 % Z6 % Z2?45. Is Z10 % Z12 % Z6 ^ Z15 % Z4 % Z12?46. Find an isomorphism from Z12 to Z4 % Z3.47. How many isomorphisms are there from Z12 to Z4 % Z3?48. Suppose that f is an isomorphism from Z3 % Z5 to Z15 and

f(2, 3) 5 2. Find the element in Z3 % Z5 that maps to 1.49. Let (a, b) belong to Zm % Zn. Prove that |(a, b)| divides lcm(m, n).50. Let G 5 {ax2 1 bx 1 c | a, b, c [ Z3}. Add elements of G as you

would polynomials with integer coefficients, except use modulo 3addition. Prove that G is isomorphic to Z3 % Z3 % Z3. Generalize.

51. Use properties of U-groups to determine all cyclic groups that haveexactly two generators.

52. Explain a way that a string of length n of the four nitrogen bases A,T, G, and C could be modeled with the external direct product of ncopies of Z2 % Z2.

53. Let p be a prime. Prove that Zp % Zp has exactly p 1 1 subgroupsof order p.

54. Give an example of an infinite non-Abelian group that has exactlysix elements of finite order.

55. Give an example to show that there exists a group with elements aand b such that |a| 5 `, |b| 5 ` and |ab| 5 2.

56. Express U(165) as an external direct product of cyclic groups ofthe form Zn.

57. Express U(165) as an external direct product of U-groups in fourdifferent ways.

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170 Groups

58. Without doing any calculations in Aut(Z20), determine how manyelements of Aut(Z20) have order 4. How many have order 2?

59. Without doing any calculations in Aut(Z720), determine how manyelements of Aut(Z720) have order 6.

60. Without doing any calculations in U(27), decide how many sub-groups U(27) has.

61. What is the largest order of any element in U(900)?62. Let p and q be odd primes and let m and n be positive integers.

Explain why U( pm) % U(qn) is not cyclic.63. Use the results presented in this chapter to prove that U(55) is

isomorphic to U(75).64. Use the results presented in this chapter to prove that U(144) is

isomorphic to U(140).65. For every n . 2, prove that U(n)2 5 {x2 | x [ U(n)} is a proper

subgroup of U(n).66. Show that U(55)3 5 {x3 | x [ U(55)} is U(55).67. Find an integer n such that U(n) contains a subgroup isomorphic to

Z5 % Z5.68. Find a subgroup of order 6 in U(700).69. Show that there is a U-group containing a subgroup isomorphic

to Z3 % Z3.70. Show that no U-group has order 14.71. Show that there is a U-group containing a subgroup isomorphic

to Z14.72. Show that no U-group is isomorphic to Z4 % Z4.73. Show that there is a U-group containing a subgroup isomorphic to

Z4 % Z4.74. Using the RSA scheme with p 5 37, q 5 73, and r 5 5, what num-

ber would be sent for the message “RM”?75. Assuming that a message has been sent via the RSA scheme with

p 5 37, q 5 73, and r 5 5, decode the received message “34.”

Computer Exercises

The geek shall inherit the earth.LEV GROSSMAN

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

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8 | External Direct Products 171

1. This software lists the elements of Us(st), where s and t are rela-tively prime. Run the program for (s, t) 5 (5, 16), (16, 5), (8, 25),(5, 9), (9, 5), (9, 10), (10, 9), and (10, 25).

2. This software computes the elements of the subgroup U(n)k 5 {xk | x [ U(n)} of U(n) and its order. Run the program for (n, k) 5(27, 3), (27, 5), (27, 7), and (27, 11). Do you see a relationshipconnecting |U(n)| and |U(n)k|, f(n), and k? Make a conjecture.Run the program for (n, k) 5 (25, 3), (25, 5), (25, 7), and (25, 11).Do you see a relationship connecting |U(n)| and |U(n)k|, f(n), andk? Make a conjecture. Run the program for (n, k) 5 (32, 2), (32,4), and (32, 8). Do you see a relationship connecting |U(n)| and|U(n)k|, f(n), and k? Make a conjecture. Is your conjecture validfor (32, 16)? If not, restrict your conjecture. Run the program for(n, k) 5 (77, 2), (77, 3), (77, 5), (77, 6), (77, 10), and (77, 15)? Doyou see a relationship among U(77, 6), U(77, 2), and U(77, 3)?What about U(77, 10) U(77, 2), and U(77, 5)? What about U(77,15), U(77, 3), and U(77, 5)? Make a conjecture. Use the theorydeveloped in this chapter about expressing U(n) as external directproducts of cyclic groups of the form Zn to analyze these groupsto verify your conjectures.

3. This software implements the algorithm given on page 160 to ex-press U(n) as an external direct product of groups of the form Zk.Run the program for n 5 3 ? 5 ? 7, 16 ? 9 ? 5, 8 ? 3 ? 25, 9 ? 5 ? 11,and 2 ? 27 ? 125.

4. This software allows you to input positive integers n1, n2, n3, . . . , nk,where k # 5, and compute the number of elements in Zn1

%

Zn 2

% ? ? ? % Znkof any specified order m. Use this software to ver-

ify the values obtained in Examples 4 and 5 and in Exercise 20.Run the software for n1 5 6, n2 5 10, n3 5 12, and m 5 6.

5. This program implements the RSA public key cryptographyscheme. The user enters two primes p and q, an r that is relativelyprime to m 5 lcm (p 2 1, q 2 1), and the message M to be sent.Then the program computes s, which is the inverse of r mod m,and the value of Mr mod pq. Also, the user can input those num-bers and have the computer raise the numbers to the s power to ob-tain the original input.

6. This software determines the order of Aut(Zp % Zp), where p is aprime. Run the software for p 5 3, 5, and 7. Is the result alwaysdivisible by p? Is the result always divisible by p 2 1? Is the resultalways divisible by p 1 1? Make a conjecture about the order ofAut(Zp % Zp) for all primes p.

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172 Groups

7. This software determines the order of Aut(Zp % Zp % Zp), where pis a prime. Run the software for p 5 3, 5, and 7. What is the highestpower of p that divides the order? What is the highest power of p 2 1that divides the order? What is the highest power of p 1 1 that di-vides the order? Make a conjecture about the order of Aut(Zp % Zp % Zp) for all primes p.

References

1. J. A. Gallian and D. Rusin, “Factoring Groups of Integers Modulo n,”Mathematics Magazine 53 (1980): 33–36.

2. D. Shanks, Solved and Unsolved Problems in Number Theory, 2nd ed.,New York: Chelsea, 1978.

3. S. Washburn, T. Marlowe, and C. Ryan, Discrete Mathematics, Reading,MA: Addison-Wesley, 1999.

Suggested Readings

Y. Cheng, “Decomposition of U-groups,” Mathematics Magazine 62(1989): 271–273.

This article explores the decomposition of U(st), where s and t arerelatively prime, in greater detail than we have provided.

David J. Devries, “The Group of Units in Zn,” Mathematics Magazine 62(1989): 340.

This article provides a simple proof that U(n) is not cyclic when n isnot of the form 1, 2, 4, pk, or 2pk, where p is an odd prime.

David R. Guichard, “When Is U(n) Cyclic? An Algebra Approach,”Mathematics Magazine 72 (1999): 139–142.

The author provides a group-theoretic proof of the fact that U(n) iscyclic if and only if n is 1, 2, 4, pk, or 2pk, where p is an odd prime.

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Leonard Adleman

LEONARD ADLEMAN grew up in San Francisco.He did not have any great ambitions for him-self and, in fact, never even thought about be-coming a mathematician. He enrolled at theUniversity of California at Berkeley intendingto be a chemist, then changed his mind andsaid he would be a doctor. Finally, he settledon a mathematics major. “I had gone through azillion things and finally the only thing thatwas left where I could get out in a reasonabletime was mathematics,” he said.

Adleman graduated in five years, in1968, “wondering what I wanted to do withmy life.” He took a job as a computer pro-grammer at the Bank of America. Then hedecided that maybe he should be a physicist,so he began taking classes at San FranciscoState College while working at the bank.Once again, Adleman lost interest. “I didn’tlike doing experiments, I liked thinkingabout things,” he said. Later, he returned toBerkeley with the aim of getting a Ph.D. incomputer science. “I thought that getting aPh.D. in computer science would at leastfurther my career,” he said.

But, while in graduate school, somethingelse happened to Adleman. He finally under-stood the true nature and compelling beautyof mathematics. He discovered, he said, thatmathematics “is less related to accountingthan it is to philosophy.”

“People think of mathematics as somekind of practical art,” Adleman said. But, headded, “the point when you become a mathe-matician is where you somehow see throughthis and see the beauty and power of mathe-matics.” Adleman got his Ph.D. in 1976 andimmediately landed a job as an assistant pro-fessor of mathematics at the MassachusettsInstitute of Technology. There he met RonaldRivest and Adi Shamir, who were trying toinvent an unbreakable public key system.They shared their excitement about the ideawith Adleman, who greeted it with a politeyawn, thinking it impractical and not very in-teresting. Nevertheless, Adleman agreed totry to break the codes Rivest and Shamir pro-posed. Rivest and Shamir invented 42 codingsystems, and each time Adleman broke thecode. Finally, on their 43rd attempt, they hitupon what is now called the RSA scheme.

Adleman’s mode of working is to findsomething that intrigues him and to dig in.He does not read mathematics journals, hesays, because he does not want to be influ-enced by other people’s ideas.

Asked what it is like to simply sit andthink for six months, Adleman responded,“That’s what a mathematician always does.Mathematicians are trained and inclined tosit and think. A mathematician can sit andthink intensely about a problem for 12 hours

“. . . Dr. Adleman [has played] a central rolein some of the most surprising, andprovocative, discoveries in theoreticalcomputer science.”

GINA KOLATA, The New York Times,

13 December 1994.

173

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174

Supplementary Exercises for Chapters 5–8

My mind rebels at stagnation. Give me problems, give me work, give me the most obstruse cryptogram, or the most intricate analysis, and I am in my own proper atmosphere.

SHERLOCK HOLMES, The Sign of Four

True/False questions for Chapters 5–8 are available on the Web at:

www.d.umn.edu/~jgallian/TF

1. A subgroup N of a group G is called a characteristic subgroup iff(N) 5 N for all automorphisms f of G. (The term characteristicwas first applied by G. Frobenius in 1895.) Prove that every sub-group of a cyclic group is characteristic.

2. Prove that the center of a group is characteristic.3. The commutator subgroup G9 of a group G is the subgroup gener-

ated by the set {x21y21xy | x, y [ G}. (That is, every element of G9has the form a1

i1a2i2 ? ? ? a k

ik, where each aj has the form x21y21xy,each ij 5 61, and k is any positive integer.) Prove that G9 is a char-acteristic subgroup of G. (This subgroup was first introduced by G. A. Miller in 1898.)

4. Prove that the property of being a characteristic subgroup is transi-tive. That is, if N is a characteristic subgroup of K and K is a char-acteristic subgroup of G, then N is a characteristic subgroup of G.

5. Let G 5 Z3 % Z3 % Z3 and let H be the subgroup of SL(3, Z3)consisting of

H 5 • £1 a b

0 1 0

0 0 1

§ † a, b c [ Z3¶ .

a day, six months straight, with perhaps justa pencil and paper.” The only prop he needs,he said, is a blackboard to stare at.

Adapted from an article by Gina Kolata,The New York Times, 13 December 1994.

For more information about Adleman,visit:

http://www.wikipedia.com

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8 | Supplementary Exercises for Chapters 5–8 175

(See Exercise 36 in Chapter 2 for the definition of multiplication.)Determine the number of elements of each order in G and H. Are Gand H isomorphic? (This exercise shows that two groups with thesame number of elements of each order need not be isomorphic.)

6. Let H and K be subgroups of a group G and let HK 5 {hk | h [ H,k [ K} and KH 5 {kh | k [ K, h [ H}. Prove that HK is a group ifand only if HK 5 KH.

7. Let H and K be subgroups of a finite group G. Prove that

.

(This exercise is referred to in Chapters 10, 11, and 24.)8. The exponent of a group is the smallest positive integer n such that

xn 5 e for all x in the group. Prove that every finite group has an ex-ponent that divides the order of the group.

9. Determine all U-groups of exponent 2.10. Suppose that H and K are subgroups of a group and that |H| and |K|

are relatively prime. Show that H y K 5 {e}.

11. Let R1 denote the multiplicative group of positive real numbers andlet T 5 {a 1 bi [ C*| a2 1 b2 5 1} be the multiplicative group ofcomplex numbers of norm 1. Show that every element of C* can beuniquely expressed in the form rz, where r [ R1 and z [ T.

12. Use a group-theoretic proof to show that Q* under multiplication isnot isomorphic to R* under multiplication.

13. Use a group-theoretic proof to show that Q under addition is notisomorphic to R under addition.

14. Prove that R under addition is not isomorphic to R* undermultiplication.

15. Show that Q1 (the set of positive rational numbers) under multipli-cation is not isomorphic to Q under addition.

16. Suppose that G 5 {e, x, x2, y, yx, yx2} is a non-Abelian group with|x| 5 3 and |y| 5 2. Show that xy 5 yx2.

17. Let p be an odd prime. Show that 1 is the only solution of xp22 5 1in U(p).

18. Let G be an Abelian group under addition. Let n be a fixed positiveinteger and let H 5 {(g, ng) | g [ G}. Show that H is a subgroup ofG % G. When G is the set of real numbers under addition, describeH geometrically.

19. Find five subgroups of Z12 % Z20 1 Z10 isomorphic to Z4 % Z5.20. Suppose that G 5 G1 % G2 % ? ? ? % Gn. Prove that Z(G) 5

Z(G1) % Z(G2) % ? ? ? % Z(Gn).

0HK 0 50H 0 0K 00H d K 0

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176 Groups

21. Exhibit four nonisomorphic groups of order 18.22. What is the order of the largest cyclic subgroup in Aut(Z720)? (Hint:

It is not necessary to consider automorphisms of Z720.)23. Let G be the group of all permutations of the positive integers. Let

H be the subset of elements of G that can be expressed as a productof a finite number of cycles. Prove that H is a subgroup of G.

24. Let H be a subgroup of G. Show that Z(G)H is a subgroup of G.25. Show that D11 % Z3 ] D3 % Z11. (This exercise is referred to in

Chapter 24.)26. Show that D33 ] D11 % Z3. (This exercise is referred to in Chapter 24.)27. Show that D33 ] D3 % Z11. (This exercise is referred to in Chapter 24.)28. Exhibit four nonisomorphic groups of order 66. (This exercise is

referred to in Chapter 24.)29. Prove that |Inn(G)| 5 1 if and only if G is Abelian.30. Prove that x100 5 1 for all x in U(1000).31. Find a subgroup of order 6 in U(450).32. List four elements of Z20 % Z5 % Z60 that form a noncyclic

subgroup.33. In S10, let b 5 (13)(17)(265)(289). Find an element in S10 that com-

mutes with b but is not a power of b.34. Prove or disprove that Z4 % Z15 < Z6 % Z10.35. Prove or disprove that D12 < Z3 % D4.36. Describe a three-dimensional solid whose symmetry group is iso-

morphic to D5.37. Let G 5 U(15) % Z10 % S5. Find the order of (2, 3, (123)(15)). Find

the inverse of (2, 3, (123)(15)).38. Let G 5 Z % Z10 and let H 5 {g [ G| |g| 5 ` or |g| 5 1}. Prove

or disprove that H is a subgroup of G.39. Let G be an infinite group of the form G1 % G2 % . . . % Gn where

each Gi is a nontrivial group and n . 1. Prove that G is not cyclic.40. For any s in Sn and any k-cycle (i1i2 . . . ik) in Sn, prove that s(i1i2 . . .

ik)s21 5 s(i1)s(i2) . . . s(ik).

41. Find an element of order 10 in A9.42. In the left regular representation for D4, write TR90

and TH in matrixform and in cycle form.

43. How many elements of order 6 are in S7?44. Prove that S3 % S4 is not isomorphic to a subgroup of S6.45. Find a permutation b such that b2 5 (13579)(268).

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8 | Supplementary Exercises for Chapters 5–8 177

46. In R % R under componentwise addition, let H 5 {(x, 3x) | x [ R}.(Note that H is the subgroup of all points on the line y 5 3x.) Showthat (2, 5) 1 H is a straight line passing through the point (2, 5) andparallel to the line y 5 3x.

47. In R % R, suppose that H is the subgroup of all points lying on aline through the origin. Show that any left coset of H is either H ora line parallel to H.

48. Let G be a group of permutations on the set {1, 2, . . . , n}. Recallthat stabG(1) 5 {a [ G | a(1) 5 1}. If g sends 1 to k, prove thatg stabG(1) 5 {b [ G | b(1) 5 k}.

49. Let H be a subgroup of G and let a, b [ G. Show that aH 5 bH ifand only if Ha21 5 Hb21.

50. Suppose that G is a finite Abelian group that does not contain asubgroup isomorphic to Zp % Zp for any prime p. Prove that G iscyclic.

51. Let p be a prime. Determine the number of elements of order p in% .

52. Show that % has exactly one subgroup isomorphic to Zp % Zp.53. Let p be a prime. Determine the number of subgroups of %

isomorphic to .54. Find a group of order 32 ? 52 ? 72 ? 28 that contains a subgroup iso-

morphic to A8.55. Let p and q be distinct odd primes. Let n 5 lcm(p 2 1, q 2 1).

Prove that xn 5 1 for all x [ U( pq).56. Prove that D6 is not isomorphic to a subgroup of S4.57. Prove that the permutations (12) and (123 . . . n) generate Sn. (That

is, every member of Sn can be expressed as some combination ofthese elements.

58. Suppose that n is even and s is an (n 2 1)-cycle in Sn. Show that sdoes not commute with any element of order 2.

59. Suppose that n is odd and s is an n-cycle in Sn. Prove that s doesnot commute with any element of order 2.

Zp2

Zp2Zp2

Zp2Zp2

Zp2Zp2

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178

Normal Subgroups and Factor Groups

It is tribute to the genius of Galois that he recognized that those subgroupsfor which the left and right cosets coincide are distinguished ones. Veryoften in mathematics the crucial problem is to recognize and to discoverwhat are the relevant concepts; once this is accomplished the job may bemore than half done.

I. N. HERSTEIN, Topics in Algebra

9

Normal SubgroupsAs we saw in Chapter 7, if G is a group and H is a subgroup of G, it is notalways true that aH 5 Ha for all a in G. There are certain situations wherethis does hold, however, and these cases turn out to be of critical impor-tance in the theory of groups. It was Galois, about 175 years ago, who firstrecognized that such subgroups were worthy of special attention.

Definition Normal Subgroup

A subgroup H of a group G is called a normal subgroup of G if aH 5Ha for all a in G. We denote this by H v G.

Many students make the mistake of thinking that “H is normal in G”means ah 5 ha for a [ G and h [ H. This is not what normality of Hmeans; rather, it means that if a [ G and h [ H, then there exist ele-ments h9 and h0 in H such that ah 5 h9a and ha 5 ah0. Think of it thisway: You can switch the order of a product of an element from the groupand an element from the normal subgroup, but you must “fudge” a bit onthe element from the normal subgroup by using h9 or h0 rather than h. (Itis possible that h9 5 h or h0 5 h, but we may not assume this.)

There are several equivalent formulations of the definition of nor-mality. We have chosen the one that is the easiest to use in applications.However, to verify that a subgroup is normal, it is usually better to useTheorem 9.1, which is a weaker version of property 7 of the lemma inChapter 7. It allows us to substitute a condition about two subgroups ofG for a condition about two cosets of G.

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9 | Normal Subgroups and Factor Groups 179

Theorem 9.1 Normal Subgroup Test

PROOF If H is normal in G, then for any x [ G and h [ H there is an h9in H such that xh 5 h9x. Thus, xhx21 5 h9, and therefore xHx21 # H.

Conversely, if xHx21 # H for all x, then, letting x 5 a, we haveaHa21 # H or aH # Ha. On the other hand, letting x 5 a21, we havea21H(a21)21 5 a21Ha # H or Ha # aH.

EXAMPLE 1 Every subgroup of an Abelian group is normal. (In thiscase, ah 5 ha for a in the group and h in the subgroup.)

EXAMPLE 2 The center Z(G) of a group is always normal. [Again,ah 5 ha for any a [ G and any h [ Z(G).]

EXAMPLE 3 The alternating group An of even permutations is a nor-mal subgroup of Sn. [Note, for example, that for (12) [ Sn and (123) [An, we have (12)(123) 2 (123)(12) but (12)(123) 5 (132)(12) and (132) [ An.]

EXAMPLE 4 The subgroup of rotations in Dn is normal in Dn. (Forany rotation r and any reflection f, we have fr 5 r21f, whereas for anyrotations r and r9, we have rr9 5 r9r.)

EXAMPLE 5 The group SL(2, R) of 2 3 2 matrices with determinant1 is a normal subgroup of GL(2, R), the group of 2 3 2 matrices withnonzero determinant. To verify this, we use the normal subgroup testgiven in Theorem 9.1. Let x [ GL(2, R) 5 G, h [ SL(2, R) 5 H andnote that det xhx21 5 (det x)(det h)(det x)21 5 (det x)(det x)21 5 1. So,xhx21 [ H, and, therefore, xHx21 # H.

EXAMPLE 6 Referring to the group table for A4 given in Table 5.1 onpage 107, we may observe that H 5 {a1, a2, a3, a4} is a normalsubgroup of A4, whereas K 5 {a1, a5, a9} is not a normal subgroupof A4. To see that H is normal, simply note that for any b in A4, bHb21 isa subgroup of order 4 and H is the only subgroup of A4 of order 4 since all other elements of A4 have order 3. Thus, bHb21 5 H. In con-trast, a2a5a 2

21 5 a7, so that a2Ka221 s K.

A subgroup H of G is normal in G if and only if xHx21 # H for all x in G.

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Factor GroupsWe have yet to explain why normal subgroups are of special significance.The reason is simple. When the subgroup H of G is normal, then the setof left (or right) cosets of H in G is itself a group—called the factor groupof G by H (or the quotient group of G by H). Quite often, one can obtaininformation about a group by studying one of its factor groups. Thismethod will be illustrated in the next section of this chapter.

Theorem 9.2 Factor Groups (O. Hölder, 1889)

PROOF Our first task is to show that the operation is well defined; thatis, we must show that the correspondence defined above from G/H 3G/H into G/H is actually a function. To do this we assume that forsome elements a, a9, b, and b9 from G, we have aH 5 a9H and bH 5b9H and verify that aHbH 5 a9Hb9H. That is, verify that abh 5 a9b9H.(This shows that the definition of multiplication depends only on thecosets and not on the coset representatives.) From aH 5 a9H and bH 5b9H , we have a9 5 ah1 and b9 5 bh2 for some h1, h2 in H, and thereforea9b9H 5 ah1bh2H 5 ah1bH 5 ah1Hb 5 aHb 5 abH. Here we have mademultiple use of associativity, property 2 of the lemma in Chapter 7, andthe fact that H v G. The rest is easy: eH 5 H is the identity; a21H is theinverse of aH; and (aHbH)cH 5 (ab)HcH 5 (ab)cH 5 a(bc)H 5aH(bc)H 5 aH(bHcH). This proves that G/H is a group.

Although it is merely a curiosity, we point out that the converse ofTheorem 9.2 is also true; that is, if the correspondence aHbH 5 abHdefines a group operation on the set of left cosets of H in G, then H isnormal in G.

The next few examples illustrate the factor group concept.

EXAMPLE 7 Let 4Z 5 {0, 64, 68, . . .}. To construct Z/4Z, we firstmust determine the left cosets of 4Z in Z. Consider the following fourcosets:

0 1 4Z 5 4Z 5 {0, 64, 68, . . .},1 1 4Z 5 {1, 5, 9, . . . ; 23, 27, 211, . . .},

Let G be a group and let H be a normal subgroup of G. The set G/H 5 {aH | a [ G} is a group under the operation (aH)(bH) 5 abH.†

†The notation G/H was first used by C. Jordan.

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9 | Normal Subgroups and Factor Groups 181

2 1 4Z 5 {2, 6, 10, . . . ; 22, 26, 210, . . .},3 1 4Z 5 {3, 7, 11, . . . ; 21, 25, 29, . . .}.

We claim that there are no others. For if k [ Z, then k 5 4q 1 r, where0 # r , 4; and, therefore, k 1 4Z 5 r 1 4q 1 4Z 5 r 1 4Z. Now thatwe know the elements of the factor group, our next job is to determinethe structure of Z/4Z. Its Cayley table is

0 1 4Z 1 1 4Z 2 1 4Z 3 1 4Z

0 1 4Z 0 1 4Z 1 1 4Z 2 1 4Z 3 1 4Z1 1 4Z 1 1 4Z 2 1 4Z 3 1 4Z 0 1 4Z2 1 4Z 2 1 4Z 3 1 4Z 0 1 4Z 1 1 4Z3 1 4Z 3 1 4Z 0 1 4Z 1 1 4Z 2 1 4Z

Clearly, then, Z/4Z L Z4. More generally, if for any n . 0 we let nZ 5{0, 6n, 62n, 63n, . . .}, then Z/nZ is isomorphic to Zn.

EXAMPLE 8 Let G 5 Z18 and let H 5 �6� 5 {0, 6, 12}. Then G/H 5{0 1 H, 1 1 H, 2 1 H, 3 1 H, 4 1 H, 5 1 H}. To illustrate how thegroup elements are combined, consider (5 1 H) 1 (4 1 H). Thisshould be one of the six elements listed in the set G/H. Well, (5 1 H) 1(4 1 H) 5 5 1 4 1 H 5 9 1 H 5 3 1 6 1 H 5 3 1 H, since H ab-sorbs all multiples of 6.

A few words of caution about notation are warranted here. When His a normal subgroup of G, the expression |aH| has two possible inter-pretations. One could be thinking of aH as a set of elements and |aH|as the size of the set; or, as is more often the case, one could be think-ing of aH as a group element of the factor group G/H and |aH| as theorder of the element aH in G/H. In Example 8, for instance, the set 3 1H has size 3, since 3 1 H 5 {3, 9, 15}. But the group element3 1 H has order 2, since (3 1 H) 1 (3 1 H) 5 6 1 H 5 0 1 H. As isusually the case when one notation has more than one meaning, the ap-propriate interpretation will be clear from the context.

EXAMPLE 9 Let _ 5 {R0, R180}, and consider the factor group ofthe dihedral group D4 (see page 31 for the multiplication table for D4)

D4/_ 5 {_, R90_, H_, D_}.

The multiplication table for D4/_ is given in Table 9.1. (Notice thateven though R90H 5 D9, we have used D_ in Table 9.1 for H_R90_because D9_ 5 D_.)

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182 Groups

D4/_ provides a good opportunity to demonstrate how a factorgroup of G is related to G itself. Suppose we arrange the heading of theCayley table for D4 in such a way that elements from the same coset of_ are in adjacent columns (Table 9.2). Then, the multiplication tablefor D4 can be blocked off into boxes that are cosets of _, and the sub-stitution that replaces a box containing the element x with the coset x_yields the Cayley table for D4/_.

For example, when we pass from D4 to D4/_, the box

in Table 9.2 becomes the element H_ in Table 9.1. Similarly, the box

becomes the element D_, and so on.

Table 9.2

R0 R180 R90 R270 H V D D9

R0 R0 R180 R90 R270 H V D D9R180 R180 R0 R270 R90 V H D9 D

R90 R90 R270 R180 R0 D9 D H VR270 R270 R90 R0 R180 D D9 V H

H H V D D9 R0 R180 R90 R270V V H D9 D R180 R0 R270 R90

D D D9 V H R270 R90 R0 R180D9 D9 D H V R90 R270 R180 R0

H V

V H

D D9

D9 D

Table 9.1

__ R90__ H__ D__

__ _ R90_ H_ D_R90__ R90_ _ D_ H_H__ H_ D_ _ R90_D__ D_ H_ R90_ _

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9 | Normal Subgroups and Factor Groups 183

Table 9.3

1 2 3 4 5 6 7 8 9 10 11 12

1 1 2 3 4 5 6 7 8 9 10 11 122 2 1 4 3 6 5 8 7 10 9 12 113 3 4 1 2 7 8 5 6 11 12 9 104 4 3 2 1 8 7 6 5 12 11 10 9

5 5 8 6 7 9 12 10 11 1 4 2 36 6 7 5 8 10 11 9 12 2 3 1 47 7 6 8 5 11 10 12 9 3 2 4 18 8 5 7 6 12 9 11 10 4 1 3 2

9 9 11 12 10 1 3 4 2 5 7 8 610 10 12 11 9 2 4 3 1 6 8 7 511 11 9 10 12 3 1 2 4 7 5 6 812 12 10 9 11 4 2 1 3 8 6 5 7

Table 9.4

1H 5H 9H

1H 1H 5H 9H5H 5H 9H 1H9H 9H 1H 5H

In this way, one can see that the formation of a factor group G/Hcauses a systematic collapse of the elements of G. In particular, all theelements in the coset of H containing a collapse to the single group el-ement aH in G/H.

EXAMPLE 10 Consider the group A4 as represented by Table 5.1on page 107. (Here i denotes the permutation ai.) Let H 5 {1, 2, 3, 4}.Then the three cosets of H are H, 5H 5 {5, 6, 7, 8}, and 9H 5 {9,10, 11, 12}. (In this case, rearrangement of the headings is unneces-sary.) Blocking off the table for A4 into boxes that are cosets of Hand replacing the boxes containing 1, 5, and 9 (see Table 9.3) withthe cosets 1H, 5H, and 9H, we obtain the Cayley table for G/H givenin Table 9.4.

This procedure can be illustrated more vividly with colors. Let’s saywe had printed the elements of H in green, the elements of 5H in red,and the elements of 9H in blue. Then, in Table 9.3, each box wouldconsist of elements of a uniform color. We could then think of

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184 Groups

the factor group as consisting of the three colors that define a grouptable isomorphic to G/H.

It is instructive to see what happens if we attempt the same proce-dure with a group G and a subgroup H that is not normal in G—that is,if we arrange the headings of the Cayley table so that the elementsfrom the same coset of H are in adjacent columns and attempt to blockoff the table into boxes that are also cosets of H to produce a Cayleytable for the set of cosets. Say, for instance, we were to take G to be A4and H 5 {1, 5, 9}. The cosets of H would be H, 2H 5 {2, 6, 10},3H 5 {3, 7, 11}, and 4H 5 {4, 8, 12}. Then the first three rows of therearranged Cayley table for A4 would be

Green Red Blue

Green Green Red BlueRed Red Blue GreenBlue Blue Green Red

1 5 9 2 6 10 3 7 11 4 8 12

1 1 5 9 2 6 10 3 7 11 4 8 125 5 9 1 8 12 4 6 10 2 7 11 39 9 1 5 11 3 7 12 4 8 10 2 6

But already we are in trouble, for blocking these off into 3 3 3 boxesyields boxes that contain elements of different cosets. Hence, it is im-possible to represent an entire box by a single element of the box in thesame way we could for boxes made from the cosets of a normal sub-group. Had we printed the rearranged table in four colors with allmembers of the same coset having the same color, we would see multi-colored boxes rather than the uniformly colored boxes produced by anormal subgroup.

In Chapter 11, we will prove that every finite Abelian group isisomorphic to a direct product of cyclic groups. In particular, anAbelian group of order 8 is isomorphic to one of Z8, Z4 % Z2, or Z2 %

Z2 % Z2. In the next two examples, we examine Abelian factor groupsof order 8 and determine the isomorphism type of each.

EXAMPLE 11 Let G 5 U(32) 5 {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21,23, 25, 27, 29, 31} and H 5 U16(32) 5 {1, 17}. Then G/H is an Abeliangroup of order 16/2 5 8. Which of the three Abelian groups of order 8is it—Z8, Z4 % Z2, or Z2 % Z2 % Z2? To answer this question, we need

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9 | Normal Subgroups and Factor Groups 185

only determine the elements of G/H and their orders. Observe that theeight cosets

1H 5 {1, 17}, 3H 5 {3, 19}, 5H 5 {5, 21}, 7H 5 {7, 23},9H 5 {9, 25}, 11H 5 {11, 27}, 13H 5 {13, 29}, 15H 5 {15, 31}

are all distinct, so that they form the factor group G/H. Clearly,(3H)2 5 9H 2 H, and so 3H has order at least 4. Thus, G/H is not Z2 % Z2 % Z2. On the other hand, direct computations show that both7H and 9H have order 2, so that G/H cannot be Z8 either, since a cyclicgroup of even order has exactly one element of order 2 (Theorem 4.4).This proves that U(32)/U16(32) L Z4 % Z2, which (not so incidentally!)is isomorphic to U(16).

EXAMPLE 12 Let G 5 U(32) and K 5 {1, 15}. Then |G/K| 5 8, andwe ask, which of the three Abelian groups of order 8 is G/K? Since(3K)4 5 81K 5 17K 2 K, |3K| 5 8. Thus, G/K L Z8.

It is crucial to understand that when we factor out by a normal sub-group H, what we are essentially doing is defining every element in Hto be the identity. Thus, in Example 9, we are making R180_ 5 _ theidentity. Likewise, R270_ 5 R90R180_ 5 R90_. Similarly, in Example 7,we are declaring any multiple of 4 to be 0 in the factor group Z/4Z. Thisis why 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z, and so on. In Example 11, wehave 3H 5 19H, since 19 5 3 ? 17 in U(32) and going to the factorgroup makes 17 the identity. Algebraists often refer to the process ofcreating the factor group G/H as “killing” H.

Applications of Factor GroupsWhy are factor groups important? Well, when G is finite and H 2 {e},G/H is smaller than G, and its structure is usually less complicated thanthat of G. At the same time, G/H simulates G in many ways. In fact, wemay think of a factor group of G as a less complicated approximationof G (similar to using the rational number 3.14 for the irrationalnumber p). What makes factor groups important is that one can oftendeduce properties of G by examining the less complicated group G/Hinstead. We illustrate this by giving another proof that A4 has no sub-group of order 6.

EXAMPLE 13 A4

Has No Subgroup of Order 6

The group A4 of even permutations on the set {1, 2, 3, 4} has no sub-group H of order 6. To see this, suppose that A4 does have a subgroup H

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186 Groups

of order 6. By Exercise 7 in this chapter, we know that H v A4. Thus,the factor group A4/H exists and has order 2. Since the order of anelement divides the order of the group, we have for all a [ A4 that a2

H 5 (aH)2 5 H. Thus, a2 [ H for all a in A4. Referring to the maindiagonal of the group table for A4 given in Table 5.1 on page 107, how-ever, we observe that A4 has nine different elements of the form a2, allof which must belong to H, a subgroup of order 6. This is clearlyimpossible, so a subgroup of order 6 cannot exist in A4.

†

The next three theorems illustrate how knowledge of a factor groupof G reveals information about G itself.

Theorem 9.3 The G/Z Theorem

PROOF Let gZ(G) be a generator of the factor group G/Z(G), and leta, b [ G. Then there exist integers i and j such that

aZ(G) 5 (gZ(G))i 5 giZ(G)

and

bZ(G) 5 (gZ(G)) j 5 gjZ(G).

Thus, a 5 gix for some x in Z(G) and b 5 gjy for some y in Z(G). It fol-lows then that

ab 5 (gix)(gjy) 5 gi(xg j)y 5 gi(gjx)y

5 (gig j)(xy) 5 (gjgi)(yx) 5 (gjy)(gix) 5 ba.

A few remarks about Theorem 9.3 are in order. First, our proof showsthat a better result is possible: If G/H is cyclic, where H is a subgroup ofZ(G), then G is Abelian. Second, in practice, it is the contrapositive ofthe theorem that is most often used—that is, if G is non-Abelian, thenG/Z(G) is not cyclic. For example, it follows immediately from thisstatement and Lagrange’s Theorem that a non-Abelian group of orderpq, where p and q are primes, must have a trivial center. Third, if G/Z(G)is cyclic, it must be trivial.

Let G be a group and let Z(G) be the center of G. If G/Z(G) is cyclic,then G is Abelian.

†How often have I said to you that when you have eliminated the impossible, whateverremains, however improbable, must be the truth. Sherlock Holmes, The Sign of Four

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9 | Normal Subgroups and Factor Groups 187

Theorem 9.4 G/Z(G) Inn(G)

PROOF Consider the correspondence from G/Z(G) to Inn(G) given byT : gZ(G) → fg [where, recall, fg(x) 5 gxg21 for all x in G]. First, weshow that T is well defined. To do this, we assume that

and verify that . (This shows that the imageof a coset of depends only on the coset itself and not on the ele-ment representing the coset.) From , we have that

belongs to Z(G). Then, for all x in G, h21gx 5 xh21g. Thus,gxg21 5 hxh21 for all x in G, and, therefore, fg 5 fh. Reversing thisargument shows that T is one-to-one, as well. Clearly, T is onto.

That T is operation-preserving follows directly from the fact thatfgfh 5 fgh for all g and h in G.

As an application of Theorems 9.3 and 9.4, we may easily determineInn(D6) without looking at Inn(D6)!

EXAMPLE 14 We know from Example 11 in Chapter 3 that|Z(D6)| 5 2. Thus, |D6 /Z (D6)| 5 6. So, by our classification of groupsof order 6 (Theorem 7.2), we know that Inn(D6) is isomorphic to D3or Z6. Now, if Inn(D6) were cyclic, then, by Theorem 9.4, D6/Z(D6)would be also. But then, Theorem 9.3 would tell us that D6 is Abelian.So, Inn(D6) is isomorphic to D3.

The next theorem demonstrates one of the most powerful proof tech-niques available in the theory of finite groups—the combined use offactor groups and induction.

Theorem 9.5 Cauchy’s Theorem for Abelian Groups

PROOF Clearly, this statement is true for the case in which G hasorder 2. We prove the theorem by using the Second Principle of Math-ematical Induction on |G|. That is, we assume that the statement is truefor all Abelian groups with fewer elements than G and use this assump-tion to show that the statement is true for G as well. Certainly, G haselements of prime order, for if |x| 5 m and m 5 qn, where q is prime,then |xn| 5 q. So let x be an element of G of some prime order q, say. If

Let G be a finite Abelian group and let p be a prime that divides theorder of G. Then G has an element of order p.

h21ggZ(G) 5 hZ(G)

Z(G)fg 5 fhgZ(G) 5 hZ(G)

For any group G, G/Z(G) is isomorphic to Inn(G).

<

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188 Groups

q 5 p, we are finished; so assume that q 2 p. Since every subgroup ofan Abelian group is normal, we may construct the factor group 5G/�x�. Then is Abelian and p divides | |, since | | 5 |G|/q. Byinduction, then, has an element—call it y�x�—of order p. The con-clusion now follows from Exercise 65.

Internal Direct ProductsAs we have seen, the external direct product provides a way of puttinggroups together into a larger group. It would be quite useful to be ableto reverse this process—that is, to be able to start with a large groupand break it down into a product of smaller groups. It is occasionallypossible to do this. To this end, suppose that H and K are subgroups ofsome group G. We define the set HK 5 {hk | h [ H, k [ K}.

EXAMPLE 15 In U(24) 5 {1, 5, 7, 11, 13, 17, 19, 23}, let H 5{1, 17} and K 5 {1, 13}. Then, HK 5 {1, 13, 17, 5}, since 5 5 17 ? 13mod 24.

EXAMPLE 16 In S3, let H 5 {(1), (12)} and K 5 {(1), (13)}. Then,HK 5 {(1), (13), (12), (12)(13)} 5 {(1), (13), (12), (132)}.

The student should be careful not to assume that the set HK is a sub-group of G; in Example 15 it is, but in Example 16 it is not.

Definition Internal Direct Product of H and K

We say that G is the internal direct product of H and K and write G 5 H 3 K if H and K are normal subgroups of G and

G 5 HK and H > K 5 {e}.

The wording of the phrase “internal direct product” is easy to justify.We want to call G the internal direct product of H and K if H and K aresubgroups of G, and if G is naturally isomorphic to the external directproduct of H and K. One forms the internal direct product by startingwith a group G and then proceeding to find two subgroups H and Kwithin G such that G is isomorphic to the external direct product of Hand K. (The definition ensures that this is the case—see Theorem 9.6.)On the other hand, one forms an external direct product by starting withany two groups H and K, related or not, and proceeding to produce thelarger group H % K. The difference between the two products is that theinternal direct product can be formed within G itself, using subgroupsof G and the operation of G, whereas the external direct product can beformed with totally unrelated groups by creating a new set and a newoperation. (See Figures 9.1 and 9.2.)

GGGG

G

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9 | Normal Subgroups and Factor Groups 189

Figure 9.1 For the internal direct product, H and K must be subgroups of the same group.

Figure 9.2 For the externaldirect product, H and K canbe any groups.

Perhaps the following analogy with integers will be useful in clari-fying the distinction between the two products of groups discussed inthe preceding paragraph. Just as we may take any (finite) collection ofintegers and form their product, we may also take any collection ofgroups and form their external direct product. Conversely, just as wemay start with a particular integer and express it as a product of cer-tain of its divisors, we may be able to start with a particular group andfactor it as an internal direct product of certain of its subgroups.

EXAMPLE 17 In D6, the dihedral group of order 12, let F denotesome reflection and let Rk denote a rotation of k degrees. Then,

D6 5 {R0, R120, R240, F, R120F, R240F} 3 {R0, R180}.

Students should be cautioned about the necessity of having all con-ditions of the definition of internal direct product satisfied to ensurethat HK L H % K. For example, if we take

G 5 S3, H 5 �(123)�, and K 5 �(12)�,

then G 5 HK, and H > K 5 {(1)}. But G is not isomorphic to H % K,since, by Theorem 8.2, H % K is cyclic, whereas S3 is not. Note that Kis not normal.

A group G can also be the internal direct product of a collection ofsubgroups.

H K

G

eH K

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190 Groups

Definition Internal Direct Product H13 H

23 ? ? ? 3 H

n

Let H1, H2, . . . , Hn be a finite collection of normal subgroups of G. Wesay that G is the internal direct product of H1, H2, . . . , Hn and writeG 5 H1 3 H2 3 ? ? ? 3 Hn, if

1. G 5 H1H2 ? ? ? Hn 5 {h1h2 ? ? ? hn | hi [ Hi}2. (H1H2 ? ? ? Hi) > Hi11 5 {e} for i 5 1, 2, . . . , n 2 1.

This definition is somewhat more complicated than the one given fortwo subgroups. The student may wonder about the motivation for it—that is, why should we want the subgroups to be normal and why is itdesirable for each subgroup to be disjoint from the product of all previ-ous ones? The reason is quite simple. We want the internal direct prod-uct to be isomorphic to the external direct product. As the next theoremshows, the conditions in the definition of internal direct product werechosen to ensure that the two products are isomorphic.

Theorem 9.6 H1 3 H2 3 ? ? ? 3 Hn L H1 % H2 % ? ? ? % Hn

PROOF We first show that the normality of the H’s together with thesecond condition of the definition guarantees that h’s from differentHi’s commute. For if hi [ Hi and hj [ Hj with i 2 j, then

(hihjhi21)hj

21 [ Hjhj21 5 Hj

and

hi(hjhi21hj

21) [ hiHi 5 Hi.

Thus, hihj hi21hj

21 [ Hi > Hj 5 {e} (see Exercise 3), and, therefore,hihj 5 hjhi. We next claim that each member of G can be expresseduniquely in the form h1h2 ? ? ? hn, where hi [ Hi. That there is at least onesuch representation is the content of condition 1 of the definition. Toprove uniqueness, suppose that g 5 h1h2 ? ? ? hn and g 5 h19 h29 ? ? ? hn9,where hi and hi9 belong to Hi for i 5 1, . . . , n. Then, using the fact thatthe h’s from different Hi’s commute, we can solve the equation

h1h2 ? ? ? hn 5 h19 h29 ? ? ? hn9 (1)

for hn9 hn21 to obtain

hn9 hn21 5 (h91)

21h1(h29)21h2 ? ? ? (h9n21)

21hn21.

If a group G is the internal direct product of a finite number ofsubgroups H1, H2, . . . , Hn, then G is isomorphic to the externaldirect product of H1, H2, . . . , Hn.

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9 | Normal Subgroups and Factor Groups 191

But then

hn9 hn21 [ H1H2 ? ? ? Hn21 > Hn 5 {e},

so that hn9 hn21 5 e and, therefore, hn9 5 hn. At this point, we can cancel

hn and hn9 from opposite sides of the equal sign in Equation (1) and repeatthe preceding argument to obtain hn21 5 h 9n21. Continuing in this fash-ion, we eventually have hi 5 hi9 for i 5 1, . . . , n. With our claim estab-lished, we may now define a function f from G to H1 % H2 % ? ? ? % Hnby f(h1h2 ? ? ? hn) 5 (h1, h2, . . . , hn). We leave to the reader the easy ver-ification that f is an isomorphism.

The next theorem provides an important application of Theorem 9.6.

Theorem 9.7 Classification of Groups of Order p2

Every group of order p2, where p is a prime, is isomorphic to orZp % Zp.

Zp2

PROOF Let G be a group of order , where p is a prime. If G has anelement of order , then G is isomorphic to . So, by Corollary 2 ofLagrange’s Theorem, we may assume that every nonidentity element ofG has order p. First we show that for any element a, the subgroup isnormal in G. If this is not the case then there is an element b in G suchthat is not in . Then and are distinct subgroups oforder p. Since is a subgroup of both and ,we have that . From this it follows that the distinctleft cosets of are , , , . . . ,

. Since must lie in one of these cosets, we may writein the form for some i and j. Cancel-

ing the terms, we obtain and therefore .This contradiction verifies our assertion that every subgroup of the form

is normal in G. To complete the proof, let x be any nonidentity ele-ment in G and y be any element of G not in . Then, by comparing or-ders and using Theorem 9.6, we see that .

As an immediate corollary of Theorem 9.7, we have the followingimportant fact.

Corollary

If G is a group of order p2, where p is a prime, then G is Abelian.

G 5 �x� 3 �y� < Zp % Zp

�x��a�

b 5 a2i2 j [ �a�e 5 aibajb21aibajb21b21 5 ai(bab21) j 5b21

b21ap21�bab21�a2�bab21�a�bab21��bab21��bab21�

�a� x �bab21� 5 5e6 �bab21��a��a� x �bab21��bab21��a��a�bab21

�a�

Zp2p2p2

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192 Groups

We mention in passing that if G 5 H1 % H2 % ? ? ? % Hn, then G canbe expressed as the internal direct product of subgroups isomorphic toH1, H2, . . . , Hn. For example, if G 5 H1 % H2, then G 5 3 ,where 5 H1 % {e} and 5 {e} % H2.

The topic of direct products is one in which notation and terminol-ogy vary widely. Many authors use H 3 K to denote both the internaldirect product and the external direct product of H and K, making nonotational distinction between the two products. A few authors defineonly the external direct product. Many people reserve the notation H % K for the situation where H and K are Abelian groups under addi-tion and call it the direct sum of H and K. In fact, we will adopt this ter-minology in the section on rings (Part 3), since rings are alwaysAbelian groups under addition.

The U-groups provide a convenient way to illustrate the precedingideas and to clarify the distinction between internal and external directproducts. It follows directly from Theorem 8.3 and its corollary andTheorem 9.6 that if m 5 n1n2 ? ? ? nk, where gcd(ni, nj) 5 1 for i 2 j, then

U(m) 5 Um/n1(m) 3 Um/n2

(m) 3 ? ? ? 3 Um/nk(m)

L U(n1) % U(n2) % ? ? ? % U(nk).

Let us return to the examples given following Theorem 8.3.

U(105) 5 U(15 ? 7) 5 U15(105) 3 U7(105)5 {1, 16, 31, 46, 61, 76} 3 {1, 8, 22, 29, 43, 64, 71, 92}L U(7) % U(15),

U(105) 5 U(5 ? 21) 5 U5(105) 3 U21(105)5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101}

3 {1, 22, 43, 64} L U(21) % U(5),

U(105) 5 U(3 ? 5 ? 7) 5 U35(105) 3 U21(105) 3 U15(105)5 {1, 71} 3 {1, 22, 43, 64} 3 {1, 16, 31, 46, 61, 76}L U(3) % U(5) % U(7).

H2H1

H2H1

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9 | Normal Subgroups and Factor Groups 193

Exercises

Understanding is a kind of ecstasy.CARL SAGAN

1. Let H 5 {(1), (12)}. Is H normal in S3?2. Prove that An is normal in Sn.3. Show that if G is the internal direct product of H1, H2, . . . , Hn and

i 2 j with 1 # i # n, 1 # j # n, then Hi > Hj 5 {e}. (This exerciseis referred to in this chapter.)

4. Let . Is H a normal sub-

group of GL(2, R)?5. Let G 5 GL(2, R) and let K be a subgroup of R*. Prove that H 5

{A [ G| det A [ K} is a normal subgroup of G.6. Viewing �3� and �12� as subgroups of Z, prove that �3�/�12� is iso-

morphic to Z4. Similarly, prove that �8�/�48� is isomorphic to Z6.Generalize to arbitrary integers k and n.

7. Prove that if H has index 2 in G, then H is normal in G. (This exer-cise is referred to in Chapters 24 and 25 and this chapter.)

8. Let H 5 {(1), (12)(34)} in A4.a. Show that H is not normal in A4.b. Referring to the multiplication table for A4 in Table 5.1 on page

107, show that, although a6H 5 a7H and a9H 5 a11H, it is nottrue that a6a9H 5 a7a11H. Explain why this proves that the leftcosets of H do not form a group under coset multiplication.

9. Let G 5 Z4 % U(4), H 5 �(2, 3)�, and K 5 �(2, 1)�. Show that G/His not isomorphic to G/K. (This shows that H L K does not implythat G/H L G/K.)

10. Prove that a factor group of a cyclic group is cyclic.11. Let H be a normal subgroup of G. If H and G/H are Abelian, must

G be Abelian?12. Prove that a factor group of an Abelian group is Abelian.13. If H is a subgroup of G and a, b [ G, prove that (ab)H 5 a(bH).14. What is the order of the element 14 1 �8� in the factor group

Z24/�8�?15. What is the order of the element 4U5(105) in the factor group

U(105)/U5(105)?16. Recall that Z(D6) 5 {R0, R180}. What is the order of the element

R60Z(D6) in the factor group D6/Z(D6)?

H 5 e ca b

0 dd ` a, b, d P R, ad 2 0 f

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194 Groups

17. Let G 5 Z/�20� and H 5 �4�/�20�. List the elements of H and G/H.18. What is the order of the factor group Z60/�15�?19. What is the order of the factor group (Z10 % U(10))/�(2, 9)�?20. Construct the Cayley table for U(20)/U5(20).21. Prove that an Abelian group of order 33 is cyclic.22. Determine the order of (Z % Z)/�(2, 2)�. Is the group cyclic?23. Determine the order of (Z % Z)/�(4, 2)�. Is the group cyclic?24. The group (Z4 % Z12)/�(2, 2)� is isomorphic to one of Z8, Z4 % Z2, or

Z2 % Z2 % Z2. Determine which one by elimination.25. Let G 5 U(32) and H 5 {1, 31}. The group G/H is isomorphic to

one of Z8, Z4 % Z2, or Z2 % Z2 % Z2. Determine which one byelimination.

26. Let G be the group of quarternions given by the table in Exercise 4of the Supplementary Exercises for Chapters 1–4 on page 91, andlet H be the subgroup {e, a2}. Is G/H isomorphic to Z4 or Z2 % Z2?

27. Let G 5 U(16), H 5 {1, 15}, and K 5 {1, 9}. Are H and K iso-morphic? Are G/H and G/K isomorphic?

28. Let G 5 Z4 % Z4, H 5 {(0, 0), (2, 0), (0, 2), (2, 2)}, and K 5 �(1, 2)�.Is G/H isomorphic to Z4 or Z2 % Z2? Is G/K isomorphic to Z4 or Z2 % Z2?

29. Prove that has no subgroup of order 18.30. Express U(165) as an internal direct product of proper subgroups

in four different ways.31. Let R* denote the group of all nonzero real numbers under multi-

plication. Let R1 denote the group of positive real numbers undermultiplication. Prove that R* is the internal direct product of R1

and the subgroup {1, 21}.32. Prove that D4 cannot be expressed as an internal direct product of

two proper subgroups.33. Let H and K be subgroups of a group G. If G 5 HK and g 5 hk,

where h [ H and k [ K, is there any relationship among |g|, |h|,and |k|? What if G 5 H 3 K?

34. In Z, let H 5 �5� and K 5 �7�. Prove that Z 5 HK. Does Z 5 H 3 K?35. Let G 5 {3a6b10c | a, b, c [ Z} under multiplication and H 5

{3a6b12c | a, b, c [ Z} under multiplication. Prove that G 5 �3� 3�6� 3 �10�, whereas H 2 �3� 3 �6� 3 �12�.

36. Determine all subgroups of R* (nonzero reals under multiplica-tion) of index 2.

A4 % Z3

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9 | Normal Subgroups and Factor Groups 195

37. Let G be a finite group and let H be a normal subgroup of G. Provethat the order of the element gH in G/H must divide the order of g in G.

38. Let H be a normal subgroup of G and let a belong to G. If the ele-ment aH has order 3 in the group G/H and |H| 510, what are thepossibilities for the order of a?

39. If H is a normal subgroup of a group G, prove that C(H), the cen-tralizer of H in G, is a normal subgroup of G.

40. An element is called a square if it can be expressed in the form b2

for some b. Suppose that G is an Abelian group and H is a sub-group of G. If every element of H is a square and every element ofG/H is a square, prove that every element of G is a square. Doesyour proof remain valid when “square” is replaced by “nth power,”where n is any integer?

41. Show, by example, that in a factor group G/H it can happen thataH 5 bH but |a| 2 |b|. (Do not use a 5 e or b 5 e.)

42. Observe from the table for A4 given in Table 5.1 on page 107 thatthe subgroup given in Example 6 of this chapter is the only sub-group of A4 of order 4. Why does this imply that this subgroupmust be normal in A4? Generalize this to arbitrary finite groups.

43. Let p be a prime. Show that if H is a subgroup of a group of order2p that is not normal, then H has order 2.

44. Show that is isomorphic to .45. Suppose that N is a normal subgroup of a finite group G and H is a

subgroup of G. If is prime, prove that H is contained in N orthat .

46. If G is a group and , prove that .47. Suppose that G is a non-Abelian group of order p3, where p is a

prime, and Z(G) 2 {e}. Prove that |Z(G)| 5 p.48. If |G| 5 pq, where p and q are primes that are not necessarily dis-

tinct, prove that |Z(G)| 5 1 or pq.49. Let N be a normal subgroup of G and let H be a subgroup of G. If

N is a subgroup of H, prove that H/N is a normal subgroup of G/Nif and only if H is a normal subgroup of G.

50. Let G be an Abelian group and let H be the subgroup consisting ofall elements of G that have finite order (See Exercise 18 in theSupplementary Exercises for Chapters 1–4). Prove that every non-identity element in G/H has infinite order.

51. Determine all subgroups of R* that have finite index.

G>Z(G) < Z2 % Z2|G : Z(G)| 5 4NH 5 G

ZG>N Z

Inn(D13)D13

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196 Groups

52. Let G 5 {61, 6i, 6j, 6k}, where i2 5 j2 5 k2 5 21, 2i 5 (21)i,12 5 (21)2 5 1, ij 5 2ji 5 k, jk 5 2kj 5 i, and ki 5 2ik 5 j.a. Construct the Cayley table for G.b. Show that H 5 {1, 21} v G.c. Construct the Cayley table for . Is isomorphic to or

?(The rules involving i, j, and k can be remembered by using the cir-cle below.

Going clockwise, the product of two consecutive elements is thethird one. The same is true for going counterclockwise, except thatwe obtain the negative of the third element.) This is the group ofquaternions that was given in another form in Exercise 4 in theSupplementary Exercises for Chapters 1–4. It was invented byWilliam Hamilton in 1843. The quaternions are used to describerotations in three-dimensional space, and they are used in physics.The quaternions can be used to extend the complex numbers in anatural way.

53. In D4, let K 5 {R0, D} and let L 5 {R0, D, D9, R180}. Show that K v

L v D4, but that K is not normal in D4. (Normality is not transitive.Compare Exercise 4, Supplementary Exercises for Chapters 5–8.)

54. Show that the intersection of two normal subgroups of G is a nor-mal subgroup of G. Generalize.

55. Let N be a normal subgroup of G and let H be any subgroup of G.Prove that NH is a subgroup of G. Give an example to show thatNH need not be a subgroup of G if neither N nor H is normal. (Thisexercise is referred to in Chapter 24.)

56. If N and M are normal subgroups of G, prove that NM is also a nor-mal subgroup of G.

57. Let N be a normal subgroup of a group G. If N is cyclic, prove thatevery subgroup of N is also normal in G. (This exercise is referredto in Chapter 24.)

58. Without looking at inner automorphisms of Dn, determine the num-ber of such automorphisms.

k j

i

Z2 % Z2

Z4G>HG>H

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9 | Normal Subgroups and Factor Groups 197

59. Let H be a normal subgroup of a finite group G and let x [ G. Ifgcd(|x|, |G/H|) 5 1, show that x [ H. (This exercise is referred toin Chapter 25.)

60. Let G be a group and let G9 be the subgroup of G generated by theset S 5 {x21y21xy | x, y [ G}. (See Exercise 3, SupplementaryExercises for Chapters 5–8, for a more complete description of G9.)a. Prove that G9 is normal in G.b. Prove that G/G9 is Abelian.c. If G/N is Abelian, prove that G9 # N.d. Prove that if H is a subgroup of G and G9 # H, then H is normal

in G.61. If N is a normal subgroup of G and |G/N| 5 m, show that xm [ N

for all x in G.62. Suppose that a group G has a subgroup of order n. Prove that the

intersection of all subgroups of G of order n is a normal subgroupof G.

63. If G is non-Abelian, show that Aut(G) is not cyclic.64. Let |G| 5 pnm, where p is prime and gcd( p, m) 5 1. Suppose that

H is a normal subgroup of G of order pn. If K is a subgroup of G oforder pk, show that K # H.

65. Suppose that H is a normal subgroup of a finite group G. If G/Hhas an element of order n, show that G has an element of order n.Show, by example, that the assumption that G is finite is necessary.(This exercise is referred to in this chapter.)

66. Recall that a subgroup N of a group G is called characteristic iff(N) 5 N for all automorphisms f of G. If N is a characteristicsubgroup of G, show that N is a normal subgroup of G.

67. In D4, let _ 5 {R0, H}. Form an operation table for the cosets _,D_, V_, and D9_. Is the result a group table? Does your answercontradict Theorem 9.2?

68. Show that S4 has a unique subgroup of order 12.69. If |G| 5 30 and |Z(G)| 5 5, what is the structure of G/Z(G)?70. If H is a normal subgroup of G and |H| 5 2, prove that H is con-

tained in the center of G.71. Prove that A5 cannot have a normal subgroup of order 2.72. Let G be a finite group and let H be an odd-order subgroup of G of

index 2. Show that the product of all the elements of G (taken inany order) cannot belong to H.

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198 Groups

73. Let G be a group and p a prime. Suppose that is asubgroup of G. Show that H is normal and that every nonidentityelement of has order p.

74. Suppose that H is a normal subgroup of G. If |H| 5 4 and gH hasorder 3 in G/H, find a subgroup of order 12 in G.

75. Let G be a group and H a subgroup of G of index 2. Show that Hcontains every element of G of odd order.

Suggested Readings

Michael Brennan and Des MacHale, “Variations on a Theme: A4 Defi-nitely Has No Subgroup of Order Six!,” Mathematics Magazine, 73(2000): 36–40.

The authors offer 11 proofs that A4 has no subgroup of order 6. Theseproofs provide a review of many of the ideas covered thus far in thistext.

J. A. Gallian, R. S. Johnson, and S. Peng. “On the Quotient Structure ofZn,” Pi Mu Epsilon Journal, 9 (1993): 524–526.

The authors determine the structure of the group (Z % Z)/�(a, b)� andrelated groups. This article can be downloaded at http://www.d.umn.edu/~jgallian/quotient_structures.pdf

Tony Rothman, “Genius and Biographers: The Fictionalization of ÉvaristeGalois,” The American Mathematical Monthly 89 (1982): 84–106.

The author convincingly argues that three of the most widely readaccounts of Galois’ life are highly fictitious.

Paul F. Zweifel, “Generalized Diatonic and Pentatonic Scales: A Group-theoretic Approach,” Perspectives of New Music, 34 (1996): 140–161.

The author discusses how group-theoretic notions such as subgroups,cosets, factor groups, and isomorphisms of Z12 and Z20 relate to musicalscales, tuning, temperament, and structure.

G>HH 5 5gp Zg [ G6

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ÉVARISTE GALOIS (pronounced gal-WAH)was born on October 25, 1811, near Paris.Although he had mastered the works ofLegendre and Lagrange at age 15, Galoistwice failed his entrance examination tol’Ecole Polytechnique. He did not knowsome basic mathematics, and he did mathe-matics almost entirely in his head, to theannoyance of the examiner.

At 18, Galois wrote his important researchon the theory of equations and submitted it tothe French Academy of Sciences for publica-tion. The paper was given to Cauchy for ref-ereeing. Cauchy, impressed by the paper,agreed to present it to the academy, but henever did. At the age of 19, Galois entered a

199

Galois at seventeen was making discover-ies of epochal significance in the theory ofequations, discoveries whose conse-quences are not yet exhausted after morethan a century.

E. T. BELL, Men of Mathematics

This French stamp was issued as part ofthe 1984 “Celebrity Series” in support ofthe Red Cross Fund.

Evariste Galois

paper of the highest quality in the competi-tion for the Grand Prize in Mathematics,given by the French Academy of Sciences.The paper was given to Fourier, who diedshortly thereafter. Galois’s paper was neverseen again.

Galois spent most of the last year and ahalf of his life in prison for revolutionary po-litical offenses. While in prison, he attemptedsuicide and prophesied that he would die in aduel. On May 30, 1832, Galois was shot in aduel and died the next day at the age of 20.

Among the many concepts introduced byGalois are normal subgroups, isomorphisms,simple groups, finite fields, and Galois theory.His work provided a method for disposingof several famous constructability problems,such as trisecting an arbitrary angle and dou-bling a cube. Galois’s entire collected worksfill only 60 pages.

To find more information about Galois,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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200200

Definition and ExamplesIn this chapter, we consider one of the most fundamental ideas ofalgebra—hom*omorphisms. The term hom*omorphism comes from theGreek words hom*o, “like,” and morphe, “form.” We will see that a ho-momorphism is a natural generalization of an isomorphism and thatthere is an intimate connection between factor groups of a group andhom*omorphisms of a group. The concept of group hom*omorphismswas introduced by Camille Jordan in 1870, in his influential bookTraité des Substitutions.

Definition Group hom*omorphism

A hom*omorphism f from a group G to a group is a mapping from G into that preserves the group operation; that is, f(ab) 5f(a)f(b) for all a, b in G.

Before giving examples and stating numerous properties ofhom*omorphisms, it is convenient to introduce an important subgroupthat is intimately related to the image of a hom*omorphism. (Seeproperty 4 of Theorem 10.1.)

Definition Kernel of a hom*omorphism

The kernel of a hom*omorphism f from a group G to a group withidentity e is the set {x [ G | f(x) 5 e}. The kernel of f is denoted byKer f.

GG

10Group hom*omorphisms

All modern theories of nuclear and electromagnetic interactions are basedon group theory.

ANDREW WATSON, New Scientist

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10 | Group hom*omorphisms 201

EXAMPLE 1 Any isomorphism is a hom*omorphism that is also ontoand one-to-one. The kernel of an isomorphism is the trivial subgroup.

EXAMPLE 2 Let R* be the group of nonzero real numbers undermultiplication. Then the determinant mapping A → det A is ahom*omorphism from GL(2, R) to R*. The kernel of the determinantmapping is SL(2, R).

EXAMPLE 3 The mapping f from R* to R*, defined by f(x) 5 |x|,is a hom*omorphism with Ker f 5 {1, 21}.

EXAMPLE 4 Let R[x] denote the group of all polynomials with realcoefficients under addition. For any f in R[x], let f 9 denote the deriva-tive of f. Then the mapping f S f 9 is a hom*omorphism from R[x] to it-self. The kernel of the derivative mapping is the set of all constant poly-nomials.

EXAMPLE 5 The mapping f from Z to Zn, defined by f(m) 5 mmod n, is a hom*omorphism (see Exercise 11 in Chapter 0). The kernelof this mapping is �n�.

EXAMPLE 6 The mapping f(x) 5 x2 from R*, the nonzero realnumbers under multiplication, to itself is a hom*omorphism, since f(ab) 5 (ab)2 5 a2b2 5 f(a)f(b) for all a and b in R*. (See Exercise 5.) The kernel is {1, –1}.

EXAMPLE 7 The mapping f(x) 5 x2 from R, the real numbersunder addition, to itself is not a hom*omorphism, since f(a 1 b) 5 (a 1 b)2 5 a2 1 2ab 1 b2, whereas f(a) 1 f(b) 5 a2 1 b2.

When defining a hom*omorphism from a group in which there areseveral ways to represent the elements, caution must be exercised to en-sure that the correspondence is a function. (The term well-defined isoften used in this context.) For example, since 3(x 1 y) 5 3x 1 3y inZ6, one might believe that the correspondence x 1 �3� S 3x from Z/�3� toZ6 is a hom*omorphism. But it is not a function, since 0 1 �3� 5 3 1�3� in Z/�3� but 3 ? 0 2 3 ? 3 in Z6.

For students who have had linear algebra, we remark that everylinear transformation is a group hom*omorphism and the nullspace isthe same as the kernel. An invertible linear transformation is a groupisomorphism.

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202 Groups

Properties of hom*omorphismsTheorem 10.1 Properties of Elements Under hom*omorphisms

Let f be a hom*omorphism from a group G to a group and let g bean element of G. Then

1. f carries the identity of G to the identity of .2. f(gn) 5 (f(g))n for all n in Z.3. If |g| is finite, then |f(g)| divides |g|.4. Ker f is a subgroup of G.5. f(a) 5 f(b) if and only if aKer f 5 bKer f.6. If f(g) 5 g9, then f21(g9) 5 {x [ G | f(x) 5 g9} 5 gKer f.

G

G

PROOF The proofs of properties 1 and 2 are identical to the proofs ofproperties 1 and 2 of isomorphisms in Theorem 6.2. To prove property 3,notice that properties 1 and 2 together with gn 5 e imply that e 5f(e) 5 f(gn) 5 (f(g))n. So, by Corollary 2 to Theorem 4.1, we have|f(g)| divides n.

By property 1 we know that Ker f is not empty. So, to prove prop-erty 4, we assume that a, b [ Ker f and show that ab21 [ Ker f.Since f(a) 5 e and f(b) 5 e, we have f(ab21) 5 f(a)f(b21) 5f(a)(f(b))21 5 ee21 5 e. So, ab21 [ Ker f.

To prove property 5, first assume that f(a) 5 f(b). Then e 5 (f(b))21f(a) 5 f(b21)f(a) 5 f(b21a), so that b21a[ Ker f. It now follows from property 5 of the lemma in Chapter 7 that bKer f 5 aKer f. Reversing this argument completes the proof.

To prove property 6, we must show that f21(g9) # gKer f and thatgKer f # f21(g9). For the first inclusion, let x [ f21(g9), so that f(x) 5 g9. Then f(g) 5 f(x) and by property 5 we have gKer f 5xKer f and therefore x [ gKer f. This completes the proof thatf21(g9) # gKer f. To prove that gKer f # f21(g9), suppose that k [Ker f. Then f(gk) 5 f(g)f(k) 5 g9e 5 g9. Thus, by definition, gk [f21(g9).

Since hom*omorphisms preserve the group operation, it should not bea surprise that they preserve many group properties.

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10 | Group hom*omorphisms 203

Theorem 10.2 Properties of Subgroups Under hom*omorphisms

PROOF First note that the proofs of properties 1, 2, and 3 are identi-cal to the proofs of properties 4, 3, and 2, respectively, of Theorem6.3, since those proofs use only the fact that an isomorphism is anoperation-preserving mapping.

To prove property 4, let f(h) [ f(H) and f(g) [ f(G). Thenf(g)f(h)f(g)21 5 f(ghg21) [ f(H), since H is normal in G.

Property 5 follows directly from property 6 of Theorem 10.1 and thefact that all cosets of Ker f 5 f21(e) have the same number of elements.

To prove property 6, let fH denote the restriction of f to the elements of H. Then fH is a hom*omorphism from H onto f(H).Suppose |Ker fH| 5 t. Then, by property 5, fH is a t-to-1 mapping. So,|f(H)|t 5 |H|.

To prove property 7, we use the One-Step Subgroup Test. Clearly,e [ f21( ), so that f21( ) is not empty. Let k1, k2 [ f21( ). Then,by the definition of f21( ), we know that f(k1), f(k2) [ . Thus,f(k2)

21 [ as well and f(k1k221) 5 f(k1)f(k2)

21 [ . So, by definition

of f21( ), we have k1k221 [ f21( ).

To prove property 8, we use the normality test given in Theorem 9.1.Note that every element in xf21( )x21 has the form xkx21, where f(k) [

. Thus, since is normal in , f(xkx21) 5 f(x)f(k)(f(x))21 [ ,and, therefore, xkx21 [ f21( ).

Finally, property 9 follows directly from property 5.K

KGKKK

KK

KKKK

KKK

Let f be a hom*omorphism from a group G to a group and let H bea subgroup of G. Then

1. f(H) 5 {f(h) | h [ H} is a subgroup of .2. If H is cyclic, then f(H) is cyclic.3. f H is Abelian, then f(H) is Abelian.4. If H is normal in G, then f(H) is normal in f(G).5. If |Ker f| 5 n, then f is an n-to-1 mapping from G onto f(G).6. If |H| 5 n, then |f(H)| divides n.7. If is a subgroup of , then f21( ) 5 {k [ G | f(k) [ }

is a subgroup of G.8. If is a normal subgroup of , then f21( ) 5 {k [ G |

f(k) [ } is a normal subgroup of G.9. If f is onto and Ker f 5 {e}, then f is an isomorphism

from G to .G

KKGK

KKGK

G

G

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204 Groups

A few remarks about Theorems 10.1 and 10.2 are in order. Studentsshould remember the various properties of these theorems in words. Forexample, properties 2 and 3 of Theorem 10.2 say that the hom*omorphicimage of a cyclic group is cyclic and the hom*omorphic image of anAbelian group is Abelian. Property 4 of Theorem 10.2 says that the ho-momorphic image of a normal subgroup of G is normal in the image ofG. Property 5 of Theorem 10.2 says that if f is a hom*omorphism fromG to , then every element of that gets “hit” by f gets hit the samenumber of times as does the identity. The set f21(g9) defined in prop-erty 6 of Theorem 10.1 is called the inverse image of g9 (or the pullbackof g9). Note that the inverse image of an element is a coset of the kerneland that every element in that coset has the same image. Similarly, theset f21( ) defined in property 7 of Theorem 10.2 is called the inverseimage of (or the pullback of ).

Property 6 of Theorem 10.1 is reminiscent of something from linearalgebra and differential equations. Recall that if x is a particular solu-tion to a system of linear equations and S is the entire solution set of thecorresponding hom*ogeneous system of linear equations, then x 1 S isthe entire solution set of the nonhom*ogeneous system. In reality, thisstatement is just a special case of property 6. Properties 1 and 6 ofTheorem 10.1 and property 5 of Theorem 10.2 are pictorially repre-sented in Figure 10.1.

The special case of property 8 of Theorem 10.2, where 5 {e}, isof such importance that we single it out.

Corollary Kernels Are Normal

The next two examples illustrate several properties of Theorems 10.1and 10.2.

EXAMPLE 8 Consider the mapping f from C* to C* given by f(x) 5 x4. Since (xy)4 5 x4y4, f is a hom*omorphism. Clearly,Ker f 5 {x | x4 5 1} 5 {1, 21, i, 2i}. So, by property 5 of Theorem10.2, we know that f is a 4-to-1 mapping. Now let’s find all elementsthat map to, say, 2. Certainly, f( ) 5 2. Then, by property 6 ofTheorem 10.1, the set of all elements that map to 2 is Ker f 5{ , 2 , i, 2 i}.4"24"24"24"2

4"2

4"2

K

KKK

GG

Let f be a group hom*omorphism from G to . Then Ker f is a nor-mal subgroup of G.

G

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10 | Group hom*omorphisms 205

Finally, we verify a specific instance of property 3 of Theorem 10.1and of property 2 and property 6 of Theorem 10.2. Let H 5�cos 30° 1 i sin 30°�. It follows from DeMoivre’s Theorem (Example 7in Chapter 0) that |H| 5 12, f(H) 5 �cos 120° 1 i sin 120°�, and |f(H)| 5 3.

EXAMPLE 9 Define f:Z12 → Z12 by f(x) 5 3x. To verify that f is ahom*omorphism, we observe that in Z12, 3(a 1 b) 5 3a 1 3b (since thegroup operation is addition modulo 12). Direct calculations show thatKer f 5 {0, 4, 8}. Thus, we know from property 5 of Theorem 10.2 thatf is a 3-to-1 mapping. Since f(2) 5 6, we have by property 6 ofTheorem 10.1 that f21(6) 5 2 1 Ker f 5 {2, 6, 10}. Notice also that �2�is cyclic and f(�2�) 5 {0, 6} is cyclic. Moreover, |2| 5 6 and |f(2)| 5|6| 5 2, so |f(2)| divides |2| in agreement with property 3 of Theorem10.1. Letting 5 {0, 6}, we see that the subgroup f21( ) 5 {0, 2, 4, 6,8, 10}. This verifies property 7 of Theorem 10.2 in this particular case.

The next example illustrates how one can easily determine all hom*o-morphisms from a cyclic group to a cyclic group.

EXAMPLE 10 We determine all hom*omorphisms from Z12 to Z30.By property 2 of Theorem 10.1, such a hom*omorphism is completelyspecified by the image of 1. That is, if 1 maps to a, then x maps to xa.Lagrange’s Theorem and property 3 of Theorem 10.1 require that |a| di-vide both 12 and 30. So, |a| 5 1, 2, 3, or 6. Thus, a 5 0, 15, 10, 20,5, or 25. This gives us a list of candidates for the hom*omorphisms. Thateach of these six possibilities yields an operation-preserving, well-defined function can now be verified by direct calculations. [Note thatgcd(12, 30) 5 6. This is not a coincidence!]

KK

φ

φ

φ φ

(g) = g9

G

G

(G)

e

φ φKer 21(e)= φ φgKer 21(g9)=

e = g1, g2,..., gn g, gg2,..., ggn

Figure 10.1

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206 Groups

EXAMPLE 11 The mapping from Sn to Z2 that takes an even permu-tation to 0 and an odd permutation to 1 is a hom*omorphism. Figure 10.2illustrates the telescoping nature of the mapping.

(12)

(12)

(12)

(23)

(13)

(13)

(13)

(13)

(12)

(23)

(1)

(1)

(1)

(1)

(123)

(132)

(123)

(132)

(123)

(123)

(132)

(132)

(1)

(132)

(1)

(123)

(12)

(13)

(23)

(13)

(23)

(12)

(23)

(23)

(13)

(23)

(12)

(23)

(12)

(13)

(132)

(123)

(123)

(123)

(132)

(132)

(1)

(1)

1

11

1

O

O

O

O

Figure 10.2 hom*omorphism from S3 to Z2.

The First Isomorphism TheoremIn Chapter 9, we showed that for a group G and a normal subgroup H,we could arrange the Cayley table of G into boxes that represented thecosets of H in G, and that these boxes then became a Cayley table forG/H. The next theorem shows that for any hom*omorphism f of G andthe normal subgroup Ker f, the same process produces a Cayley tableisomorphic to the hom*omorphic image of G. Thus, hom*omorphisms,like factor groups, cause a systematic collapse of a group to a simplerbut closely related group. This can be likened to viewing a groupthrough the reverse end of a telescope—the general features of thegroup are present, but the apparent size is diminished. The important

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10 | Group hom*omorphisms 207

relationship between hom*omorphisms and factor groups given below isoften called the Fundamental Theorem of Group hom*omorphisms.

Theorem 10.3 First Isomorphism Theorem (Jordan, 1870)

PROOF Let us use c to denote the correspondence gKerf S f(g).That c is well defined (that is, the correspondence is independent ofthe particular coset representative chosen) and one-to-one followsdirectly from property 5 of Theorem 10.1. To show that c is operation-preserving, observe that c(xKer f yKer f) 5 c(xyKer f) 5 f(xy) 5f(x) f(y) 5 c(xKer f)c(yKer f).

The next corollary follows directly from Theorem 10.3, property 1 ofTheorem 10.2, and Lagrange’s Theorem.

Corollary

EXAMPLE 10 To illustrate Theorem 10.3 and its proof, consider thehom*omorphism f from D4 to itself given by

R0 R180 R90 R270 H V D D9

R0 H R180 V

Then Ker f 5 {R0, R180}, and the mapping c in Theorem 10.3 isR0Ker f S R0, R90Ker f S H, HKer f S R180, DKer f S V. It isstraight-forward to verify that the mapping c is an isomorphism.

Mathematicians often give a pictorial representation of Theorem10.3, as follows:

G (G)φ

φ

φ

γ ψ

G/Ker

Let f be a group hom*omorphism from G to . Then the mappingfrom G/Ker f to f(G), given by gKer f → f(g), is an isomorphism.In symbols, G/Ker f L f(G).

G

If f is a hom*omorphism from a finite group G to , then |f(G)|divides |G| and | |.G

G

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208 Groups

where g:G S G/Ker f is defined as g(g) 5 gKer f. The mapping gis called the natural mapping from G to G/Ker f. Our proof ofTheorem 10.3 shows that cg 5 f. In this case, one says that the pre-ceding diagram is commutative.

As a consequence of Theorem 10.3, we see that all hom*omorphic im-ages of G can be determined using G. We may simply consider the variousfactor groups of G. For example, we know that the hom*omorphic image ofan Abelian group is Abelian because the factor group of an Abelian groupis Abelian. We know that the number of hom*omorphic images of a cyclicgroup G of order n is the number of divisors of n, since there is exactly onesubgroup of G (and therefore one factor group of G) for each divisor of n.(Be careful: The number of hom*omorphisms of a cyclic group of order nneed not be the same as the number of divisors of n, since different hom*o-morphisms can have the same image.)

An appreciation for Theorem 10.3 can be gained by looking at a fewexamples.

EXAMPLE 13 Z/8N9 LL ZN

Consider the mapping from Z to Zn defined in Example 5. Clearly, itskernel is �n�. So, by Theorem 10.3, Z/�n� L Zn.

EXAMPLE 14 The Wrapping Function

Recall the wrapping function W from trigonometry. The real numberline is wrapped around a unit circle in the plane centered at (0, 0) withthe number 0 on the number line at the point (1, 0), the positive realsin the counterclockwise direction and the negative reals in theclockwise direction (see Figure 10.3). The function W assigns to eachreal number a the point a radians from (1, 0) on the circle. This map-ping is a hom*omorphism from the group R under addition onto thecircle group (the group of complex numbers of magnitude 1 undermultiplication). Indeed, it follows from elementary facts of trigonom-etry that W(x) 5 cos x 1 i sin x and W(x 1 y) 5 W(x)W(y). Since W isperiodic of period 2p, Ker W 5 �2p�. So, from the First IsomorphismTheorem, we see that R/�2p� is isomorphic to the circle group.

Figure 10.3

W(3)

W(2)

W(0)

W(1)

(0, 0)

(1, 0)

W(21)

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10 | Group hom*omorphisms 209

Our next example is a theorem that is used repeatedly in Chapters 24and 25.

EXAMPLE 15 The N/C Theorem

Let H be a subgroup of a group G. Recall that the normalizer of H inG is N(H) 5 {x [ G | xHx21 5 H} and the centralizer of H in G is C(H) 5 {x [ G | xhx21 5 h for all h in H}. Consider the mapping fromN(H) to Aut(H) given by g S fg, where fg is the inner automorphism ofH induced by g [that is, fg(h) 5 ghg21 for all h in H]. This mapping is ahom*omorphism with kernel C(H). So, by Theorem 10.3, N(H)/C(H) isisomorphic to a subgroup of Aut(H).

As an application of the N/C Theorem, we will show that everygroup of order 35 is cyclic.

EXAMPLE 16 Let G be a group of order 35. By Lagrange’sTheorem, every nonidentity element of G has order 5, 7, or 35. Ifsome element has order 35, G is cyclic. So we may assume that allnonidentity elements have order 5 or 7. However, not all suchelements can have order 5, since elements of order 5 come 4 at a time(if |x| 5 5, then |x2| 5 |x3| 5 |x4| 5 5) and 4 does not divide 34.Similarly, since 6 does not divide 34, not all nonidentity elements canhave order 7. So, G has elements of order 7 and order 5. Since G hasan element of order 7, it has a subgroup of order 7. Let us call it H. Infact, H is the only subgroup of G of order 7, for if K is another sub-group of G of order 7, we have by Exercise 7 of the SupplementaryExercises for Chapters 5–8 that |HK| 5 |H||K|/|H > K| 5 7 ? 7/1 5 49.But, of course, this is impossible in a group of order 35. Since for everya in G, aHa21 is also a subgroup of G of order 7 (see Exercise 1 of theSupplementary Exercises for Chapters 1–4), we must have aHa21 5 H.So, N(H) 5 G. Since H has prime order, it is cyclic and thereforeAbelian. In particular, C(H) contains H. So, 7 divides |C(H)| and|C(H)| divides 35. It follows, then, that C(H) 5 G or C(H) 5 H. IfC(H) 5 G, then we may obtain an element x of order 35 by letting x 5 hk, where h is a nonidentity element of H and k has order 5. On theother hand, if C(H) 5 H, then |C(H)| 5 7 and |N(H)/C(H)| 5 35/7 5 5.However, 5 does not divide |Aut(H)| 5 |Aut(Z7)| 5 6. This contradic-tion shows that G is cyclic.

The corollary of Theorem 10.2 says that the kernel of every hom*o-morphism of a group is a normal subgroup of the group. We concludethis chapter by verifying that the converse of this statement is also true.

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210 Groups

Theorem 10.4 Normal Subgroups Are Kernels

PROOF Define g:G S G/N by g(g) 5 gN. (This mapping is called thenatural hom*omorphism from G to G/N.) Then, g(xy) 5 (xy)N 5 xNyN 5g(x)g(y). Moreover, g [ Ker g if and only if gN 5 g(g) 5 N, which istrue if and only if g [ N (see property 2 of the lemma in Chapter 7).

Examples 13, 14, and 15 illustrate the utility of the First IsomorphismTheorem. But what about hom*omorphisms in general? Why would onecare to study a hom*omorphism of a group? The answer is that, just aswas the case with factor groups of a group, hom*omorphic images of agroup tell us some of the properties of the original group. One measureof the likeness of a group and its hom*omorphic image is the size of thekernel. If the kernel of the hom*omorphism of group G is the identity,then the image of G tells us everything (group theoretically) about G (thetwo being isomorphic). On the other hand, if the kernel of the hom*omor-phism is G itself, then the image tells us nothing about G. Between thesetwo extremes, some information about G is preserved and some is lost.The utility of a particular hom*omorphism lies in its ability to preservethe group properties we want, while losing some inessential ones. In thisway, we have replaced G by a group less complicated (and therefore eas-ier to study) than G; but, in the process, we have saved enough informa-tion to answer questions that we have about G itself. For example, if G isa group of order 60 and G has a hom*omorphic image of order 12 that iscyclic, then we know from properties 5, 7, and 8 of Theorem 10.2 that Ghas normal subgroups of orders 5, 10, 15, 20, 30, and 60. To illustratefurther, suppose we are asked to find an infinite group that is the union ofthree proper subgroups. Instead of attempting to do this directly, we firstmake the problem easier by finding a finite group that is the unionof three proper subgroups. Observing that Z2 % Z2 is the union of H1 5�1, 0�, H2 5 �0, 1�, and H3 5 �1, 1�, we have found our finite group. Nowall we need do is think of an infinite group that has Z2 % Z2 as a hom*o-morphic image and pull back H1, H2, and H3, and our original problem issolved. Clearly, the mapping from Z2 % Z2 % Z onto Z2 % Z2 given byf(a, b, c) 5 (a, b) is such a mapping, and therefore Z2 % Z2 % Z is theunion of f21(H1) 5 {(a, 0, c,) | a [ Z2, c [ Z}, f21(H2) 5 {(0, b, c) | b[ Z2, c [ Z}, and f21(H3) 5 {(a, a, c) | a [ Z2, c [ Z}.

Every normal subgroup of a group G is the kernel of a hom*omor-phism of G. In particular, a normal subgroup N is the kernel of the mapping g S gN from G to G/N.

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10 | Group hom*omorphisms 211

Although an isomorphism is a special case of a hom*omorphism, thetwo concepts have entirely different roles. Whereas isomorphismsallow us to look at a group in an alternative way, hom*omorphisms act asinvestigative tools. The following analogy between hom*omorphismsand photography may be instructive.† A photograph of a person cannottell us the person’s exact height, weight, or age. Nevertheless, we maybe able to decide from a photograph whether the person is tall or short,heavy or thin, old or young, male or female. In the same way, a hom*o-morphic image of a group gives us some information about the group.

In certain branches of group theory, and especially in physics andchemistry, one often wants to know all hom*omorphic images of a groupthat are matrix groups over the complex numbers (these are called grouprepresentations). Here, we may carry our analogy with photography onestep further by saying that this is like wanting photographs of a personfrom many different angles (front view, profile, head-to-toe view, close-up, etc.), as well as x-rays! Just as this composite information from thephotographs reveals much about the person, several hom*omorphic im-ages of a group reveal much about the group.

Exercises

The greater the difficulty, the more glory in surmounting it. Skillful pilotsgain their reputation from storms and tempests.

EPICURUS

1. Prove that the mapping given in Example 2 is a hom*omorphism.2. Prove that the mapping given in Example 3 is a hom*omorphism.3. Prove that the mapping given in Example 4 is a hom*omorphism.4. Prove that the mapping given in Example 11 is a hom*omorphism.5. Let R* be the group of nonzero real numbers under multiplication,

and let r be a positive integer. Show that the mapping that takes xto xr is a hom*omorphism from R* to R* and determine the kernel.Which values of r yield an isomorphism?

6. Let G be the group of all polynomials with real coefficients under ad-dition. For each f in G, let �f denote the antiderivative of f that passesthrough the point (0, 0). Show that the mapping f S �f from G to G isa hom*omorphism. What is the kernel of this mapping? Is this map-ping a hom*omorphism if �f denotes the antiderivative of f that passesthrough (0, 1)?

†All perception of truth is the detection of an analogy. Henry David Thoreau, Journal.

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212 Groups

7. If f is a hom*omorphism from G to H and s is a hom*omorphismfrom H to K, show that is a hom*omorphism from G to K. Howare Ker f and Ker related? If f and s are onto and G is finite,describe [Ker : Ker f] in terms of |H| and |K|.

8. Let G be a group of permutations. For each s in G, define

sgn(s) 5

Prove that sgn is a hom*omorphism from G to the multiplicativegroup {11, 21}. What is the kernel? Why does this hom*omor-phism allow you to conclude that is a normal subgroup of ofindex 2?

9. Prove that the mapping from G % H to G given by (g, h) S g is ahom*omorphism. What is the kernel? This mapping is called theprojection of G % H onto G.

10. Let G be a subgroup of some dihedral group. For each x in G, define

f(x) 5

Prove that f is a hom*omorphism from G to the multiplicativegroup . What is the kernel?

11. Prove that (Z % Z )/(�(a, 0)� 3 �(0, b)�) is isomorphic to Za % Zb.12. Suppose that k is a divisor of n. Prove that Zn/�k� L Zk.13. Prove that (A % B)/(A % {e}) L B.14. Explain why the correspondence x → 3x from Z12 to Z10 is not a ho-

momorphism.15. Suppose that f is a hom*omorphism from Z30 to Z30 and Ker f 5

{0, 10, 20}. If f(23) 5 9, determine all elements that map to 9.16. Prove that there is no hom*omorphism from Z8 % Z2 onto Z4 % Z4.17. Prove that there is no hom*omorphism from Z16 % Z2 onto Z4 % Z4.18. Can there be a hom*omorphism from Z4 % Z4 onto Z8? Can there be

a hom*omorphism from Z16 onto Z2 % Z2? Explain your answers.19. Suppose that there is a hom*omorphism f from Z17 to some group

and that f is not one-to-one. Determine f.20. How many hom*omorphisms are there from Z20 onto Z8? How many

are there to Z8?21. If f is a hom*omorphism from Z30 onto a group of order 5, deter-

mine the kernel of f.

511,216

e11 if x is a rotation,

21 if x is a reflection.

SnAn

e11 if s is an even permutation,

21 if s is an odd permutation.

sf

sfsf

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10 | Group hom*omorphisms 213

22. Suppose that f is a hom*omorphism from a finite group G onto and that has an element of order 8. Prove that G has an elementof order 8. Generalize.

23. Suppose that f is a hom*omorphism from Z36 to a group of order 24.a. Determine the possible hom*omorphic images.b. For each image in part a, determine the corresponding kernel of f.

24. Suppose that f: Z50 S Z15 is a group hom*omorphism with f(7) 5 6.a. Determine f(x).b. Determine the image of f.c. Determine the kernel of f.d. Determine f21(3). That is, determine the set of all elements

that map to 3.25. How many hom*omorphisms are there from Z20 onto Z10? How

many are there to Z10?26. Determine all hom*omorphisms from Z4 to Z2 % Z2.27. Determine all hom*omorphisms from Zn to itself.28. Suppose that f is a hom*omorphism from S4 onto Z2. Determine

Ker f. Determine all hom*omorphisms from S4 to Z2.29. Suppose that there is a hom*omorphism from a finite group G onto

Z10. Prove that G has normal subgroups of indexes 2 and 5.30. Suppose that f is a hom*omorphism from a group G onto Z6 % Z2

and that the kernel of f has order 5. Explain why G must have nor-mal subgroups of orders 5, 10, 15, 20, 30, and 60.

31. Suppose that f is a hom*omorphism from U(30) to U(30) and that Ker f 5 {1, 11}. If f(7) 5 7, find all elements of U(30) thatmap to 7.

32. Find a hom*omorphism f from U(30) to U(30) with kernel {1, 11}and f(7) 5 7.

33. Suppose that f is a hom*omorphism from U(40) to U(40) and that Ker f 5 {1, 9, 17, 33}. If f(11) 5 11, find all elements of U(40)that map to 11.

34. Find a hom*omorphism f from U(40) to U(40) with kernel {1, 9,17, 33} and f(11) 5 11.

35. Prove that the mapping f: Z % Z S Z given by (a, b) S a 2 b is ahom*omorphism. What is the kernel of f? Describe the set f21(3)(that is, all elements that map to 3).

36. Suppose that there is a hom*omorphism f from Z % Z to a group Gsuch that f((3, 2)) 5 a and f((2, 1)) 5 b. Determine f((4, 4)) interms of a and b. Assume that the operation of G is addition.

GG

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214 Groups

37. Prove that the mapping x S x6 from C* to C* is a hom*omorphism.What is the kernel?

38. For each pair of positive integers m and n, we can define a hom*o-morphism from Z to Zm % Zn by x S (x mod m, x mod n). What isthe kernel when (m, n) 5 (3, 4)? What is the kernel when (m, n) 5(6, 4)? Generalize.

39. (Second Isomorphism Theorem) If K is a subgroup of G and N isa normal subgroup of G, prove that K/(K > N) is isomorphic to KN/N.

40. (Third Isomorphism Theorem) If M and N are normal subgroups ofG and N # M, prove that (G/N)/(M/N) L G/M.

41. Let f(d) denote the Euler phi function of d (see page 79). Showthat the number of hom*omorphisms from Zn to Zk is Sf(d), wherethe sum runs over all common divisors d of n and k. [It followsfrom number theory that this sum is actually gcd(n, k).]

42. Let k be a divisor of n. Consider the hom*omorphism from U(n) toU(k) given by x S x mod k. What is the relationship between thishom*omorphism and the subgroup Uk(n) of U(n)?

43. Determine all hom*omorphic images of D4 (up to isomorphism).44. Let N be a normal subgroup of a finite group G. Use the theorems

of this chapter to prove that the order of the group element gN inG/N divides the order of g.

45. Suppose that G is a finite group and that Z10 is a hom*omorphicimage of G. What can we say about |G|? Generalize.

46. Suppose that Z10 and Z15 are both hom*omorphic images of a finitegroup G. What can be said about |G|? Generalize.

47. Suppose that for each prime p, Zp is the hom*omorphic image of agroup G. What can we say about |G|? Give an example of such agroup.

48. (For students who have had linear algebra.) Suppose that x is aparticular solution to a system of linear equations and that S is theentire solution set of the corresponding hom*ogeneous system oflinear equations. Explain why property 6 of Theorem 10.1 guaran-tees that x 1 S is the entire solution set of the nonhom*ogeneoussystem. In particular, describe the relevant groups and the hom*o-morphism between them.

49. Let N be a normal subgroup of a group G. Use property 7 ofTheorem 10.2 to prove that every subgroup of G/N has the formH/N, where H is a subgroup of G. (This exercise is referred to inChapter 24.)

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10 | Group hom*omorphisms 215

50. Show that a hom*omorphism defined on a cyclic group is com-pletely determined by its action on a generator of the group.

51. Use the First Isomorphism Theorem to prove Theorem 9.4.

52. Let a and b be group hom*omorphisms from G to and let H 5

{g [ G | a(g) 5 b(g)}. Prove or disprove that H is a subgroup of G.53. Let Z[x] be the group of polynomials in x with integer coefficients

under addition. Prove that the mapping from Z[x] into Z given byf(x) S f(3) is a hom*omorphism. Give a geometric description ofthe kernel of this hom*omorphism. Generalize.

54. Prove that the mapping from R under addition to GL(2, R) thattakes x to

is a group hom*omorphism. What is the kernel of the hom*omorphism?55. Suppose there is a hom*omorphism from G onto Z2 % Z2. Prove

that G is the union of three proper normal subgroups.56. If H and K are normal subgroups of G and H > K 5 {e}, prove that

G is isomorphic to a subgroup of G/H % G/K.57. Suppose that H and K are distinct subgroups of G of index 2. Prove

that H > K is a normal subgroup of G of index 4 and that G/(H > K)is not cyclic.

58. Suppose that the number of hom*omorphisms from G to H is n.How many hom*omorphisms are there from G to H % H % ? ? ? % H(s terms)? When H is Abelian, how many hom*omorphisms are therefrom G % G % ? ? ? % G (s terms) to H?

59. Prove that every group of order 77 is cyclic.60. Determine all hom*omorphisms from Z onto S3. Determine all

hom*omorphisms from Z to S3.61. Suppose G is an Abelian group under addition with the property

that for every positive integer n the set nG 5{ng|g [ G} 5 G.Show that every proper subgroup of G is properly contained in aproper subgroup of G. Name two familiar groups that satisfy thehypothesis.

62. Let p be a prime. Determine the number of hom*omorphisms frominto .ZpZp % Zp

f

c cos x sin x

2sin x cos xd

G

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216 Groups

Computer Exercise

A computer lets you make more mistakes faster than any invention inhuman history—with the possible exceptions of handguns and tequila.

MITCH RATLIFFE

Software for the computer exercise in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software determines the hom*omorphisms from Zm to Zn. (Recall that a hom*omorphism from Zm is completely determined bythe image of 1.) Run the program for m 5 20 with various choicesfor n. Run the program for m 5 15 with various choices for n. Whatrelationship do you see between m and n and the number of hom*o-morphisms from Zm to Zn? For each choice of m and n, observe thesmallest positive image of 1. Try to see the relationship between thisimage and the values of m and n. What relationship do you see be-tween the smallest positive image of 1 and the other images of 1?Test your conclusions with other choices of m and n.

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217

Camille Jordan

CAMILLE JORDAN was born into a well-to-dofamily on January 5, 1838, in Lyons, France.Like his father, he graduated from the ÉcolePolytechnique and became an engineer.Nearly all of his 120 research papers inmathematics were written before his retire-ment from engineering in 1885. From 1873until 1912, Jordan taught simultaneously atthe École Polytechnique and at the Collegeof France.

In the great French tradition, Jordan wasa universal mathematician who published innearly every branch of mathematics. Amongthe concepts named after him are the Jordancanonical form in matrix theory, the Jordancurve theorem from topology, and theJordan-Hölder theorem from group theory.

His classic book Traité des Substitutions,published in 1870, was the first to be de-voted solely to group theory and its applica-tions to other branches of mathematics.

Another book that had great influenceand set a new standard for rigor was hisCours d’analyse. This book gave the firstclear definitions of the notions of volumeand multiple integral. Nearly 100 years afterthis book appeared, the distinguishedmathematician and mathematical historianB. L. van der Waerden wrote, “For me, everysingle chapter of the Cours d’analyse is apleasure to read.” Jordan died in Paris onJanuary 22, 1922.

To find more information about Jordan,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

Although these contributions [to analysis and topology] would have beenenough to rank Jordan very high amonghis mathematical contemporaries, it ischiefly as an algebraist that he reachedcelebrity when he was barely thirty; andduring the next forty years he wasuniversally regarded as the undisputedmaster of group theory.

J. DIEUDONNÉ, Dictionary of

Scientific Biography

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218

Fundamental Theorem of FiniteAbelian Groups

The Fundamental TheoremIn this chapter, we present a theorem that describes to an algebraist’seye (that is, up to isomorphism) all finite Abelian groups in a stan-dardized way. Before giving the proof, which is long and difficult, wediscuss some consequences of the theorem and its proof. The firstproof of the theorem was given by Leopold Kronecker in 1858.

Theorem 11.1 Fundamental Theorem of Finite Abelian Groups

11

By a small sample we may judge of the whole piece.MIGUEL DE CERVANTES, Don Quixote

Every finite Abelian group is a direct product of cyclic groups ofprime-power order. Moreover, the number of terms in the productand the orders of the cyclic groups are uniquely determined by thegroup.

Since a cyclic group of order n is isomorphic to Zn, Theorem 11.1shows that every finite Abelian group G is isomorphic to a group ofthe form

Zp1n1 % Zp2

n2 % ? ? ? % Zpknk,

where the pi’s are not necessarily distinct primes and the prime-powers p1

n1, p2n2, . . . , pk

nk are uniquely determined by G. Writing agroup in this form is called determining the isomorphism class of G.

The Isomorphism Classes of Abelian Groups

The Fundamental Theorem is extremely powerful. As an application,we can use it as an algorithm for constructing all Abelian groups of anyorder. Let’s look at groups whose orders have the form pk, where p is

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11 | Fundamental Theorem of Finite Abelian Groups 219

prime and k # 4. In general, there is one group of order pk for each setof positive integers whose sum is k (such a set is called a partition of k);that is, if k can be written as

k 5 n1 1 n2 1 ? ? ? 1 nt,

where each ni is a positive integer, then

Zpn1 % Zp

n2 % ? ? ? % Zpnt

is an Abelian group of order pk.

Possible direct Order of G Partitions of k products for G

p 1 Zp

p2 2 Zp2

1 1 1 Zp % Zp

p3 3 Zp3

2 1 1 Zp2 % Zp

1 1 1 1 1 Zp % Zp % Zp

p4 4 Zp4

3 1 1 Zp3 % Zp

2 1 2 Zp2 % Zp2

2 1 1 1 1 Zp2 % Zp % Zp

1 1 1 1 1 1 1 Zp % Zp % Zp % Zp

Furthermore, the uniqueness portion of the Fundamental Theoremguarantees that distinct partitions of k yield distinct isomorphismclasses. Thus, for example, Z9 % Z3 is not isomorphic to Z3 % Z3 % Z3.A reliable mnemonic for comparing external direct products is the can-cellation property: If A is finite, then

A % B L A % C if and only if B L C (see [1]).

Thus Z4 % Z4 is not isomorphic to Z4 % Z2 % Z2 because Z4 is not isomorphic to Z2 % Z2.

To appreciate fully the potency of the Fundamental Theorem, contrastthe ease with which the Abelian groups of order pk, k # 4, weredetermined with the corresponding problem for non-Abelian groups.Even a description of the two non-Abelian groups of order 8 is a chal-lenge (see Chapter 26), and a description of the nine non-Abeliangroups of order 16 is well beyond the scope of this text.

Now that we know how to construct all the Abelian groups of prime-power order, we move to the problem of constructing all Abelian

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220 Groups

groups of a certain order n, where n has two or more distinct primedivisors. We begin by writing n in prime-power decomposition form n 5 p1

n1p2n2 ? ? ? pk

nk. Next, we individually form all Abelian groups oforder p1

n1, then p2n 2, and so on, as described earlier. Finally, we form all

possible external direct products of these groups. For example, let n 51176 5 23 ? 3 ? 72. Then, the complete list of the distinct isomorphismclasses of Abelian groups of order 1176 is

Z8 % Z3 % Z49,Z4 % Z2 % Z3 % Z49,Z2 % Z2 % Z2 % Z3 % Z49,Z8 % Z3 % Z7 % Z7,Z4 % Z2 % Z3 % Z7 % Z7,Z2 % Z2 % Z2 % Z3 % Z7 % Z7.

If we are given any particular Abelian group G of order 1176, thequestion we want to answer about G is: Which of the preceding six iso-morphism classes represents the structure of G? We can answer thisquestion by comparing the orders of the elements of G with the orders ofthe elements in the six direct products, since it can be shown that two fi-nite Abelian groups are isomorphic if and only if they have the samenumber of elements of each order. For instance, we could determinewhether G has any elements of order 8. If so, then G must be isomorphicto the first or fourth group above, since these are the only ones with ele-ments of order 8. To narrow G down to a single choice, we now needonly check whether or not G has an element of order 49, since the firstproduct above has such an element, whereas the fourth one does not.

What if we have some specific Abelian group G of order p1n1p2

n2

? ? ? pknk, where the pi’s are distinct primes? How can G be expressed as

an internal direct product of cyclic groups of prime-power order? Forsimplicity, let us say that the group has 2n elements. First, we mustcompute the orders of the elements. After this is done, pick an elementof maximum order 2r, call it a1. Then �a1� is one of the factors in thedesired internal direct product. If G 2 �a1�, choose an element a2 ofmaximum order 2s such that s # n 2 r and none of a2, a2

2, a24, . . . ,

a22 s21

is in �a1�. Then �a2� is a second direct factor. If n 2 r 1 s, selectan element a3 of maximum order 2t such that t # n 2 r 2 s and none ofa3, a3

2, a34, . . . , a3

2 t21is in �a1� 3 �a2� 5 {a1

ia2j | 0 # i , 2r, 0 #

j , 2s}. Then �a3� is another direct factor. We continue in this fashionuntil our direct product has the same order as G.

A formal presentation of this algorithm for any Abelian group G ofprime-power order pn is as follows.

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11 | Fundamental Theorem of Finite Abelian Groups 221

Element 1 8 12 14 18 21 27 31 34 38 44 47 51 53 57 64

Order 1 4 4 2 4 4 4 4 4 4 4 4 2 4 4 2

From the table of orders, we can instantly rule out all but Z4 % Z4 and Z4 % Z2 % Z2 as possibilities. Finally, we observe that since this lattergroup has a subgroup isomorphic to Z2 % Z2 % Z2, it has more thanthree elements of order 2, and therefore we must have G L Z4 % Z4.

Expressing G as an internal direct product is even easier. Pick an el-ement of maximum order, say the element 8. Then �8� is a factor in theproduct. Next, choose a second element, say a, so that a has order 4 anda and a2 are not in �8� 5 {1, 8, 64, 57}. Since 12 has this property, wehave G 5 �8� 3 �12�.

Greedy Algorithm for an Abelian Group of Order pn

1. Compute the orders of the elements of the group G.2. Select an element a1 of maximum order and define G1 5 �a1�.

Set i 5 1.3. If |G| 5 |Gi|, stop. Otherwise, replace i by i 1 1.4. Select an element ai of maximum order pk such that pk #

|G|/|Gi21| and none of ai , aip, ai

p 2,. . . , ai

pk21is in Gi21, and define

Gi 5 Gi21 3 �ai�.5. Return to step 3.

In the general case where |G| 5 p1n 1p2

n 2 ? ? ? pkn k, we simply use the

algorithm to build up a direct product of order p1n 1, then another of

order p2n 2, and so on. The direct product of all of these pieces is the

desired factorization of G. The following example is small enough thatwe can compute the appropriate internal and external direct productsby hand.

EXAMPLE 1 Let G 5 {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51,53, 57, 64} under multiplication modulo 65. Since G has order 16, weknow it is isomorphic to one of

Z16,Z8 % Z2,Z4 % Z4,

Z4 % Z2 % Z2,Z2 % Z2 % Z2 % Z2.

To decide which one, we dirty our hands to calculate the orders of theelements of G.

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222 Groups

Example 1 illustrates how quickly and easily one can write an Abeliangroup as a direct product given the orders of the elements of the group.But calculating all those orders is certainly not an appealing prospect!The good news is that, in practice, a combination of theory and calcula-tion of the orders of a few elements will usually suffice.

EXAMPLE 2 Let G 5 {1, 8, 17, 19, 26, 28, 37, 44, 46, 53, 62,64, 71, 73, 82, 89, 91, 98, 107, 109, 116, 118, 127, 134} under multi-plication modulo 135. Since G has order 24, it is isomorphic to one of

Z8 % Z3 L Z24,Z4 % Z2 % Z3 L Z12 % Z2,

Z2 % Z2 % Z2 % Z3 L Z6 % Z2 % Z2.

Consider the element 8. Direct calculations show that 86 5 109 and 812 51. (Be sure to mod as you go. For example, 83 mod 135 5 512 mod 135 5 107, so compute 84 as 8 ? 107 rather than 8 ? 512.) But now weknow G. Why? Clearly, |8| 5 12 rules out the third group in the list. Atthe same time, |109| 5 2 5 |134| (remember, 134 5 21 mod 135) im-plies that G is not Z24 (see Theorem 4.4). Thus, G L Z12 % Z2, and G 5�8� 3 �134�.

Rather than express an Abelian group as a direct product of cyclicgroups of prime-power orders, it is often more convenient to combinethe cyclic factors of relatively prime order, as we did in Example 2, toobtain a direct product of the form Zn1

% Zn2% ? ? ? % Znk

, where ni di-vides ni21. For example, Z4 % Z4 % Z2 % Z9 % Z3 % Z5 would be writtenas Z180 % Z12 % Z2 (see Exercise 11). The algorithm above is easilyadapted to accomplish this by replacing step 4 by 49: select an element aiof maximum order m such that m # |G|/|Gi21| and none of ai, ai

2, . . . ,ai

m21 is in Gi21, and define Gi 5 Gi21 3 �ai�.As a consequence of the Fundamental Theorem of Finite Abelian

Groups, we have the following corollary, which shows that the converseof Lagrange’s Theorem is true for finite Abelian groups.

Corollary Existence of Subgroups of Abelian Groups

If m divides the order of a finite Abelian group G, then G has asubgroup of order m.

It is instructive to verify this corollary for a specific case. Let us saythat G is an Abelian group of order 72 and we wish to produce a subgroup

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11 | Fundamental Theorem of Finite Abelian Groups 223

of order 12. According to the Fundamental Theorem, G is isomorphic toone of the following six groups:

Z8 % Z9, Z8 % Z3 % Z3,Z4 % Z2 % Z9, Z4 % Z2 % Z3 % Z3,Z2 % Z2 % Z2 % Z9, Z2 % Z2 % Z2 % Z3 % Z3.

Obviously, Z8 % Z9 L Z72 and Z4 % Z2 % Z3 % Z3 L Z12 % Z6 bothhave a subgroup of order 12. To construct a subgroup of order 12 in Z4% Z2 % Z9, we simply piece together all of Z4 and the subgroup of order3 in Z9; that is, {(a, 0, b) | a [ Z4, b [ {0, 3, 6}}. A subgroup of order12 in Z8 % Z3 % Z3 is given by {(a, b, 0) | a [ {0, 2, 4, 6}, b [ Z3}. Ananalogous procedure applies to the remaining cases and indeed to anyfinite Abelian group.

Proof of the Fundamental TheoremBecause of the length and complexity of the proof of the FundamentalTheorem of Finite Abelian Groups, we will break it up into a series oflemmas.

Lemma 1

PROOF It is an easy exercise to prove that H and K are subgroups of G(see Exercise 29 in Chapter 3). Because G is Abelian, to prove that G 5H 3 K we need only prove that G 5 HK and H > K 5 {e}. Since wehave gcd(m, pn) 5 1, there are integers s and t such that 1 5 sm 1 tpn.For any x in G, we have x 5 x1 5 xsm1tpn

5 xsmxtpnand, by Corollary 4

of Lagrange’s Theorem (Theorem 7.1), xsm [ H and x tpn[ K. Thus,

G 5 HK. Now suppose that some x [ H > K. Then xpn 5 e 5 xm and,by Corollary 2 to Theorem 4.1, |x| divides both pn and m. Since p doesnot divide m, we have |x| 5 1 and, therefore, x 5 e.

To prove the second assertion of the lemma, note that pnm 5|HK| 5 |H||K|/|H > K| 5 |H||K| (see Exercise 7 in the SupplementaryExercises for Chapters 5–8). It follows from Theorem 9.5 andCorollary 2 to Theorem 4.1 that p does not divide |K| and therefore |H| 5 pn.

Given an Abelian group G with |G| 5 p1n1p2

n2 ? ? ? pknk, where the

p’s are distinct primes, we let G(pi) denote the set {x [ G | x pini

5 e}.

Let G be a finite Abelian group of order pnm, where p is a prime thatdoes not divide m. Then G 5 H 3 K, where H 5 {x [ G | x p

n

5 e}and K 5 {x [ G | xm 5 e}. Moreover, |H| 5 pn.

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224 Groups

It then follows immediately from Lemma 1 and induction that G 5G(p1) 3 G(p2) 3 ? ? ? 3 G(pk) and |G(pi)| 5 pi

n i. Hence, we turn ourattention to groups of prime-power order.

Lemma 2

PROOF We denote |G| by pn and induct on n. If n 5 1, then G 5�a� 3 �e�. Now assume that the statement is true for all Abeliangroups of order pk, where k , n. Among all the elements of G, choosea of maximal order pm. Then x pm 5 e for all x in G. We may assumethat G 2 �a�, for otherwise there is nothing to prove. Now, among allthe elements of G, choose b of smallest order such that b o �a�. Weclaim that �a� > �b� 5 {e}. Since |b p| 5 |b|/p, we know that b p [ �a�by the manner in which b was chosen. Say b p 5 ai. Notice that e 5b pm 5 (b p) pm21 5 (ai) pm21, so |ai| # pm21. Thus, ai is not a generator of�a� and, therefore, by Corollary 3 to Theorem 4.2, gcd(pm, i) 2 1.This proves that p divides i, so that we can write i 5 pj. Then bp 5ai 5 apj. Consider the element c 5 a2jb. Certainly, c is not in �a�, forif it were, b would be, too. Also, cp 5 a2jpb p 5 a2ib p 5 b2pb p 5 e.Thus, we have found an element c of order p such that c o �a�. Sinceb was chosen to have smallest order such that b o �a�, we concludethat b also has order p. It now follows that �a� > �b� 5 {e} becauseany nonidentity element of the intersection would generate �b� andthus contradict b o �a�.

Now consider the factor group 5 G/�b�. To simplify the notation,we let denote the coset x�b� in . If | | , |a| 5 pm, then pm21

5 . Thismeans that (a�b�) pm21

5 a pm21�b� 5 �b�, so that apm21

[ �a� > �b� 5 {e},contradicting the fact that |a| 5 pm. Thus, | | 5 |a| 5 pm, and therefore

is an element of maximal order in . By induction, we know that can be written in the form � � 3 for some subgroup of . Let K bethe pullback of under the natural hom*omorphism from G to (thatis, K 5 {x [ G | [ }). We claim that �a� > K 5 {e}. For if x [ �a�> K, then [ � � > 5 { } 5 �b� and x [ �a� > �b� 5 {e}. It nowfollows from an order argument (see Exercise 33) that G 5 �a�K, andtherefore G 5 �a� 3 K.

Lemma 2 and induction on the order of the group now give thefollowing.

eKaxKx

GKGKKa

GGaa

eaaGxG

Let G be an Abelian group of prime-power order and let a be anelement of maximal order in G. Then G can be written in the form�a� 3 K.

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11 | Fundamental Theorem of Finite Abelian Groups 225

Lemma 3

Let us pause to determine where we are in our effort to prove theFundamental Theorem of Finite Abelian Groups. The remark followingLemma 1 shows that G 5 G( p1) 3 G( p2) 3 ? ? ? 3 G( pn), where each G( pi) is a group of prime-power order, and Lemma 3 shows that each ofthese factors is an internal direct product of cyclic groups. Thus, we haveproved that G is an internal direct product of cyclic groups of prime-power order. All that remains to be proved is the uniqueness of the factors.Certainly the groups G( pi) are uniquely determined by G, since theycomprise the elements of G whose orders are powers of pi. So we mustprove that there is only one way (up to isomorphism and rearrangement offactors) to write each G( pi) as an internal direct product of cyclic groups.

Lemma 4

PROOF We proceed by induction on |G|. Clearly, the case where |G| 5p is true. Now suppose that the statement is true for all Abelian groupsof order less than |G|. For any Abelian group L, the set Lp 5 {x p | x [L} is a subgroup of L (see Exercise 15 in the Supplementary Exercisesfor Chapters 1– 4) and, by Theorem 9.5, is a proper subgroup if pdivides |L|. It follows that Gp 5 H1

p 3 H2p 3 ? ? ? 3 Hm9

p, and Gp 5K1

p 3 K2p 3 ? ? ? 3 Kn9

p, where m9 is the largest integer i such that |Hi| . p, and n9 is the largest integer j such that |Kj| . p. (This ensuresthat our two direct products for G p do not have trivial factors.) Since |G p|, |G|, we have, by induction, m9 5 n9 and |Hi

p| 5 |Kip| for i 5 1, . . . ,

m9. Since |Hi| 5 p|Hip|, this proves that |Hi| 5 |Ki| for all i 5 1, . . . , m9.

All that remains to be proved is that the number of Hi of order p equalsthe number of Ki of order p; that is, we must prove that m 2 m9 5 n 2 n9(since n9 5 m9). This follows directly from the facts that |H1||H2| ? ? ?|Hm9|p

m2m9 5 |G| 5 |K1||K2| ? ? ? |Kn9|pn2n9, |Hi| 5 |Ki|, and m9 5 n9.

A finite Abelian group of prime-power order is an internal directproduct of cyclic groups.

Suppose that G is a finite Abelian group of prime-power order. If G 5 H1 3 H2 3 ? ? ? 3 Hm and G 5 K1 3 K2 3 ? ? ? 3 Kn, where theH’s and K’s are nontrivial cyclic subgroups with |H1| $ |H2| $ ? ? ? $|Hm| and |K1| $ |K2| $ ? ? ? $ |Kn|, then m 5 n and |Hi| 5 |Ki|for all i.

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226 Groups

Exercises

You know it ain’t easy, you know how hard it can be.JOHN LENNON AND PAUL MCCARTNEY,

“The Ballad of John and Yoko”

1. What is the smallest positive integer n such that there are two non-isomorphic groups of order n? Name the two groups.

2. What is the smallest positive integer n such that there are threenonisomorphic Abelian groups of order n? Name the three groups.

3. What is the smallest positive integer n such that there are exactlyfour nonisomorphic Abelian groups of order n? Name the fourgroups.

4. Calculate the number of elements of order 2 in each of Z16, Z8 % Z2,Z4 % Z4, and Z4 % Z2 % Z2. Do the same for the elements of order 4.

5. Prove that any Abelian group of order 45 has an element of order 15.Does every Abelian group of order 45 have an element of order 9?

6. Show that there are two Abelian groups of order 108 that have ex-actly one subgroup of order 3.

7. Show that there are two Abelian groups of order 108 that have ex-actly four subgroups of order 3.

8. Show that there are two Abelian groups of order 108 that have ex-actly 13 subgroups of order 3.

9. Suppose that G is an Abelian group of order 120 and that G hasexactly three elements of order 2. Determine the isomorphism classof G.

10. Find all Abelian groups (up to isomorphism) of order 360.11. Prove that every finite Abelian group can be expressed as the

(external) direct product of cyclic groups of orders n1, n2, . . . , nt,where ni11 divides ni for i 5 1, 2, . . . , t 2 1. (This exercise is re-ferred to in this chapter and in Chapter 22.)

12. Suppose that the order of some finite Abelian group is divisible by10. Prove that the group has a cyclic subgroup of order 10.

13. Show, by example, that if the order of a finite Abelian group is di-visible by 4, the group need not have a cyclic subgroup of order 4.

14. On the basis of Exercises 12 and 13, draw a general conclusionabout the existence of cyclic subgroups of a finite Abelian group.

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11 | Fundamental Theorem of Finite Abelian Groups 227

15. How many Abelian groups (up to isomorphism) are therea. of order 6?b. of order 15?c. of order 42?d. of order pq, where p and q are distinct primes?e. of order pqr, where p, q, and r are distinct primes?f. Generalize parts d and e.

16. How does the number (up to isomorphism) of Abelian groups oforder n compare with the number (up to isomorphism) of Abeliangroups of order m wherea. n 5 32 and m 5 52?b. n 5 24 and m 5 54?c. n 5 pr and m 5 qr, where p and q are prime?d. n 5 pr and m 5 prq, where p and q are distinct primes?e. n 5 pr and m 5 prq2, where p and q are distinct primes?

17. The symmetry group of a nonsquare rectangle is an Abelian groupof order 4. Is it isomorphic to Z4 or Z2 % Z2?

18. Verify the corollary to the Fundamental Theorem of FiniteAbelian Groups in the case that the group has order 1080 and thedivisor is 180.

19. The set {1, 9, 16, 22, 29, 53, 74, 79, 81} is a group under multipli-cation modulo 91. Determine the isomorphism class of this group.

20. Suppose that G is a finite Abelian group that has exactly one sub-group for each divisor of |G|. Show that G is cyclic.

21. Characterize those integers n such that the only Abelian groups oforder n are cyclic.

22. Characterize those integers n such that any Abelian group of ordern belongs to one of exactly four isomorphism classes.

23. Refer to Example 1 in this chapter and explain why it is unneces-sary to compute the orders of the last five elements listed to deter-mine the isomorphism class of G.

24. Let G 5 {1, 7, 17, 23, 49, 55, 65, 71} under multiplication modulo 96. Express G as an external and an internal direct product of cyclicgroups.

25. Let G 5 {1, 7, 43, 49, 51, 57, 93, 99, 101, 107, 143, 149, 151, 157,193, 199} under multiplication modulo 200. Express G as an exter-nal and an internal direct product of cyclic groups.

26. The set G 5 {1, 4, 11, 14, 16, 19, 26, 29, 31, 34, 41, 44} is a groupunder multiplication modulo 45. Write G as an external and an in-ternal direct product of cyclic groups of prime-power order.

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228 Groups

27. Suppose that G is an Abelian group of order 9. What is the maxi-mum number of elements (excluding the identity) of which oneneeds to compute the order to determine the isomorphism class ofG? What if G has order 18? What about 16?

28. Suppose that G is an Abelian group of order 16, and in computingthe orders of its elements, you come across an element of order 8and two elements of order 2. Explain why no further computationsare needed to determine the isomorphism class of G.

29. Let G be an Abelian group of order 16. Suppose that there are ele-ments a and b in G such that |a| 5 |b| 5 4 and a2 2 b2. Determinethe isomorphism class of G.

30. Prove that an Abelian group of order 2n (n $ 1) must have an oddnumber of elements of order 2.

31. Without using Lagrange’s Theorem, show that an Abelian group ofodd order cannot have an element of even order.

32. Let G be the group of all n 3 n diagonal matrices with 61 diago-nal entries. What is the isomorphism class of G?

33. Prove the assertion made in the proof of Lemma 2 that G 5 �a�K.34. Suppose that G is a finite Abelian group. Prove that G has order pn,

where p is prime, if and only if the order of every element of G is apower of p.

35. Dirichlet’s Theorem says that, for every pair of relatively prime in-tegers a and b, there are infinitely many primes of the form at 1 b.Use Dirichlet’s Theorem to prove that every finite Abelian group isisomorphic to a subgroup of a U-group.

36. Determine the isomorphism class of Aut(Z2 % Z3 % Z5).37. Give an example to show that Lemma 2 is false if G is non-Abelian.

Computer Exercises

The purpose of computation is insight, not numbers.RICHARD HAMMING

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software lists the isomorphism classes of all finite Abeliangroups of any particular order n. Run the program for n 5 16, 24,512, 2048, 441000, and 999999.

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11 | Fundamental Theorem of Finite Abelian Groups 229

2. This software determines how many integers in a given interval arethe order of exactly one Abelian group, of exactly two Abeliangroups, and so on, up to exactly nine Abelian groups. Run your pro-gram for the integers up to 1000. Then from 10001 to 11000. Thenchoose your own interval of 1000 consecutive integers. Is theremuch difference in the results?

3. This software expresses a U-group as an internal direct product ofsub-groups H1 3 H2 3 ? ? ? 3 Hi, where |Hi| divides |Hi21|. Run theprogram for the groups U(32), U(80), and U(65).

Reference

1. R. Hirshon, “On Cancellation in Groups,” American MathematicalMonthly 76 (1969): 1037–1039.

Suggested Readings

J. A. Gallian, “Computers in Group Theory,” Mathematics Magazine49 (1976): 69–73.

This paper discusses several computer-related projects in group theorydone by undergraduate students.

J. Kane, “Distribution of Orders of Abelian Groups,” Mathematics Magazine49 (1976): 132–135.

In this article, the author determines the percentages of integers k between 1 and n, for sufficiently large n, that have exactly one isomorphism class ofAbelian groups of order k, exactly two isomorphism classes of Abeliangroups of order k, and so on, up to 13 isomorphism classes.

G. Mackiw, “Computing in Abstract Algebra,” The College MathematicsJournal 27 (1996): 136–142.

This article explains how one can use computer software to implement thealgorithm given in this chapter for expressing an Abelian group as an inter-nal direct product.

Suggested Website

To find more information about the development of group theory, visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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230 Groups

Supplementary Exercises for Chapters 9–11

Every prospector drills many a dry hole, pulls out his rig, and moves on.JOHN L. HESS

True/false questions for Chapters 9–11 are available on the Web at:

http://www.d.umn.edu/,jgallian/TF

1. Suppose that H is a subgroup of G and that each left coset of H inG is some right coset of H in G. Prove that H is normal in G.

2. Use a factor group-induction argument to prove that a finiteAbelian group of order n has a subgroup of order m for every posi-tive divisor m of n.

3. Let diag(G) 5 {(g, g) | g [ G}. Prove that diag(G) v G % G ifand only if G is Abelian. When G is finite, what is the index ofdiag(G) in G % G?

4. Let H be any group of rotations in Dn. Prove that H is normal in Dn.5. Prove that Inn(G) v Aut(G).6. Let H be a subgroup of G. Prove that H is a normal subgroup if and

only if, for all a and b in G, ab [ H implies ba [ H.7. The factor group GL(2, R)/SL(2, R) is isomorphic to some very

familiar group. What is the group?8. Let k be a divisor of n. The factor group (Z/�n�)/(�k�/�n�) is isomor-

phic to some very familiar group. What is the group?9. Let

under matrix multiplication.a. Find .b. Prove that Z(H) is isomorphic to Q under addition.c. Prove that H/Z(H) is isomorphic to Q % Q.d. Are your proofs for parts a and b valid when Q is replaced by

R? Are they valid when Q is replaced by Zp, where p is prime?10. Prove that D4/Z(D4) is isomorphic to Z2 % Z2.11. Prove that Q/Z under addition is an infinite group in which every

element has finite order.12. Show that the intersection of any collection of normal subgroups of

a group is a normal subgroup.

Z(H)

H 5 • £

1

a

1

b

c

1

§ † a, b, c [ Q ¶

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11 | Supplementary Exercises for Chapters 9–11 231

13. Let n . 1 be a fixed integer and let G be a group. If the set H 5{x [ G| |x| 5 n} together with the identity forms a subgroup ofG, prove that it is a normal subgroup of G. In the case where sucha subgroup exists, what can be said about n? Give an example ofa non-Abelian group that has such a subgroup. Give an exampleof a group G and a prime n for which the set H together with theidentity is not a subgroup.

14. Show that Q/Z has a unique subgroup of order n for each positiveinteger n.

15. If H and K are normal Abelian subgroups of a group and H > K 5{e}, prove that HK is Abelian.

16. Let G be a group of odd order. Prove that the mapping x S x2 fromG to itself is one-to-one.

17. Suppose that G is a group of permutations on some set. If |G| 5 60and orbG(5) 5 {1, 5}, prove that stabG(5) is normal in G.

18. Suppose that G 5 H 3 K and that N is a normal subgroup of H.Prove that N is normal in G.

19. Show that there is no hom*omorphism from Z8 % Z2 % Z2 onto Z4 % Z4.

20. Show that there is no hom*omorphism from A4 onto a group oforder 2, 4, or 6, but that there is a hom*omorphism from A4 onto agroup of order 3.

21. Let H be a normal subgroup of S4 of order 4. Prove that S4/H is iso-morphic to S3.

22. Suppose that f is a hom*omorphism of U(36), Ker f 5 {1, 13, 25},and f(5) 5 17. Determine all elements that map to 17.

23. Let n 5 2m, where m is odd. How many elements of order 2 does Dn/Z(Dn) have? How many elements are in the subgroup�R360/n�/Z(Dn)? How do these numbers compare with the numberof elements of order 2 in Dm?

24. Suppose that H is a normal subgroup of a group G of odd order andthat |H| 5 5. Show that H # Z(G).

25. Let G be an Abelian group and let n be a positive integer. Let Gn 5{g | gn 5 e} and Gn 5 {gn | g [ G}. Prove that G/Gn is isomorphicto Gn.

26. Let R1 denote the multiplicative group of positive reals and let T 5{a 1 bi [ C| a2 1 b2 5 1} be the multiplicative group of complexnumbers of norm 1. Show that C* is the internal direct product of R1

and T.

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232 Groups

27. Let G be a finite group and let p be a prime. If p2 . |G|, show thatany subgroup of order p is normal in G.

28. Let G 5 Z % Z and H 5 {(x, y)| x and y are even integers}. Showthat H is a subgroup of G. Determine the order of G/H. To whichfamiliar group is G/H isomorphic?

29. Let n be a positive integer. Prove that every element of order n inQ/Z is contained in �1/n 1 Z�.

30. (1997 Putnam Competition) Let G be a group and let f : G S G bea function such that

f(g1)f(g2)f(g3) 5 f(h1)f(h2)f(h3)

whenever g1g2g3 5 e 5 h1h2h3. Prove that there exists an element ain G such that c(x) 5 af(x) is a hom*omorphism.

31. Prove that every hom*omorphism from Z % Z into Z has the form (x, y) S ax 1 by, where a and b are integers.

32. Prove that every hom*omorphism from Z % Z into Z % Z has theform (x, y) S (ax 1 by, cx 1 dy), where a, b, c, and d are integers.

33. Prove that Q/Z is not isomorphic to a proper subgroup of itself.34. Prove that for each positive integer n, the group Q/Z has exactly

f(n) elements of order n (f is the Euler phi function).35. Show that any group with more than two elements has an automor-

phism other than the identity mapping.36. A proper subgroup H of a group G is called maximal if there is no

subgroup K such that H , K , G. Prove that Q under addition hasno maximal subgroups.

37. Let G be the group of quaternions as given in Exercise 4 of theSupplementary Exercises for Chapters 1–4 and . Determinewhether is isomorphic to or . Is isomorphic to asubgroup of G?

38. Write the dihedral group as and let .

Prove that N is normal in . Given that de-termine whether is cyclic.D8>N

F1N 5 5F1, F4, F3, F26D8

N 5 5R0, R90, R180, R2706R315, F1, F2, F3, F4, F5, F6, F7, F8

5R0, R45, R90, R135, R180, R225, R270,D8

G>HZ2 % Z2Z4G>H H 5 �a2�

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11 | Supplementary Exercises for Chapters 9–11 233

39. Let G be the group where a, b [ R, and

| where x [ R . Show that H is a subgroup of G. Is

H a normal subgroup of G? Justify your answer.40. Find a subgroup H of Zp2 % Zp2 such that (Zp2 % Zp2)/H is isomorphic

to Zp % Zp.41. Recall that H is a characteristic subgroup of K if for

every automorphism of K. Prove that if H is a characteristic sub-group of K, and K is a normal subgroup of G, then H is a normalsubgroup of G.

ff(H) 5 H

fH 5 e c1 x

0 1d

fb ? 0e c1 a

0 bd |

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P A R T 3

Rings

235

For online student resources, visit this textbook’s website athttp://college.hmco.com/PIC/gallian7e

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Introduction to Rings

Example is the school of mankind, and they will learn at no other.EDMUND BURKE, On a Regicide Peace

237

12

†The term ring was first applied in 1897 by the German mathematician David Hilbert(1862–1943).

Motivation and DefinitionMany sets are naturally endowed with two binary operations: additionand multiplication. Examples that quickly come to mind are the inte-gers, the integers modulo n, the real numbers, matrices, and polynomi-als. When considering these sets as groups, we simply used additionand ignored multiplication. In many instances, however, one wishes totake into account both addition and multiplication. One abstract con-cept that does this is the concept of a ring.† This notion was originatedin the mid-nineteenth century by Richard Dedekind, although its firstformal abstract definition was not given until Abraham Fraenkel pre-sented it in 1914.

Definition Ring

A ring R is a set with two binary operations, addition (denoted by a 1 b) and multiplication (denoted by ab), such that for all a, b, c in R:

1. a 1 b 5 b 1 a.2. (a 1 b) 1 c 5 a 1 (b 1 c).3. There is an additive identity 0. That is, there is an element 0 in R

such that a 1 0 5 a for all a in R.4. There is an element 2a in R such that a 1 (2a) 5 0.5. a(bc) 5 (ab)c.6. a(b 1 c) 5 ab 1 ac and (b 1 c) a 5 ba 1 ca.

So, a ring is an Abelian group under addition, also having an asso-ciative multiplication that is left and right distributive over addition.Note that multiplication need not be commutative. When it is, we saythat the ring is commutative. Also, a ring need not have an identity

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under multiplication. A unity (or identity) in a ring is a nonzero elementthat is an identity under multiplication. A nonzero element of a com-mutative ring with unity need not have a multiplicative inverse. When itdoes, we say that it is a unit of the ring. Thus, a is a unit if a21 exists.

The following terminology and notation are convenient. If a and bbelong to a commutative ring R and a is nonzero, we say that a dividesb (or that a is a factor of b) and write a | b, if there exists an element cin R such that b 5 ac. If a does not divide b, we write a B b.

Recall that if a is an element from a group under the operation ofaddition and n is a positive integer, na means a 1 a 1 ? ? ? 1 a, wherethere are n summands. When dealing with rings, this notation can causeconfusion, since we also use juxtaposition for the ring multiplication.When there is the potential for confusion, we will use n ? a to meana 1 a 1 ? ? ? 1 a (n summands).

For an abstraction to be worthy of study, it must have many diverseconcrete realizations. The following list of examples shows that thering concept is pervasive.

Examples of RingsEXAMPLE 1 The set Z of integers under ordinary addition and

multiplication is a commutative ring with unity 1. The units of Z are1 and 21.

EXAMPLE 2 The set Zn 5 {0, 1, . . . , n 2 1} under addition andmultiplication modulo n is a commutative ring with unity 1. The set ofunits is U(n).

EXAMPLE 3 The set Z[x] of all polynomials in the variable x withinteger coefficients under ordinary addition and multiplication is acommutative ring with unity f(x) 5 1.

EXAMPLE 4 The set M2(Z) of 2 3 2 matrices with integer entries

is a noncommutative ring with unity .

EXAMPLE 5 The set 2Z of even integers under ordinary additionand multiplication is a commutative ring without unity.

EXAMPLE 6 The set of all continuous real-valued functions of areal variable whose graphs pass through the point (1, 0) is a commuta-tive ring without unity under the operations of pointwise addition and

c1 0

0 1d

238 Rings

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multiplication [that is, the operations ( f 1 g)(a) 5 f(a) 1 g(a) and ( fg)(a) 5 f(a)g(a)].

EXAMPLE 7 Let R1, R2, . . . , Rn be rings. We can use these to con-struct a new ring as follows. Let

R1 % R2 % ? ? ? % Rn 5 {(a1, a2, . . . , an) | ai [ Ri}

and perform componentwise addition and multiplication; that is, define

(a1, a2, . . . , an) 1 (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn)

and

(a1, a2, . . . , an)(b1, b2, . . . , bn) 5 (a1b1, a2b2, . . . , anbn).

This ring is called the direct sum of R1, R2, . . . , Rn.

Properties of RingsOur first theorem shows how the operations of addition and multiplica-tion intertwine. We use b 2 c to denote b 1 (2c).

Theorem 12.1 Rules of Multiplication

PROOF We will prove rules 1 and 2 and leave the rest as easy exercises(see Exercise 11). To prove statements such as those in Theorem 12.1, weneed only “play off ” the distributive property against the fact that R is agroup under addition with additive identity 0. Consider rule 1. Clearly,

0 1 a0 5 a0 5 a(0 1 0) 5 a0 1 a0.

So, by cancellation, 0 5 a0. Similarly, 0a 5 0.

Let a, b, and c belong to a ring R. Then

1. a0 5 0a 5 0.2. a(2b) 5 (2a)b 5 2(ab).3. (2a)(2b) 5 ab.†

4. a(b 2 c) 5 ab 2 ac and (b 2 c)a 5 ba 2 ca.

Furthermore, if R has a unity element 1, then

5. (21)a 5 2a.6. (21)(21) 5 1.

12 | Introduction to Rings 239

†Minus times minus is plus.The reason for this we need not discuss.W. H. Auden

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To prove rule 2, we observe that a(2b) 1 ab 5 a(2b 1 b) 5a0 5 0. So, adding 2(ab) to both sides yields a(2b) 5 2(ab). The re-mainder of rule 2 is done analogously.

Recall that in the case of groups, the identity and inverses are unique.The same is true for rings, provided that these elements exist. The proofsare identical to the ones given for groups and therefore are omitted.

Theorem 12.2 Uniqueness of the Unity and Inverses

Many students have the mistaken tendency to treat a ring as if it werea group under multiplication. It is not. The two most common errors arethe assumptions that ring elements have multiplicative inverses—theyneed not—and that a ring has a multiplicative identity—it need not. Forexample, if a, b, and c belong to a ring, a 2 0 and ab 5 ac, we cannotconclude that b 5 c. Similarly, if a2 5 a, we cannot conclude that a 5 0or 1 (as is the case with real numbers). In the first place, the ring neednot have multiplicative cancellation, and in the second place, the ringneed not have a multiplicative identity. There is an important class ofrings wherein multiplicative identities exist and for which multiplica-tive cancellation holds. This class is taken up in the next chapter.

SubringsIn our study of groups, subgroups played a crucial role. Subrings, theanalogous structures in ring theory, play a much less prominent role thantheir counterparts in group theory. Nevertheless, subrings are important.

Definition Subring

A subset S of a ring R is a subring of R if S is itself a ring with theoperations of R.

Just as was the case for subgroups, there is a simple test for subrings.

Theorem 12.3 Subring Test

A nonempty subset S of a ring R is a subring if S is closed undersubtraction and multiplication—that is, if a 2 b and ab are in Swhenever a and b are in S.

If a ring has a unity, it is unique. If a ring element has a multipli-cative inverse, it is unique.

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PROOF Since addition in R is commutative and S is closed under sub-traction, we know by the One-Step Subgroup Test (Theorem 3.1) that Sis an Abelian group under addition. Also, since multiplication in R isassociative as well as distributive over addition, the same is true formultiplication in S. Thus, the only condition remaining to be checkedis that multiplication is a binary operation on S. But this is exactly whatclosure means.

We leave it to the student to confirm that each of the following ex-amples is a subring.

EXAMPLE 8 {0} and R are subrings of any ring R. {0} is called thetrivial subring of R.

EXAMPLE 9 {0, 2, 4} is a subring of the ring Z6, the inte-gers modulo 6. Note that although 1 is the unity in Z6, 4 is the unity in{0, 2, 4}.

EXAMPLE 10 For each positive integer n, the set

nZ 5 {0, 6n, 62n, 63n, . . .}

is a subring of the integers Z.

EXAMPLE 11 The set of Gaussian integers

Z[i] 5 {a 1 bi | a, b [ Z}

is a subring of the complex numbers C.

EXAMPLE 12 Let R be the ring of all real-valued functions of a sin-gle real variable under pointwise addition and multiplication. The sub-set S of R of functions whose graphs pass through the origin forms asubring of R.

EXAMPLE 13 The set

of diagonal matrices is a subring of the ring of all 2 3 2 matrices over Z.

We can picture the relationship between a ring and its various sub-rings by way of a subring lattice diagram. In such a diagram, any ringis a subring of all the rings that it is connected to by one or more up-ward lines. Figure 12.1 shows the relationships among some of therings we have already discussed.

e ca 0

0 bd 0 a, b [ Z f

12 | Introduction to Rings 241

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Figure 12.1 Partial subring lattice diagram of C

In the next several chapters, we will see that many of the fundamen-tal concepts of group theory can be naturally extended to rings. In par-ticular, we will introduce ring hom*omorphisms and factor rings.

Exercises

There is no substitute for hard work.THOMAS ALVA EDISON, Life

1. Give an example of a finite noncommutative ring. Give an exam-ple of an infinite noncommutative ring that does not have a unity.

2. The ring {0, 2, 4, 6, 8} under addition and multiplication modulo10 has a unity. Find it.

3. Give an example of a subset of a ring that is a subgroup underaddition but not a subring.

4. Show, by example, that for fixed nonzero elements a and b in aring, the equation ax 5 b can have more than one solution. Howdoes this compare with groups?

5. Prove Theorem 12.2.6. Find an integer n that shows that the rings Zn need not have the fol-

lowing properties that the ring of integers has.a. a2 5 a implies a 5 0 or a 5 1.b. ab 5 0 implies a 5 0 or b 5 0.c. ab 5 ac and a 2 0 imply b 5 c.Is the n you found prime?

C

QQ(√2) = {a 1 b√2 | a, b [ Q}

Z

R

5Z 2Z 3Z

6Z4Z

8Z 12Z 18Z

10Z

7Z

9Z

Z[i] = {a 1 bi | a, b [ Z}

242 Rings

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7. Show that the three properties listed in Exercise 6 are valid for Zp,where p is prime.

8. Show that a ring is commutative if it has the property that ab 5 caimplies b 5 c when a 2 0.

9. Prove that the intersection of any collection of subrings of a ring Ris a subring of R.

10. Verify that Examples 8 through 13 in this chapter are as stated.11. Prove parts 3 through 6 of Theorem 12.1.12. Let a, b, and c be elements of a commutative ring, and suppose that

a is a unit. Prove that b divides c if and only if ab divides c.13. Describe all the subrings of the ring of integers.14. Let a and b belong to a ring R and let m be an integer. Prove that

m ? (ab) 5 (m ? a)b 5 a(m ? b).15. Show that if m and n are integers and a and b are elements from a

ring, then (m ? a)(n ? b) 5 (mn) ? (ab). (This exercise is referred toin Chapters 13 and 15.)

16. Show that if n is an integer and a is an element from a ring, then n ? (2a) 5 2(n ? a).

17. Show that a ring that is cyclic under addition is commutative.18. Let a belong to a ring R. Let S 5 {x [ R | ax 5 0}. Show that S is

a subring of R.19. Let R be a ring. The center of R is the set {x [ R | ax 5 xa for all

a in R}. Prove that the center of a ring is a subring.20. Describe the elements of M2(Z) (see Example 4) that have multi-

plicative inverses.21. Suppose that R1, R2, . . . , Rn are rings that contain nonzero ele-

ments. Show that R1 % R2 % ? ? ? % Rn has a unity if and only ifeach Ri has a unity.

22. Let R be a commutative ring with unity and let U(R) denote the setof units of R. Prove that U(R) is a group under the multiplication ofR. (This group is called the group of units of R.)

23. Determine U(Z[i]) (see Example 11).24. If R1, R2, . . . , Rn are commutative rings with unity, show that

U(R1 % R2 % ? ? ? % Rn) 5 U(R1) % U(R2) % ? ? ? % U(Rn).25. Determine U(Z[x]). (This exercise is referred to in Chapter 17.)26. Determine U(R[x]).27. Show that a unit of a ring divides every element of the ring.28. In Z6, show that 4 | 2; in Z8, show that 3 | 7; in Z15, show that 9 | 12.

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29. Suppose that a and b belong to a commutative ring R with unity. Ifa is a unit of R and b2 5 0, show that a 1 b is a unit of R.

30. Suppose that there is an integer n . 1 such that xn 5 x for all elementsx of some ring. If m is a positive integer and am 5 0 for some a, showthat a 5 0.

31. Give an example of ring elements a and b with the properties thatab 5 0 but ba 2 0.

32. Let n be an integer greater than 1. In a ring in which xn 5 x for all x,show that ab 5 0 implies ba 5 0.

33. Suppose that R is a ring such that x3 5 x for all x in R. Prove that6x 5 0 for all x in R.

34. Suppose that a belongs to a ring and a4 5 a2. Prove that a2n 5 a2

for all n $ 1.35. Find an integer n . 1 such that an 5 a for all a in Z6. Do the same

for Z10. Show that no such n exists for Zm when m is divisible by thesquare of some prime.

36. Let m and n be positive integers and let k be the least common mul-tiple of m and n. Show that mZ > nZ 5 kZ.

37. Explain why every subgroup of Zn under addition is also a subringof Zn.

38. Is Z6 a subring of Z12?39. Suppose that R is a ring with unity 1 and a is an element of R such

that a2 5 1. Let S 5 {ara | r [ R}. Prove that S is a subring of R.Does S contain 1?

40. Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5

. Prove or disprove that R is a subring

of M2(Z).41. Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5

. Prove or disprove that R is a sub-

ring of M2(Z).

42. Let R 5 . Prove or disprove that R is a subring

of M2(Z).43. Let R 5 Z % Z % Z and S 5 {(a, b, c) [ R | a 1 b 5 c}. Prove or

disprove that S is a subring of R.44. Suppose that there is a positive even integer n such that an 5 a for

all elements a of some ring. Show that 2a 5 a for all a in the ring.

e ca a

b bd ` a, b [ Z f

e c a a 2 b

a 2 b bd ` a, b [ Z f

e c a a 1 b

a 1 b bd ` a, b [ Z f

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45. Let R be a ring with unity 1. Show that S 5 {n ? 1| n [ Z} is a sub-ring of R.

46. Show that 2Z < 3Z is not a subring of Z.47. Determine the smallest subring of Q that contains 1/2. (That is,

find the subring S with the property that S contains 1/2 and, if T isany subring containing 1/2, then T contains S.)

48. Determine the smallest subring of Q that contains 2/3.49. Let R be a ring. Prove that a2 2 b2 5 (a 1 b)(a 2 b) for all a, b in

R if and only if R is commutative.50. Suppose that R is a ring and that a2 5 a for all a in R. Show that R

is commutative. [A ring in which a2 5 a for all a is called aBoolean ring, in honor of the English mathematician George Boole(1815–1864).]

51. Give an example of a Boolean ring with four elements. Give an ex-ample of an infinite Boolean ring.

Computer Exercises

Theory is the general; experiments are the soldiers.LEONARDO DA VINCI

Software for the computer exercises in this chapter is available at the website:

http://www.d.umn.edu/~jgallian

1. This software finds all solutions to the equation x2 1 y2 5 0 in Zp.Run the software for all odd primes up to 37. Make a conjectureabout the existence of nontrivial solutions in Zp ( p a prime) and theform of p.

2. Let Zn[i] 5 {a 1 bi| a, b belong to Zn, i2 5 21} (the Gaussian inte-gers modulo n). This software finds the group of units of this ringand the order of each element of the group. Run the program for n 5 3, 7, 11, and 23. Is the group of units cyclic for these cases? Tryto guess a formula for the order of the group of units of Zn[i] as afunction of n when n is a prime and n mod 4 5 3. Run the programfor n 5 9 and 27. Are the groups cyclic? Try to guess a formula forthe order when n 5 3k. Run the program for n 5 5, 13, 17, and 29.Is the group cyclic for these cases? What is the largest order of anyelement in the group? Try to guess a formula for the order of thegroup of units of Zn[i] as a function of n when n is a prime andn mod 4 5 1. Try to guess a formula for the largest order of any

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element in the group of units of Zn[i] as a function of n when n is aprime and n mod 4 5 1. On the basis of the orders of the elementsof the group of units, try to guess the isomorphism class of thegroup. Run the program for n 5 25. Is this group cyclic? Based onthe number of elements in this group and the orders of the elements,try to guess the isomorphism class of the group.

3. This software determines the isomorphism class of the group ofunits of Zn[i]. Run the program for n 5 5, 13, 17, 29, and 37. Makea conjecture. Run the program for n 5 3, 7, 11, 19, 23, and 31.Make a conjecture. Run the program for n 5 5, 25, and 125. Makea conjecture. Run the program for n 5 13 and 169. Make a conjec-ture. Run the program for n 5 3, 9, and 27. Make a conjecture. Runthe program for n 5 7 and 49. Make a conjecture. Run the programfor n 5 11 and 121. Make a conjecture. Make a conjecture aboutthe case where n 5 pk where p is a prime and p mod 4 5 1. Makea conjecture about the case where n 5 pk where p is a prime andp mod 4 5 3.

4. This software determines the order of the group of units in the ringof 2 3 2 matrices over Zn (that is, the group GL(2, Zn)) and the sub-group SL(2, Zn). Run the program for n 5 2, 3, 5, 7, 11, and 13.What relationship do you see between the order of GL(2, Zn) and theorder of SL(2, Zn) in these cases? Run the program for n 516, 27,25, and 49. Make a conjecture about the relationship between the or-der of GL(2, Zn) and the order of SL(2, Zn) when n is a power of aprime. Run the program for n 5 32. (Notice that when you run theprogram for n 5 32, the table shows the orders for all divisors of 32greater than 1.) How do the orders of the two groups change eachtime you increase the power of 2 by 1? Run the program for n 5 27.How do the orders of the two groups change each time you increasethe power of 3 by 1? Run the program for n 5 25. How do the ordersof the two groups change when you increase the power of 5 by 1?Make a conjecture about the relationship between |SL(2, Zpi)| and|SL(2, Zpi11)|. Make a conjecture about the relationship between|GL(2, Zpt)| and |GL(2, Zpi11)|. Run the program for n 5 12, 15, 20,21, and 30. Make a conjecture about the order of GL(2, Zn) in termsof the orders of GL(2, Zs) and GL(2, Zt) where n 5 st and s and t arerelatively prime. (Notice that when you run the program for st, thetable shows the values for st, s, and t.) For each value of n, is the or-der of SL(2, Zn) divisible by n? Is it divisible by n 1 1? Is it divisibleby n 2 1?

5. In the ring Zn, this software finds the number of solutions to theequation x2 5 21. Run the program for all primes between 3 and 29.

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How does the answer depend on the prime? Make a conjecture aboutthe number of solutions when n is a prime greater than 2. Run theprogram for the squares of all primes between 3 and 29. Make a con-jecture about the number of solutions when n is the square of aprime greater than 2. Run the program for the cubes of primes be-tween 3 and 29. Make a conjecture about the number of solutionswhen n is any power of an odd prime. Run the program for n 5 2, 4,8, 16, and 32. Make a conjecture about the number of solutionswhen n is a power of 2. Run the program for n 5 12, 20, 24, 28,and 36. Make a conjecture about the number of solutions when n is amultiple of 4. Run the program for various cases where n 5 pq andn 5 2pq where p and q are odd primes. Make a conjecture about thenumber of solutions when n 5 pq or n 5 2pq where p and q are oddprimes. What relationship do you see among the numbers of solu-tions for n 5 p, n 5 q, and n 5 pq? Run the program for variouscases where n 5 pqr and n 5 2pqr where p, q, and r are odd primes.Make a conjecture about the number of solutions when n 5 pqr orn 5 2pqr where p, q, and r are odd primes. What relationship do yousee among the numbers of solutions when n 5 p, n 5 q, and n 5 rand the case that n 5 pqr?

6. This software determines the number of solutions to the equationX2 5 2I where X is a 2 3 2 matrix with entries from Zn and I is theidentity. Run the program for n 5 32. Make a conjecture about thenumber of solutions when n 5 2k where k . 1. Run the programfor n 5 3, 11, 19, 23, and 31. Make a conjecture about the numberof solutions when n is a prime of the form 4q 1 3. Run the pro-gram for n 5 27 and 49. Make a conjecture about the number ofsolutions when n has the form pi where p is a prime of the form4q 1 3. Run the program for n 5 5, 13, 17, 29, and 37. Make aconjecture about the number of solutions when n is a prime of theform 4q 1 1. Run the program for n 5 6, 10, 14, 22, 15, 21, 33, 39,30, 42. What seems to be the relationship between the number ofsolutions for a given n and the number of solutions for the primepower factors of n?

Suggested Reading

D. B. Erickson, “Orders for Finite Noncommutative Rings,” AmericanMathematical Monthly 73 (1966): 376–377.

In this elementary paper, it is shown that there exists a noncommutative ringof order m . 1 if and only if m is divisible by the square of a prime.

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I. N. Herstein

A whole generation of textbooks and an entiregeneration of mathematicians, myself included,have been profoundly influenced by that text[Herstein’s Topics in Algebra].

GEORGIA BENKART

I. N. HERSTEIN was born on March 28, 1923,in Poland. His family moved to Canadawhen he was seven. He grew up in a poor andtough environment, on which he commentedthat in his neighborhood you became either agangster or a college professor. During hisschool years he played football, hockey, golf,tennis, and pool. During this time he workedas a steeplejack and as a barber at a fair.Herstein received a B.S. degree from theUniversity of Manitoba, an M.A. from theUniversity of Toronto, and, in 1948, a Ph.D.degree from Indiana University under the su-pervision of Max Zorn. Before permanentlysettling at the University of Chicago in 1962,he held positions at the University of Kansas,the Ohio State University, the University ofPennsylvania, and Cornell University.

Herstein wrote more than 100 researchpapers and a dozen books. Although hisprincipal interest was noncommutative ring

theory, he also wrote papers on finitegroups, linear algebra, and mathematicaleconomics. His textbook Topics in Algebra,first published in 1964, dominated the fieldfor 20 years and has become a classic.Herstein had great influence through histeaching and his collaboration with col-leagues. He had 30 Ph.D. students, andtraveled and lectured widely. His nonmath-ematical interests included languages andart. He spoke Italian, Hebrew, Polish, andPortuguese. Herstein died on February 9,1988, after a long battle with cancer.

To find more information about Herstein,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

248

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249

Integral Domains13

Definition and ExamplesTo a certain degree, the notion of a ring was invented in an attempt toput the algebraic properties of the integers into an abstract setting. Aring is not the appropriate abstraction of the integers, however, for toomuch is lost in the process. Besides the two obvious properties of com-mutativity and existence of a unity, there is one other essential featureof the integers that rings in general do not enjoy—the cancellationproperty. In this chapter, we introduce integral domains—a particularclass of rings that have all three of these properties. Integral domainsplay a prominent role in number theory and algebraic geometry.

Definition Zero-Divisors

A zero-divisor is a nonzero element a of a commutative ring R suchthat there is a nonzero element b [ R with ab 5 0.

Definition Integral Domain

An integral domain is a commutative ring with unity and no zero-divisors.

Thus, in an integral domain, a product is 0 only when one of thefactors is 0; that is, ab 5 0 only when a 5 0 or b 5 0. The followingexamples show that many familiar rings are integral domains and somefamiliar rings are not. For each example, the student should verify theassertion made.

EXAMPLE 1 The ring of integers is an integral domain.

Don’t just read it! Ask your own questions, look for your own examples,discover your own proofs. Is the hypothesis necessary? Is the conversetrue? What happens in the classical special case? Where does the proof use the hypothesis?

PAUL HALMOS

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250 Rings

EXAMPLE 2 The ring of Gaussian integers Z[i] 5 {a 1 bi | a, b [ Z}is an integral domain.

EXAMPLE 3 The ring Z[x] of polynomials with integer coefficientsis an integral domain.

EXAMPLE 4 The ring Z[ ] 5 {a 1 b | a, b [ Z} is an integraldomain.

EXAMPLE 5 The ring Zp of integers modulo a prime p is an integraldomain.

EXAMPLE 6 The ring Zn of integers modulo n is not an integral do-main when n is not prime.

EXAMPLE 7 The ring M2(Z) of 2 3 2 matrices over the integers isnot an integral domain.

EXAMPLE 8 Z % Z is not an integral domain.

What makes integral domains particularly appealing is that they havean important multiplicative group-theoretic property, in spite of the factthat the nonzero elements need not form a group under multiplication.This property is cancellation.

Theorem 13.1 Cancellation

"2"2

Let a, b, and c belong to an integral domain. If a 2 0 and ab 5 ac,then b 5 c.

PROOF From ab 5 ac, we have a(b 2 c) 5 0. Since a 2 0, we musthave b 2 c 5 0.

Many authors prefer to define integral domains by the cancellationproperty—that is, as commutative rings with unity in which the cancel-lation property holds. This definition is equivalent to ours.

FieldsIn many applications, a particular kind of integral domain called a fieldis necessary.

Definition Field

A field is a commutative ring with unity in which every nonzeroelement is a unit.

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13 | Integral Domains 251

To verify that every field is an integral domain, observe that if a andb belong to a field with a 2 0 and ab 5 0, we can multiply both sidesof the last expression by a21 to obtain b 5 0.

It is often helpful to think of ab21 as a divided by b. With this inmind, a field can be thought of as simply an algebraic system that is closed under addition, subtraction, multiplication, and division(except by 0). We have had numerous examples of fields: the complexnumbers, the real numbers, the rational numbers. The abstract theoryof fields was initiated by Heinrich Weber in 1893. Groups, rings, andfields are the three main branches of abstract algebra. Theorem 13.2says that, in the finite case, fields and integral domains are the same.

Theorem 13.2 Finite Integral Domains Are Fields

A finite integral domain is a field.

PROOF Let D be a finite integral domain with unity 1. Let a be anynonzero element of D. We must show that a is a unit. If a 5 1, a is itsown inverse, so we may assume that a 2 1. Now consider the followingsequence of elements of D: a, a2, a3, . . . . Since D is finite, there mustbe two positive integers i and j such that i . j and ai 5 a j. Then, by can-cellation, ai2j 5 1. Since a 2 1, we know that i 2 j . 1, and we haveshown that ai2j21 is the inverse of a.

Corollary Zp Is a Field

For every prime p, Zp, the ring of integers modulo p, is a field.

PROOF According to Theorem 13.2, we need only prove that Zp hasno zero-divisors. So, suppose that a, b [ Zp and ab 5 0. Then ab 5 pkfor some integer k. But then, by Euclid’s Lemma (see Chapter 0), pdivides a or p divides b. Thus, in Zp, a 5 0 or b 5 0.

Putting the preceding corollary together with Example 6, we see thatZn is a field if and only if n is prime. In Chapter 22, we will describehow all finite fields can be constructed. For now, we give one exampleof a finite field that is not of the form Zp.

EXAMPLE 9 Field with Nine ElementsLet Z3[i] 5 {a 1 bi | a, b [ Z3}

5 {0, 1, 2, i, 1 1 i, 2 1 i, 2i, 1 1 2i, 2 1 2i},

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252 Rings

where i2 5 21. This is the ring of Gaussian integers modulo 3. Ele-ments are added and multiplied as in the complex numbers, except thatthe coefficients are reduced modulo 3. In particular, 21 5 2. Table 13.1is the multiplication table for the nonzero elements of Z3[i].

Table 13.1 Multiplication Table for Z3[i]*

1 2 i 1 1 i 2 1 i 2i 1 1 2i 2 1 2i

1 1 2 i 1 1 i 2 1 i 2i 1 1 2i 2 1 2i2 2 1 2i 2 1 2i 1 1 2i i 2 1 i 1 1 ii i 2i 2 2 1 i 2 1 2i 1 1 1 i 1 1 2i1 1 i 1 1 i 2 1 2i 2 1 i 2i 1 1 1 2i 2 i2 1 i 2 1 i 1 1 2i 2 1 2i 1 i 1 1 i 2i 22i 2i i 1 1 1 2i 1 1 i 2 2 1 2i 2 1 i1 1 2i 1 1 2i 2 1 i 1 1 i 2 2i 2 1 2i i 12 1 2i 2 1 2i 1 1 i 1 1 2i i 2 2 1 i 1 2i

EXAMPLE 10 Let Q[ ] 5 {a 1 b | a, b [ Q}. It is easy to seethat Q[ ] is a ring. Viewed as an element of R, the multiplicative in-verse of any nonzero element of the form a 1 b is simply 1/(a 1b ). To verify that Q[ ] is a field, we must show that 1/(a 1 b )can be written in the form c 1 d . In high school algebra, this processis called “rationalizing the denominator.” Specifically,

.

(Note that a 1 b 2 0 guarantees that a 2 b 2 0.)

Characteristic of a RingNote that for any element x in Z3[i], we have 3x 5 x 1 x 1 x 5 0, sinceaddition is done modulo 3. Similarly, in the subring {0, 3, 6, 9} of Z12,we have 4x 5 x 1 x 1 x 1 x 5 0 for all x. This observation motivatesthe following definition.

Definition Characteristic of a Ring

The characteristic of a ring R is the least positive integer n such thatnx 5 0 for all x in R. If no such integer exists, we say that R has char-acteristic 0. The characteristic of R is denoted by char R.

Thus, the ring of integers has characteristic 0, and Zn has character-istic n. An infinite ring can have a nonzero characteristic. Indeed, the

"2"2

1

a 1 b"25

1

a 1 b"2 a 2 b"2

a 2 b"25

a

a2 2 2b2 2b

a2 2 2b2 "2

"2"2"2"2

"2"2

"2"2

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13 | Preliminaries 253

Let R be a ring with unity 1. If 1 has infinite order under addition,then the characteristic of R is 0. If 1 has order n under addition,then the characteristic of R is n.

PROOF If 1 has infinite order, then there is no positive integer n suchthat n ? 1 5 0, so R has characteristic 0. Now suppose that 1 has addi-tive order n. Then n ? 1 5 0, and n is the least positive integer with thisproperty. So, for any x in R, we have

n ? x 5 x 1 x 1 ? ? ? 1 x (n summands)5 1x 1 1x 1 ? ? ? 1 1x (n summands)5 (1 1 1 1 ? ? ? 1 1)x (n summands)5 (n ? 1)x 5 0x 5 0.

Thus, R has characteristic n.

In the case of an integral domain, the possibilities for the character-istic are severely limited.

Theorem 13.4 Characteristic of an Integral Domain

The characteristic of an integral domain is 0 or prime.

PROOF By Theorem 13.3, it suffices to show that if the additive orderof 1 is finite, it must be prime. Suppose that 1 has order n and that n 5 st,where 1 # s, t # n. Then, by Exercise 15 in Chapter 12,

0 5 n ? 1 5 (st) ? 1 5 (s ? 1)(t ? 1).

So, s ? 1 5 0 or t ? 1 5 0. Since n is the least positive integer with theproperty that n ? 1 5 0, we must have s 5 n or t 5 n. Thus, n is prime.

We conclude this chapter with a brief discussion of polynomialswith coefficients from a ring—a topic we will consider in detail in

ring Z2[x] of all polynomials with coefficients in Z2 has characteristic 2.(Addition and multiplication are done as for polynomials with ordinaryinteger coefficients except that the coefficients are reduced modulo 2.)When a ring has a unity, the task of determining the characteristic issimplified by Theorem 13.3.

Theorem 13.3 Characteristic of a Ring with Unity

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254 Rings

later chapters. The existence of zero-divisors in a ring causes unusualresults when one is finding zeros of polynomials with coefficients inthe ring. Consider, for example, the equation x2 2 4x 1 3 5 0. In theintegers, we could find all solutions by factoring

x2 2 4x 1 3 5 (x 2 3)(x 2 1) 5 0

and setting each factor equal to 0. But notice that when we say we canfind all solutions in this manner, we are using the fact that the only wayfor a product to equal 0 is for one of the factors to be 0—that is, we areusing the fact that Z is an integral domain. In Z12, there are many pairs ofnonzero elements whose products are 0: 2 ? 6 5 0, 3 ? 4 5 0, 4 ? 6 5 0,6 ? 8 5 0, and so on. So, how do we find all solutions of x2 2 4x 1 3 5 0in Z12? The easiest way is simply to try every element! Upon doing so,we find four solutions: x 5 1, x 5 3, x 5 7, and x 5 9. Observe that wecan find all solutions of x2 2 4x 1 3 5 0 over Z11 or Z13, say, by settingthe two factors x 2 3 and x 2 1 equal to 0. Of course, the reason thisworks for these rings is that they are integral domains. Perhaps this willconvince you that integral domains are particularly advantageous rings.Table 13.2 gives a summary of some of the rings we have introduced andtheir properties.

Table 13.2 Summary of Rings and Their Properties

IntegralRing Form of Element Unity Commutative Domain Field Characteristic

Z k 1 Yes Yes No 0

Zn, n composite k 1 Yes No No n

Zp, p prime k 1 Yes Yes Yes p

Z[x] anxn 1 ? ? ? 1 f(x) 5 1 Yes Yes No 0

a1x 1 a0

nZ, n . 1 nk None Yes No No 0

M2(Z) No No No 0

M2(2Z) None No No No 0

Z[i] a 1 bi 1 Yes Yes No 0

Z3[i] a 1 bi; a, b [ Z3 1 Yes Yes Yes 3

Z[ ] a 1 b ; a, b [ Z 1 Yes Yes No 0

Q[ ] a 1 b ; a, b [ Q 1 Yes Yes Yes 0

Z % Z (a, b) (1, 1) Yes No No 0

"2"2"2"2

c2a 2b

2c 2dd

c1 0

0 1dca b

c dd

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13 | Integral Domains 255

Exercises

It looked absolutely impossible. But it so happens that you go on worryingaway at a problem in science and it seems to get tired, and lies down andlets you catch it.

WILLIAM LAWRENCE BRAGG†

1. Verify that Examples 1 through 8 are as claimed.2. Which of Examples 1 through 5 are fields?3. Show that a commutative ring with the cancellation property

(under multiplication) has no zero-divisors.4. List all zero-divisors in Z20. Can you see a relationship between the

zero-divisors of Z20 and the units of Z20?5. Show that every nonzero element of Zn is a unit or a zero-divisor.6. Find a nonzero element in a ring that is neither a zero-divisor nor a

unit.7. Let R be a finite commutative ring with unity. Prove that every

nonzero element of R is either a zero-divisor or a unit. What hap-pens if we drop the “finite” condition on R?

8. Describe all zero-divisors and units of Z % Q % Z.9. Let d be an integer. Prove that Z[ ] 5 {a 1 b | a, b [ Z} is

an integral domain. (This exercise is referred to in Chapter 18.)10. In Z7, give a reasonable interpretation for the expressions 1/2,

22/3, , and 21/6.11. Give an example of a commutative ring without zero-divisors that

is not an integral domain.12. Find two elements a and b in a ring such that both a and b are zero-

divisors, a 1 b 2 0, and a 1 b is not a zero-divisor.13. Let a belong to a ring R with unity and suppose that an 5 0 for

some positive integer n. (Such an element is called nilpotent.)Prove that 1 2 a has a multiplicative inverse in R. [Hint: Consider(1 2 a)(1 1 a 1 a2 1 ? ? ? 1 an21).]

14. Show that the nilpotent elements of a commutative ring form asubring.

15. Show that 0 is the only nilpotent element in an integral domain.16. A ring element a is called an idempotent if a2 5 a. Prove that the

only idempotents in an integral domain are 0 and 1.

"23

"d"d

†Bragg, at age 24, won the Nobel Prize for the invention of x-ray crystallography. Heremains the youngest person ever to receive the Nobel Prize.

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256 Rings

17. Let a and b be idempotents in a commutative ring. Show that eachof the following is also an idempotent: ab, , ,

.18. Prove that if a is a ring idempotent, then for all positive inte-

gers n.19. Determine all ring elements that are both nilpotents and idempotents.20. Find a zero-divisor in Z5[i] 5 {a 1 bi | a, b [ Z5}.21. Find an idempotent in Z5[i] 5 {a 1 bi | a, b [ Z5}.22. Find all units, zero-divisors, idempotents, and nilpotent elements

in Z3 % Z6.23. Determine all elements of a ring that are both units and idempotents.24. Let R be the set of all real-valued functions defined for all real

numbers under function addition and multiplication.a. Determine all zero-divisors of R.b. Determine all nilpotent elements of R.c. Show that every nonzero element is a zero-divisor or a unit.

25. (Subfield Test) Let F be a field and let K be a subset of F with atleast two elements. Prove that K is a subfield of F if, for any a, b (b 2 0) in K, a 2 b and ab21 belong to K.

26. Let d be a positive integer. Prove that Q[ ] 5 {a 1 b |a, b [ Q} is a field.

27. Let R be a ring with unity 1. If the product of any pair of nonzeroelements of R is nonzero, prove that ab 5 1 implies ba 5 1.

28. Let R 5 {0, 2, 4, 6, 8} under addition and multiplication modulo10. Prove that R is a field.

29. Formulate the appropriate definition of a subdomain (that is, a“sub” integral domain). Let D be an integral domain with unity 1.Show that P 5 { | n [ Z} (that is, all integral multiples of 1)is a subdomain of D. Show that P is contained in every subdomainof D. What can we say about the order of P?

30. Prove that there is no integral domain with exactly six elements. Canyour argument be adapted to show that there is no integral domainwith exactly four elements? What about 15 elements? Use these ob-servations to guess a general result about the number of elements ina finite integral domain.

31. Let F be a field of order 2n. Prove that char F 5 2.32. Determine all elements of an integral domain that are their own in-

verses under multiplication.33. Characterize those integral domains for which 1 is the only ele-

ment that is its own multiplicative inverse.

n ? 1

"d"d

an 5 aa 1 b 2 2ab

a 1 b 2 aba 2 ab

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13 | Integral Domains 257

34. Determine all integers for which ! is a zero-divisorin .

35. Suppose that a and b belong to an integral domain.a. If a5 5 b5 and a3 5 b3, prove that a 5 b.b. If am 5 bm and an 5 bn, where m and n are positive integers that

are relatively prime, prove that a 5 b.36. Find an example of an integral domain and distinct positive inte-

gers m and n such that am 5 bm and an 5 bn, but a 2 b.37. If a is an idempotent in a commutative ring, show that 1 2 a is also

an idempotent.38. Construct a multiplication table for Z2[i], the ring of Gaussian in-

tegers modulo 2. Is this ring a field? Is it an integral domain?39. The nonzero elements of Z3[i] form an Abelian group of order 8 un-

der multiplication. Is it isomorphic to Z8, Z4 % Z2, or Z2 % Z2 % Z2?

40. Show that Z 7[ ] 5 {a 1 b | a, b [ Z 7} is a field. For anypositive integer k and any prime p, determine a necessary and suf-ficient condition for Zp[ ] 5 {a 1 b | a, b [ Zp} to be a field.

41. Show that a finite commutative ring with no zero-divisors and atleast two elements has a unity.

42. Suppose that a and b belong to a commutative ring and ab is azero-divisor. Show that either a or b is a zero-divisor.

43. Suppose that R is a commutative ring without zero-divisors. Showthat all the nonzero elements of R have the same additive order.

44. Suppose that R is a commutative ring without zero-divisors. Showthat the characteristic of R is 0 or prime.

45. Let x and y belong to a commutative ring R with prime character-istic p.a. Show that (x 1 y)p 5 xp 1 yp.b. Show that, for all positive integers n, (x 1 y)pn

5 xpn1 ypn

.c. Find elements x and y in a ring of characteristic 4 such that

(x 1 y)4 2 x4 1 y4. (This exercise is referred to in Chapter 20.)

46. Let R be a commutative ring with unity 1 and prime characteristic.If a [ R is nilpotent, prove that there is a positive integer k such that(1 1 a)k 5 1.

47. Show that any finite field has order pn, where p is a prime. Hint: Usefacts about finite Abelian groups. (This exercise is referred to inChapter 22.)

48. Give an example of an infinite integral domain that has character-istic 3.

49. Let R be a ring and let M2(R) be the ring of 2 3 2 matrices with entriesfrom R. Explain why these two rings have the same characteristic.

"k"k

"3"3

Zn

(n 2 1)n . 1

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258 Rings

50. Let R be a ring with m elements. Show that the characteristic of Rdivides m.

51. Explain why a finite ring must have a nonzero characteristic.52. Find all solutions of x2 2 x 1 2 5 0 over Z3[i]. (See Example 9.)53. Consider the equation x2 2 5x 1 6 5 0.

a. How many solutions does this equation have in Z7?b. Find all solutions of this equation in Z8.c. Find all solutions of this equation in Z12.d. Find all solutions of this equation in Z14.

54. Find the characteristic of Z4 % 4Z.55. Suppose that R is an integral domain in which 20 ? 1 5 0 and

12 ? 1 5 0. (Recall that n ? 1 means the sum 1 1 1 1 ? ? ? 1 1 withn terms.) What is the characteristic of R?

56. In a commutative ring of characteristic 2, prove that the idempo-tents form a subring.

57. Describe the smallest subfield of the field of real numbers that con-tains . (That is, describe the subfield K with the property that Kcontains and if F is any subfield containing , then F con-tains K.)

58. Let F be a finite field with n elements. Prove that xn21 5 1 for allnonzero x in F.

59. Let F be a field of prime characteristic p. Prove that K 5 {x [ F |xp 5 x} is a subfield of F.

60. Suppose that a and b belong to a field of order 8 and that a2 1 ab 1b2 5 0. Prove that a 5 0 and b 5 0. Do the same when the field hasorder 2n with n odd.

61. Let F be a field of characteristic 2 with more than two elements.Show that (x 1 y)3 2 x3 1 y3 for some x and y in F.

62. Suppose that F is a field with characteristic not 2, and that thenonzero elements of F form a cyclic group under multiplication.Prove that F is finite.

63. Suppose that D is an integral domain and that f is a nonconstantfunction from D to the nonnegative integers such that f(xy) 5f(x)f(y). If x is a unit in D, show that f(x) 5 1.

64. Let F be a field of order 32. Show that the only subfields of F areF itself and {0, 1}.

65. Suppose that F is a field with 27 elements. Show that for everyelement , .5a 5 2aa [ F

"2"2"2

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13 | Integral Domains 259

Computer Exercises

The basic unit of mathematics is conjecture.ARNOLD ROSS

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software lists the idempotents (see Exercise 16 for the defini-tion) in Zn. Run the program for various values of n. Use these datato make conjectures about the number of idempotents in Zn as afunction of n. For example, how many idempotents are there whenn is a prime power? What about when n is divisible by exactly twodistinct primes? In the case where n is of the form pq where p andq are primes, can you see a relationship between the two idempo-tents that are not 0 and 1? Can you see a relationship between thenumber of idempotents for a given n and the number of distinctprime divisors of n?

2. This software lists the nilpotent elements (see Exercise 13 for thedefinition) in Zn. Run the program for various values of n. Usethese data to make conjectures about the number of nilpotent ele-ments in Zn as a function of n.

3. This software determines which rings of the form Zp[i] are fields.Run the program for all primes up to 37. From these data, make aconjecture about the form of the primes that yield a field.

4. This software finds the idempotents in Zn[i] 5 {a 1 bi | a, b [ Zn}(Gaussian integers modulo n). Run the software for n 5 4, 8, 16, and32. Make a conjecture about the number of idempotents when n 5 2k.Run the software for n 5 13, 17, 29, and 37. What do these values ofn have in common? Make a conjecture about the number of idempo-tents for these n. Run the software for n 5 7, 11, 19, 23, 31, and 43.What do these values of n have in common? Make a conjecture aboutthe number of idempotents for these n.

5. This software finds the nilpotent elements in Zn[i] 5 {a 1 bi | a, b [ Zn}. Run the software for n 5 4, 8, 16, and 32. Make a con-jecture about the number of nilpotent elements when n 5 2k. Run thesoftware for n 5 3, 5, 7, 11, 13, and 17. What do these values of nhave in common? Make a conjecture about the number of nilpotentelements for these n. Run the program for n 5 9. Do you need torevise the conjecture you made based on n 5 3, 5, 7, 11, 13, and 17?Run the software for n 5 9, 25, and 49. What do these values

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260 Rings

of n have in common? Make a conjecture about the number of nilpo-tent elements for these n. Run the program for n 5 81. Do you needto revise the conjecture you made based on n 59, 25, and 49?What do these values of n have in common? Make a conjectureabout the number of nilpotent elements for these n. Run the pro-gram for n 5 27. Do you need to revise the conjecture you madebased on n 5 9, 25, and 49? Run your program for n 5 125 (thismay take a few seconds). On the basis of all of your data for this ex-ercise, make a single conjecture in the case that n 5 pk where p isany prime. Run the program for n 5 6, 15, and 21. Make a conjec-ture. Run the program for 12, 20, 28, and 45. Make a conjecture.Run the program for 36 and 100 (this may take a few minutes). Onthe basis of all your data for this exercise, make a single conjecturethat covers all integers n . 1.

6. This software determines the zero-divisors in Zn[i] 5 {a 1 bi | a,b [ Zn}. Use the software to formulate and test conjectures about thenumber of zero-divisors in Zn[i] based on various conditions of n.

Suggested Readings

Eric Berg, “A Family of Fields,” Pi Mu Epsilon 9 (1990): 154–155.

In this article, the author uses properties of logarithms and exponents to define recursively an infinite family of fields starting with the realnumbers.

N. A. Khan, “The Characteristic of a Ring,” American Mathematical Monthly70 (1963): 736.

Here it is shown that a ring has nonzero characteristic n if and only if n is the maximum of the orders of the elements of R.

K. Robin McLean, “Groups in Modular Arithmetic,” The MathematicalGazette 62 (1978): 94–104.

This article explores the interplay between various groups of integers un-der multiplication modulo n and the ring Zn. It shows how to constructgroups of integers in which the identity is not obvious; for example, 1977is the identity of the group {1977, 5931} under multiplication modulo7908.

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261

Nathan Jacobson

NATHAN JACOBSON was born on September 8,1910, in Warsaw, Poland. After arriving inthe United States in 1917, Jacobson grew upin Alabama, Mississippi, and Georgia, wherehis father owned small clothing stores. Hereceived a B.A. degree from the University ofAlabama in 1930 and a Ph.D. from Princetonin 1934. After brief periods as a professor atBryn Mawr, the University of Chicago, theUniversity of North Carolina, and JohnsHopkins, Jacobson accepted a position atYale, where he remained until his retirementin 1981.

Jacobson’s principal contributions to al-gebra were in the areas of rings, Lie algebras,and Jordan algebras. In particular, he devel-oped structure theories for these systems. Hewas the author of nine books and numerousarticles, and he had 33 Ph.D. students.

Few mathematicians have been as produc-tive over such a long career or have had asmuch influence on the profession as hasProfessor Jacobson.

Citation for the Steele Prize

for Lifetime Achievement

Jacobson held visiting positions inFrance, India, Italy, Israel, China, Australia,and Switzerland. Among his many honorswere the presidency of the AmericanMathematical Society, memberships in theNational Academy of Sciences and theAmerican Academy of Arts and Sciences, aGuggenheim Fellowship, and an honorarydegree from the University of Chicago.Jacobson died on December 5, 1999, at theage of 89.

To find more information about Jacobson,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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262

Ideals and Factor Rings

The secret of science is to ask the right questions, and it is the choice ofproblem more than anything else that marks the man of genius in thescientific world.

SIR HENRY TIZARD IN C. P. SNOW,

A postscript to Science and Government

14

IdealsNormal subgroups play a special role in group theory—they permit usto construct factor groups. In this chapter, we introduce the analogousconcepts for rings—ideals and factor rings.

Definition Ideal

A subring A of a ring R is called a (two-sided) ideal of R if for every r [ R and every a [ A both ra and ar are in A.

So, a subring A of a ring R is an ideal of R if A “absorbs” elementsfrom R—that is, if rA 5 {ra| a [ A} # A and Ar 5 {ar| a [ A} # Afor all r [ R.

An ideal A of R is called a proper ideal of R if A is a proper subsetof R. In practice, one identifies ideals with the following test, which isan immediate consequence of the definition of ideal and the subringtest given in Theorem 12.3.

Theorem 14.1 Ideal Test

A nonempty subset A of a ring R is an ideal of R if

1. a 2 b [ A whenever a, b [ A.2. ra and ar are in A whenever a [ A and r [ R.

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14 | Ideals and Factor Rings 263

EXAMPLE 1 For any ring R, {0} and R are ideals of R. The ideal {0}is called the trivial ideal.

EXAMPLE 2 For any positive integer n, the set nZ 5 {0, 6n,62n, . . .} is an ideal of Z.

EXAMPLE 3 Let R be a commutative ring with unity and let a [ R.The set �a� 5 {ra | r [ R} is an ideal of R called the principal idealgenerated by a. (Notice that �a� is also the notation we used for the cyclic subgroup generated by a. However, the intended meaningwill always be clear from the context.) The assumption that R is com-mutative is necessary in this example (see Exercise 29 in the Sup-plementary Exercises for Chapters 12–14).

EXAMPLE 4 Let R[x] denote the set of all polynomials with real co-efficients and let A denote the subset of all polynomials with constantterm 0. Then A is an ideal of R[x] and A 5 �x�.

EXAMPLE 5 Let R be a commutative ring with unity and let a1,a2, . . . , an belong to R. Then I 5 �a1, a2, . . . , an� 5 {r1a1 1 r2a2 1? ? ? 1 rnan | ri [ R} is an ideal of R called the ideal generated by a1,a2, . . . , an. The verification that I is an ideal is left as an easy exercise(Exercise 3).

EXAMPLE 6 Let Z[x] denote the ring of all polynomials with inte-ger coefficients and let I be the subset of Z[x] of all polynomials witheven constant terms. Then I is an ideal of Z[x] and I 5 �x, 2� (seeExercise 37).

EXAMPLE 7 Let R be the ring of all real-valued functions of a realvariable. The subset S of all differentiable functions is a subring of Rbut not an ideal of R.

Factor RingsLet R be a ring and let A be an ideal of R. Since R is a group under addi-tion and A is a normal subgroup of R, we may form the factor groupR/A 5 {r 1 A | r [ R}. The natural question at this point is: How maywe form a ring of this group of cosets? The addition is already taken careof, and, by analogy with groups of cosets, we define the product of twocosets of s 1 A and t 1 A as st 1 A. The next theorem shows that this de-finition works as long as A is an ideal, and not just a subring, of R.

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264 Rings

Theorem 14.2 Existence of Factor Rings

PROOF We know that the set of cosets forms a group under addition.Once we know that multiplication is indeed a binary operation on thecosets, it is trivial to check that the multiplication is associative andthat multiplication is distributive over addition. Hence, the proof boilsdown to showing that multiplication is well defined if and only if A isan ideal of R. To do this, let us suppose that A is an ideal and let s 1 A 5s9 1 A and t 1 A 5 t9 1 A. Then we must show that st 1 A 5 s9t9 1 A.Well, by definition, s 5 s9 1 a and t 5 t9 1 b, where a and b belongto A. Then

st 5 (s9 1 a)(t9 1 b) 5 s9t9 1 at9 1 s9b 1 ab,

and so

st 1 A 5 s9t9 1 at9 1 s9b 1 ab 1 A 5 s9t9 1 A,

since A absorbs at9 1 s9b 1 ab. Thus, multiplication is well definedwhen A is an ideal.

On the other hand, suppose that A is a subring of R that is not anideal of R. Then there exist elements a [ A and r [ R such that ar o Aor ra o A. For convenience, say ar o A. Consider the elements a 1 A 50 1 A and r 1 A. Clearly, (a 1 A)(r 1 A) 5 ar 1 A but (0 1 A) ?(r 1 A) 5 0 ? r 1 A 5 A. Since ar 1 A 2 A, the multiplication is notwell defined and the set of cosets is not a ring.

Let’s look at a few factor rings.

EXAMPLE 8 Z/4Z 5 {0 1 4Z, 1 1 4Z, 2 1 4Z, 3 1 4Z}. To see howto add and multiply, consider 2 1 4Z and 3 1 4Z.

(2 1 4Z) 1 (3 1 4Z) 5 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z,(2 1 4Z)(3 1 4Z) 5 6 1 4Z 5 2 1 4 1 4Z 5 2 1 4Z.

One can readily see that the two operations are essentially modulo 4arithmetic.

Let R be a ring and let A be a subring of R. The set of cosets {r 1 A |r [ R} is a ring under the operations (s 1 A) 1 (t 1 A) 5 s 1 t 1 Aand (s 1 A)(t 1 A) 5 st 1 A if and only if A is an ideal of R.

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14 | Ideals and Factor Rings 265

EXAMPLE 9 2Z/6Z 5 {0 1 6Z, 2 1 6Z, 4 1 6Z}. Here the opera-tions are essentially modulo 6 arithmetic. For example, (4 1 6Z) 1(4 1 6Z) 5 2 1 6Z and (4 1 6Z)(4 1 6Z) 5 4 1 6Z.

Here is a noncommutative example of an ideal and factor ring.

EXAMPLE 10 Let R 5 and let I be the

subset of R consisting of matrices with even entries. It is easy to show that I is indeed an ideal of R (Exercise 21). Consider the factor ring R/I. The interesting question about this ring is: What is its size?

We claim R/I has 16 elements; in fact, R/I 5 .

An example illustrates the typical situation. Which of the 16 elements

is ? Well, observe that

, since an ideal absorbs its own elements.

The general case is left to the reader (Exercise 23).

EXAMPLE 11 Consider the factor ring of the Gaussian integersR 5 Z[i]/�2 2 i�. What does this ring look like? Of course, the elementsof R have the form a 1 bi 1 �2 2 i�, where a and b are integers, but theimportant question is: What do the distinct cosets look like? The factthat 2 2 i 1 �2 2 i� 5 0 1 �2 2 i� means that when dealing with cosetrepresentatives, we may treat 2 2 i as equivalent to 0, so that 2 5 i. Forexample, the coset 3 1 4i 1 �2 2 i� 5 3 1 8 1 �2 2 i� 5 11 1 �2 2 i�.Similarly, all the elements of R can be written in the form a 1 �2 2 i�,where a is an integer. But we can further reduce the set of distinct cosetrepresentatives by observing that when dealing with coset representa-tives, 2 5 i implies (by squaring both sides) that 4 5 21 or 5 5 0.Thus, the coset 3 1 4i 1 �2 2 i� 5 11 1 �2 2 i� 5 1 1 5 1 5 1 �2 2 i� 51 1 �2 2 i�. In this way, we can show that every element of R is equal toone of the following cosets: 0 1 �2 2 i�, 1 1 �2 2 i�, 2 1 �2 2 i�, 3 1�2 2 i�, 4 1 �2 2 i�. Is any further reduction possible? To demonstratethat there is not, we will show that these five cosets are distinct. It suf-fices to show that 1 1 �2 2 i� has additive order 5. Since 5(1 1 �2 2 i�) 55 1 �2 2 i� 5 0 1 �2 2 i�, 1 1 �2 2 i� has order 1 or 5. If the order isactually 1, then 1 1 �2 2 i� 5 0 1 �2 2 i�, so 1 [ �2 2 i�. Thus, 1 5(2 2 i) (a 1 bi) 5 2a 1 b 1 (2a 1 2b)i for some integers a and b. Butthis equation implies that 1 5 2a 1 b and 0 5 2a 1 2b, and solving these

c6 8

4 24d 1 I 5 c1 0

1 1d 1 I

c7 8

5 23d 1 I 5 c1 0

1 1d 1c7 8

5 23d 1 I

ecr1 r2

r3 r4d1I 0 ri[ 50, 16f

e c a1 a2

a3 a4d ` ai P Z f

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266 Rings

simultaneously yields b 5 1/5, which is a contradiction. It should beclear that the ring R is essentially the same as the field Z5.

EXAMPLE 12 Let R[x] denote the ring of polynomials with real co-efficients and let �x2 1 1� denote the principal ideal generated by x2 1 1; that is,

�x2 1 1� 5 {f(x)(x2 1 1) | f(x) [ R[x]}.

Then

R[x]/�x2 1 1� 5 {g(x) 1 �x2 1 1� | g(x) [ R[x]}5 {ax 1 b 1 �x2 1 1� | a, b [ R}.

To see this last equality, note that if g(x) is any member of R[x], thenwe may write g(x) in the form q(x)(x2 1 1) 1 r(x), where q(x) is thequotient and r(x) is the remainder upon dividing g(x) by x2 1 1. Inparticular, r(x) 5 0 or the degree of r(x) is less than 2, so that r(x) 5ax 1 b for some a and b in R. Thus,

g(x) 1 �x2 1 1� 5 q(x)(x2 1 1) 1 r(x) 1 �x2 1 1�

5 r(x) 1 �x2 1 1�,

since the ideal �x2 1 1� absorbs the term q(x)(x2 1 1).How is multiplication done? Since

x2 1 1 1 �x2 1 1� 5 0 1 �x2 1 1�,

one should think of x2 1 1 as 0 or, equivalently, as x2 5 21. So, forexample,

(x 1 3 1 �x2 1 1�) ? (2x 1 5 1 �x2 1 1�)5 2x2 1 11x 1 15 1 �x2 1 1� 5 11x 1 13 1 �x2 1 1�.

In view of the fact that the elements of this ring have the form ax 1b 1 �x2 1 1�, where x2 1 �x2 1 1� 5 21 1 �x2 1 1�, it is perhaps notsurprising that this ring turns out to be algebraically the same ring asthe ring of complex numbers. This observation was first made byCauchy in 1847.

Examples 11 and 12 illustrate one of the most important applica-tions of factor rings—the construction of rings with highly desirableproperties. In particular, we shall show how one may use factor ringsto construct integral domains and fields.

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14 | Ideals and Factor Rings 267

Prime Ideals and Maximal IdealsDefinition Prime Ideal, Maximal Ideal

A prime ideal A of a commutative ring R is a proper ideal of R suchthat a, b [ R and ab [ A imply a [ A or b [ A. A maximal ideal of acommutative ring R is a proper ideal of R such that, whenever B is anideal of R and A # B # R, then B 5 A or B 5 R.

So, the only ideal that properly contains a maximal ideal is the en-tire ring. The motivation for the definition of a prime ideal comes fromthe integers.

EXAMPLE 13 Let n be an integer greater than 1. Then, in the ring ofintegers, the ideal nZ is prime if and only if n is prime (Exercise 9).({0} is also a prime ideal of Z.)

EXAMPLE 14 The lattice of ideals of Z36 (Figure 14.1) shows thatonly �2� and �3� are maximal ideals.

EXAMPLE 15 The ideal �x2 1 1� is maximal in R[x]. To see this, as-sume that A is an ideal of R[x] that properly contains �x2 1 1�. We willprove that A 5 R[x] by showing that A contains some nonzero realnumber c. [This is the constant polynomial h(x) 5 c for all x.] Then 1 5(1/c)c [ A and therefore, by Exercise 15, A 5 R[x]. To this end, letf(x) [ A, but f(x) o �x2 1 1�. Then

f(x) 5 q(x)(x2 1 1) 1 r(x),

where r(x) 2 0 and the degree of r(x) is less than 2. It follows that r(x) 5 ax 1 b, where a and b are not both 0, and

ax 1 b 5 r(x) 5 f(x) 2 q(x)(x2 1 1) [ A.

Figure 14.1

<2>

<4>

<12> <18>

<6>

<3>

<9>

<0>

Z36

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268 Rings

Thus,

a2x2 2 b2 5 (ax 1 b)(ax 2 b) [ A and a2(x2 1 1) [ A.

So,

0 2 a2 1 b2 5 (a2x2 1 a2) 2 (a2x2 2 b2) [ A.

EXAMPLE 16 The ideal �x2 1 1� is not prime in Z2[x], since it con-tains (x 1 1)2 5 x2 1 2x 1 1 5 x2 1 1 but does not contain x 1 1.

The next two theorems are useful for determining whether a particu-lar ideal is prime or maximal.

Theorem 14.3 R/A Is an Integral Domain If and Only If A Is Prime

PROOF Suppose that R/A is an integral domain and ab [ A. Then (a 1 A)(b 1 A) 5 ab 1 A 5 A, the zero element of the ring R/A. So,either a 1 A 5 A or b 1 A 5 A; that is, either a [ A or b [ A. Hence,A is prime.

To prove the other half of the theorem, we first observe that R/A is acommutative ring with unity for any proper ideal A. Thus, our task issimply to show that when A is prime, R/A has no zero-divisors. So, sup-pose that A is prime and (a 1 A)(b 1 A) 5 0 1 A 5 A. Then ab [ Aand, therefore, a [ A or b [ A. Thus, one of a 1 A or b 1 A is the zerocoset in R/A.

For maximal ideals, we can do even better.

Theorem 14.4 R/A Is a Field If and Only If A Is Maximal

PROOF Suppose that R/A is a field and B is an ideal of R that properlycontains A. Let b [ B but b o A. Then b 1 A is a nonzero elementof R/A and, therefore, there exists an element c 1 A such that (b 1 A) ? (c 1 A) 5 1 1 A, the multiplicative identity of R/A. Since b [ B, we have bc [ B. Because

1 1 A 5 (b 1 A)(c 1 A) 5 bc 1 A,

Let R be a commutative ring with unity and let A be an ideal of R.Then R/A is a field if and only if A is maximal.

Let R be a commutative ring with unity and let A be an ideal of R.Then R/A is an integral domain if and only if A is prime.

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14 | Ideals and Factor Rings 269

we have 1 2 bc [ A , B. So, 1 5 (1 2 bc) 1 bc [ B. By Exercise 15,B 5 R. This proves that A is maximal.

Now suppose that A is maximal and let b [ R but b o A. It sufficesto show that b 1 A has a multiplicative inverse. (All other propertiesfor a field follow trivially.) Consider B 5 {br 1 a | r [ R, a [ A}. Thisis an ideal of R that properly contains A (Exercise 25). Since A is maxi-mal, we must have B 5 R. Thus, 1 [ B, say, 1 5 bc 1 a9, where a9 [ A.Then

1 1 A 5 bc 1 a9 1 A 5 bc 1 A 5 (b 1 A)(c 1 A).

When a commutative ring has a unity, it follows from Theorems14.3 and 14.4 that a maximal ideal is a prime ideal. The next exampleshows that a prime ideal need not be maximal.

EXAMPLE 17 The ideal �x� is a prime ideal in Z[x] but not a maxi-mal ideal in Z[x]. To verify this, we begin with the observation that�x� 5 {f(x) [ Z[x] | f(0) 5 0} (see Exercise 29). Thus, if g(x)h(x) [ �x�,then g(0)h(0) 5 0. And since g(0) and h(0) are integers, we have g(0) 5 0or h(0) 5 0.

To see that �x� is not maximal, we simply note that �x� , �x, 2� ,Z[x] (see Exercise 37).

Exercises

1. Verify that the set defined in Example 3 is an ideal.2. Verify that the set A in Example 4 is an ideal and that A 5 �x�.3. Verify that the set I in Example 5 is an ideal and that if J is any

ideal of R that contains a1, a2, . . . , an, then I # J. (Hence, �a1,a2, . . . , an� is the smallest ideal of R that contains a1, a2, . . . , an.)

4. Find a subring of Z % Z that is not an ideal of Z % Z.5. Let S 5 {a 1 bi | a, b [ Z, b is even}. Show that S is a subring of

Z[i], but not an ideal of Z[i].6. Find all maximal ideals in

a. Z8, b. Z10, c. Z12, d. Zn.7. Let a belong to a commutative ring R. Show that aR 5 {ar | r [ R} is

an ideal of R. If R is the ring of even integers, list the elements of 4R.8. Prove that the intersection of any set of ideals of a ring is an ideal.

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Text not available due to copyright restrictions

270 Rings

9. If n is an integer greater than 1, show that �n� 5 nZ is a prime idealof Z if and only if n is prime. (This exercise is referred to in thischapter.)

10. If A and B are ideals of a ring, show that the sum of A and B, A 1 B 5{a 1 b | a [ A, b [ B}, is an ideal.

11. In the ring of integers, find a positive integer a such thata. �a� 5 �2� 1 �3�,b. �a� 5 �6� 1 �8�,c. �a� 5 �m� 1 �n�.

12. If A and B are ideals of a ring, show that the product of A and B,AB 5 {a1b1 1 a2b2 1 ? ? ? 1 anbn | ai [ A, bi [ B, n a positiveinteger}, is an ideal.

13. Find a positive integer a such thata. �a� 5 �3��4�,b. �a� 5 �6��8�,c. �a� 5 �m��n�.

14. Let A and B be ideals of a ring. Prove that AB # A > B.15. If A is an ideal of a ring R and 1 belongs to A, prove that A 5 R.

(This exercise is referred to in this chapter.)16. If A and B are ideals of a commutative ring R with unity and A 1 B 5 R,

show that A > B 5 AB.17. If an ideal I of a ring R contains a unit, show that I 5 R.18. Suppose that in the ring Z the ideal �35� is a proper ideal of J and J

is a proper ideal of I. What are the possibilities for J? What are thepossibilities for I?

19. Give an example of a ring that has exactly two maximal ideals.20. Suppose that R is a commutative ring and . If I is an ideal

of R and , prove that I is a maximal ideal.21. Let R and I be as described in Example 10. Prove that I is an ideal

of R.22. Let I 5 �2�. Prove that I[x] is not a maximal ideal of Z[x] even

though I is a maximal ideal of Z.23. Verify the claim made in Example 10 about the size of R/I.24. Give an example of a commutative ring that has a maximal ideal

that is not a prime ideal.25. Show that the set B in the latter half of the proof of Theorem 14.4

is an ideal of R. (This exercise is referred to in this chapter.)26. If R is a commutative ring with unity and A is a proper ideal of R,

show that R/A is a commutative ring with unity.27. Prove that the only ideals of a field F are {0} and F itself.

|I| 5 10|R| 5 30

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14 | Ideals and Factor Rings 271

28. Show that R[x]/�x2 1 1� is a field.29. In Z[x], the ring of polynomials with integer coefficients, let I 5

{ f (x) [ Z [x] | f (0) 5 0}. Prove that I 5 �x�. (This exercise is re-ferred to in this chapter and in Chapter 15.)

30. Show that A 5 {(3x, y) | x, y [ Z} is a maximal ideal of Z % Z.Generalize. What happens if 3x is replaced by 4x? Generalize.

31. Let R be the ring of continuous functions from R to R. Show that A 5 { f [ R | f (0) 5 0} is a maximal ideal of R.

32. Let R 5 Z8 % Z30. Find all maximal ideals of R, and for each max-imal ideal I, identify the size of the field R/I.

33. How many elements are in Z[i]/�3 1 i�? Give reasons for youranswer.

34. In Z[x], the ring of polynomials with integer coefficients, let I 5{ f (x) [ Z[x] | f (0) 5 0}. Prove that I is not a maximal ideal.

35. In Z % Z, let I 5 {(a, 0) | a [ Z}. Show that I is a prime ideal butnot a maximal ideal.

36. Let R be a ring and let I be an ideal of R. Prove that the factor ringR/I is commutative if and only if rs 2 sr [ I for all r and s in R.

37. In Z[x], let I 5 { f(x) [ Z[x] | f (0) is an even integer}. Prove that I 5 �x, 2�. Is I a prime ideal of Z[x]? Is I a maximal ideal? Howmany elements does Z[x]/I have? (This exercise is referred to inthis chapter.)

38. Prove that I 5 �2 1 2i� is not a prime ideal of Z[i]. How manyelements are in Z[i]/I? What is the characteristic of Z[i]/I?

39. In Z5[x], let I 5 �x2 1 x 1 2�. Find the multiplicative inverse of 2x 13 1 I in Z5[x]/I.

40. Let R be a ring and let p be a fixed prime. Show that Ip 5 {r [ R |additive order of r is a power of p} is an ideal of R.

41. An integral domain D is called a principal ideal domain if everyideal of D has the form �a� 5 {ad | d [ D} for some a in D. Showthat Z is a principal ideal domain. (This exercise is referred to inChapter 18.)

42. Let and

. If S is an ideal of R, what can you say about r and t?

43. If R and S are principal ideal domains, prove that R % S is a princi-

pal ideal ring.

is even fS 5 e c r s

0 td 0 r, s, t [ Z, sR 5 e ca b

0 dd 0 a, b, d [ Z6

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272 Rings

44. Let a and b belong to a commutative ring R. Prove that {x [ R |ax [ bR} is an ideal.

45. Let R be a commutative ring and let A be any subset of R. Showthat the annihilator of A, Ann(A) 5 {r [ R | ra 5 0 for all a in A},is an ideal.

46. Let R be a commutative ring and let A be any ideal of R. Show thatthe nil radical of A, N(A) 5 {r [ R | r n [ A for some positive in-teger n (n depends on r)}, is an ideal of R. [N(�0�) is called the nilradical of R.]

47. Let R 5 Z27. Finda. N(�0�), b. N(�3�), c. N(�9�).

48. Let R 5 Z36. Finda. N(�0�), b. N(�4�), c. N(�6�).

49. Let R be a commutative ring. Show that R/N(�0�) has no nonzeronilpotent elements.

50. Let A be an ideal of a commutative ring. Prove that N(N(A)) 5 N(A).51. Let Z2[x] be the ring of all polynomials with coefficients in Z2 (that

is, coefficients are 0 or 1, and addition and multiplication of coef-ficients are done modulo 2). Show that Z2[x]/�x2 1 x 1 1� is afield.

52. List the elements of the field given in Exercise 51, and make an ad-dition and multiplication table for the field.

53. Show that Z3[x]/�x2 1 x 1 1� is not a field.54. Let R be a commutative ring without unity, and let a [ R. Describe

the smallest ideal I of R that contains a (that is, if J is any ideal thatcontains a, then I # J).

55. Let R be the ring of continuous functions from R to R. Let A 5{ f [ R | f (0) is an even integer}. Show that A is a subring of R,but not an ideal of R.

56. Show that Z[i]/�1 2 i� is a field. How many elements does thisfield have?

57. If R is a principal ideal domain and I is an ideal of R, prove thatevery ideal of R/I is principal (see Exercise 41).

58. How many elements are in Z5[i]/�1 1 i�?59. Let R be a commutative ring with unity that has the property that

a2 5 a for all a in R. Let I be a prime ideal in R. Show that |R/I| 5 2.60. Let R be a commutative ring with unity, and let I be a proper ideal

with the property that every element of R that is not in I is a unit of R.Prove that I is the unique maximal ideal of R.

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61. Let I0 5 { f(x) [ Z[x] | f(0) 5 0}. For any positive integer n, showthat there exists a sequence of strictly increasing ideals such thatI0 , I1 , I2 , ? ? ? , In , Z[x].

62. Let R 5 {(a1, a2, a3, . . .)}, where each ai [ Z. Let I 5 {(a1, a2,a3, . . . )}, where only a finite number of terms are nonzero. Provethat I is not a principal ideal of R.

63. Let R be a commutative ring with unity and let a, b [ R. Show that �a, b�, the smallest ideal of R containing a and b, is I 5 {ra 1 sb|r, s [ R}. That is, show that I contains a and b and that any idealthat contains a and b also contains I.

Computer Exercises

What is the common denominator of intellectual accomplishment? In math,science, economics, history, or any other subject, the answer is the same:great thinkers notice patterns.

DAVID NIVEN, PSYCHOLOGIST

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software determines the number of elements in the ring Z[i]/�a 1 bi� (where i2 5 21). Run the program for several cases andformulate a conjecture based on your data.

2. This software determines the characteristic of the ring Z[i]/�a 1 bi�(where i2 5 21). Let d 5 gcd(a, b). Run the program for severalcases with d 5 1 and formulate a conjecture based on your data.Run the program for several cases with d . 1 and formulate a con-jecture in terms of a, b, and d based on your data. Does the formulayou found for d . 1 also work in the case that d 5 1?

3. This software determines when the ring Z[i]/�a 1 bi� (where i2 5 21)is isomorphic to the ring Za21b2. Run the program for several casesand formulate a conjecture based on your data.

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274

Richard Dedekind

This stamp was issued by East Germany in 1981 to commemorate the 150thanniversary of Dedekind’s birth. Noticethat it features the representation of anideal as the product of powers of primeideals.

Richard Dedekind was not only a mathematician, but one of the whollygreat in the history of mathematics, nowand in the past, the last hero of a greatepoch, the last pupil of Gauss, for fourdecades himself a classic, from whoseworks not only we, but our teachers andthe teachers of our teachers, have drawn.

EDMUND LANDAU,Commemorative Address

to the Royal Society of Göttingen

RICHARD DEDEKIND was born on October 6,1831, in Brunswick, Germany, the birth-place of Gauss. Dedekind was the youngestof four children of a law professor. His earlyinterests were in chemistry and physics, buthe obtained a doctor’s degree in mathemat-ics at the age of 21 under Gauss at theUniversity of Göttingen. Dedekind contin-ued his studies at Göttingen for a few years,and in 1854 he began to lecture there.

Dedekind spent the years 1858–1862 as aprofessor in Zürich. Then he accepted a po-sition at an institute in Brunswick where hehad once been a student. Although thisschool was less than university level,Dedekind remained there for the next 50years. He died in Brunswick in 1916.

During his career, Dedekind made numer-ous fundamental contributions to mathemat-ics. His treatment of irrational numbers,“Dedekind cuts,” put analysis on a firm,logical foundation. His work on uniquefactorization led to the modern theory ofalgebraic numbers. He was a pioneer in thetheory of rings and fields. The notion ofideals as well as the term itself are attributedto Dedekind. Mathematics historian MorrisKline has called him “the effective founderof abstract algebra.”

To find more information aboutDedekind, visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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275

EMMY NOETHER was born on March 23,1882, in Germany. When she entered theUniversity of Erlangen, she was one ofonly two women among the 1000 students.Noether completed her doctorate in 1907.

In 1916, Noether went to Göttingen and,under the influence of David Hilbert andFelix Klein, became interested in generalrelativity. While there, she made a majorcontribution to physics with her theoremthat whenever there is a symmetry in nature,there is also a conservation law, and viceversa. Hilbert tried unsuccessfully to obtaina faculty appointment at Göttingen forNoether, saying, “I do not see that the sex ofthe candidate is an argument against her ad-mission as Privatdozent. After all, we are auniversity and not a bathing establishment.”

It was not until she was 38 that Noether’strue genius revealed itself. Over the next13 years, she used an axiomatic method todevelop a general theory of ideals and non-commutative algebras. With this abstracttheory, Noether was able to weld togethermany important concepts. Her approach waseven more important than the individual re-sults. Hermann Weyl said of Noether, “Sheoriginated above all a new and epoch-mak-ing style of thinking in algebra.”

With the rise of Hitler in 1933, Noether, aJew, fled to the United States and took a po-sition at Bryn Mawr College. She died sud-denly on April 14, 1935, following an oper-ation.

To find more information about Noether,visit:

http://www-groups.dcs.st-and.ac.uk/~history/

In the judgment of the most competentliving mathematicians, Fräulein Noetherwas the most significant creative mathe-matical genius thus far produced since thehigher education of women began. In therealm of algebra, in which the most giftedmathematicians have been busy for cen-turies, she discovered methods which haveproved of enormous importance in the de-velopment of the present-day youngergeneration of mathematicians.

ALBERT EINSTEIN, The New York Times

Emmy Noether

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276 Rings

Supplementary Exercises for Chapters 12–14

If at first you do succeed—try to hide your astonishment.HARRY F. BANKS

True/false questions for Chapters 12–14 are available on the Web at:

http://www.d.umn.edu/~jgallian/TF

1. Find all idempotent elements in Z10, Z20, and Z30. (Recall that a isidempotent if a2 5 a.)

2. If m and n are relatively prime integers greater than 1, prove thatZmn has at least two idempotents besides 0 and 1.

3. Suppose that R is a ring in which a2 5 0 implies a 5 0. Show thatR has no nonzero nilpotent elements. (Recall that b is nilpotent ifbn 5 0 for some positive integer n.)

4. Let R be a commutative ring with more than one element. Provethat if for every nonzero element a of R we have aR 5 R, then R isa field.

5. Let A, B, and C be ideals of a ring R. If AB # C and C is a primeideal of R, show that A # C or B # C. (Compare this with Euclid’sLemma in Chapter 0.)

6. Show, by example, that the intersection of two prime ideals neednot be a prime ideal.

7. Let R denote the ring of real numbers. Determine all ideals of R % R.What happens if R is replaced by any field F?

8. Determine all factor rings of Z.9. Suppose that n is a square-free positive integer (that is, n is not

divisible by the square of any prime). Prove that Zn has no nonzeronilpotent elements.

10. Let R be a commutative ring with unity. Suppose that a is a unitand b is nilpotent. Show that a 1 b is a unit. (Hint: See Exercise 29in Chapter 12.)

11. Let A, B, and C be subrings of a ring R. If A # B < C, show thatA # B or A # C.

12. For any element a in a ring R, define �a� to be the smallest ideal ofR that contains a. If R is a commutative ring with unity, show that�a� 5 aR 5 {ar | r [ R}. Show, by example, that if R is commuta-tive but does not have a unity, then �a� and aR may be different.

13. Let R be a ring with unity. Show that �a� 5 {s1at1 1 s2at2 1 ? ? ? 1snatn | si, ti [ R and n is a positive integer}.

14. Show that Zn[x] has characteristic n.

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14 | Supplementary Exercises for Chapters 12–14 277

15. Let A and B be ideals of a ring R. If A > B 5 {0}, show that ab 5 0when a [ A and b [ B.

16. Show that the direct sum of two integral domains is not an integraldomain.

17. Consider the ring R 5 {0, 2, 4, 6, 8, 10} under addition and multi-plication modulo 12. What is the characteristic of R?

18. What is the characteristic of Zm % Zn? Generalize.19. Let R be a commutative ring with unity. Suppose that the only

ideals of R are {0} and R. Show that R is a field.20. Suppose that I is an ideal of J and that J is an ideal of R. Prove that

if I has a unity, then I is an ideal of R. (Be careful not to assume thatthe unity of I is the unity of R. It need not be—see Exercise 2 inChapter 12.)

21. Recall that an idempotent element b in a ring is one with the propertythat b2 5 b. Find a nontrivial idempotent (that is, not 0 and not 1) inQ[x]/�x4 1 x2�.

22. In a principal ideal domain, show that every nontrivial prime idealis a maximal ideal.

23. Find an example of a commutative ring R with unity such that a,b [ R, a 2 b, and an 5 bn and am 5 bm, where n and m are positiveintegers that are relatively prime. (Compare with Exercise 35, part b,in Chapter 13.)

24. Let Q( ) denote the smallest subfield of R that contains Q and. [That is, Q( ) is the subfield with the property that Q( )

contains Q and and if F is any subfield containing Q and ,then F contains Q( ).] Describe the elements of Q( ).

25. Let R be an integral domain with nonzero characteristic. If A is aproper ideal of R, show that R/A has the same characteristic as R.

26. Let F be a field of order pn. Determine the group isomorphismclass of F under the operation addition.

27. If R is a finite commutative ring with unity, prove that every primeideal of R is a maximal ideal of R.

28. Let R be a noncommutative ring and let C(R) be the center of R(see Exercise 19 in Chapter 12). Prove that the additive group ofR/C(R) is not cyclic.

29. Let

R 5 e ca b

c dd ` a, b, c, d [ Z2 f

3"23"2

3"23"2

3"23"23"2

3"2

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278 Rings

with ordinary matrix addition and multiplication modulo 2. Show that

is not an ideal of R. (Hence, in Exercise 7 in Chapter 14, the com-mutativity assumption is necessary.)

30. If R is an integral domain and A is a proper ideal of R, must R/A bean integral domain?

31. Let A 5 {a 1 bi | a, b [ Z, a mod 2 5 b mod 2}. How many ele-ments does Z[i]/A have? Show that A is a maximal ideal of Z[i].

32. Suppose that R is a commutative ring with unity such that for eacha in R there is a positive integer n greater than 1 (n depends on a)such that an 5 a. Prove that every prime ideal of R is a maximalideal of R.

33. State a “finite subfield test”; that is, state conditions that guaranteethat a finite subset of a field is a subfield.

34. Let F be a finite field with more than two elements. Prove that thesum of all of the elements of F is 0.

35. Show that if there are nonzero elements a and b in Zn such that a2 1b2 5 0, then the ring Zn[i] 5 {x 1 yi | x, y [ Zn} has zero-divisors.Use this fact to find a zero-divisor in Z13[i].

36. Suppose that R is a ring with no zero-divisors and that R contains anonzero element b such that b2 5 b. Show that b is a unity for R.

37. Find the characteristic of Z[i]/�2 1 i�.38. Show that the characteristic of Z[i]/�a 1 bi� divides a2 1 b2.39. Show that 4x2 1 6x 1 3 is a unit in Z8[x].40. For any commutative ring R, R[x, y] is the ring of polynomials in x

and y with coefficients in R (that is, R[x, y] consists of all finite sumsof terms of the form axiyj, where a [ R and i and j are nonnegativeintegers). (For example, x4 2 3x2y 2 y3 [ Z[x, y].) Prove that �x, y�is a prime ideal in Z[x, y] but not a maximal ideal in Z[x, y].

41. Prove that �x, y� is a maximal ideal in Z5[x, y].42. Prove that �2, x, y� is a maximal ideal in Z[x, y].43. Let R and S be rings. Prove that (a, b) is nilpotent in R % S if and

only if both a and b are nilpotent.44. Let R and S be commutative rings. Prove that (a, b) is a zero-divisor

in R % S if and only if a or b is a zero-divisor or exactly one of a orb is 0.

A 5 e c1 0

0 0d r ` r [ R f

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14 | Supplementary Exercises for Chapters 12–14 279

45. Determine all idempotents in Zp k, where p is a prime.46. Let R be a commutative ring with unity 1. Show that a is an idem-

potent if and only if there exists an element b in R such that ab 5 0and a 1 b 5 1.

47. Let Zn[ ] 5 {a 1 b | a, b [ Zn}. Define addition and multi-plication as in Z[ ], except that modulo n arithmetic is used tocombine the coefficients. Show that Z3[ ] is a field but Z7[ ]is not.

48. Let p be a prime. Prove that every zero-divisor in is a nilpotentelement.

49. If x is a nilpotent element in a commutative ring R, prove that rx isnilpotent for all r in R.

Zpn

"2"2"2

"2"2

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280

Ring hom*omorphisms

If there is one central idea which is common to all aspects of modern algebra it is the notion of hom*omorphism.

I. N. HERSTEIN, Topics in Algebra

Definition and ExamplesIn our work with groups, we saw that one way to discover informationabout a group is to examine its interaction with other groups by way ofhom*omorphisms. It should not be surprising to learn that this conceptextends to rings with equally profitable results.

Just as a group hom*omorphism preserves the group operation, a ringhom*omorphism preserves the ring operations.

Definitions Ring hom*omorphism, Ring Isomorphism

A ring hom*omorphism f from a ring R to a ring S is a mapping fromR to S that preserves the two ring operations; that is, for all a, b in R,

f(a 1 b) 5 f(a) 1 f(b) and f(ab) 5 f(a)f(b).

A ring hom*omorphism that is both one-to-one and onto is called aring isomorphism.

As is the case for groups, in the preceding definition the operationson the left of the equal signs are those of R, whereas the operations onthe right of the equal signs are those of S.

Again as with group theory, the roles of isomorphisms and hom*omor-phisms are entirely distinct. An isomorphism is used to show that tworings are algebraically identical; a hom*omorphism is used to simplify aring while retaining certain of its features.

A schematic representation of a ring hom*omorphism is given inFigure 15.1. The dashed arrows indicate the results of performing thering operations.

The following examples illustrate ring hom*omorphisms. The readershould supply the missing details.

15

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15 | Ring hom*omorphisms 281

Figure 15.1

EXAMPLE 1 For any positive integer n, the mapping k S k mod n isa ring hom*omorphism from Z onto Zn (see Exercise 11 in Chapter 0).This mapping is called the natural hom*omorphism from Z to Zn.

EXAMPLE 2 The mapping a 1 bi S a 2 bi is a ring isomorphismfrom the complex numbers onto the complex numbers (see Exercise 25in Chapter 6).

EXAMPLE 3 Let R[x] denote the ring of all polynomials with realcoefficients. The mapping f (x) S f (1) is a ring hom*omorphism fromR[x] onto R.

EXAMPLE 4 The correspondence f: x S 5x from Z4 to Z10

is a ring hom*omorphism. Although showing that f(x 1 y) 5f(x) 1 f(y) appears to be accomplished by the simple statement that5(x 1 y) 5 5x 1 5y, we must bear in mind that the addition on the left isdone modulo 4, whereas the addition on the right and the multiplicationon both sides are done modulo 10. An analogous difficulty arises in show-ing that f preserves multiplication. So, to verify that f preserves both op-erations, we write x 1 y 5 4q1 1 r1 and xy 5 4q2 1 r2, where 0 # r1 , 4and 0 # r2 , 4. Then f(x 1 y) 5 f(r1) 5 5r1 5 5(x 1 y 2 4q1) 5 5x 15y 2 20q1 5 5x 1 5y 5 f(x) 1 f(y) in Z10. Similarly, using the fact that5 ? 5 5 5 in Z10, we have f(xy) 5 f(r2) 5 5r2 5 5(xy 2 4q2) 5 5xy 220q2 5 (5 ? 5)xy 5 5x5y 5 f(x)f(y) in Z10.

EXAMPLE 5 We determine all ring hom*omorphisms from Z12 to Z30.By Example 10 in Chapter 10, the only group hom*omorphisms from Z12

to Z30 are x S ax, where a 5 0, 15, 10, 20, 5, or 25. But, since 1 ? 1 5 1in Z12, we must have a ? a 5 a in Z30. This requirement rules out 20 and 5as possibilities for a. Finally, simple calculations show that each of the re-maining four choices does yield a ring hom*omorphism.

a

b

a . b (a) . (b)a 1 b

φ

φ

φ φ

(a) 1 (b)φ φ

φ

φ

φ

SR

(a)

φ (b)

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282 Rings

EXAMPLE 6 Let R be a commutative ring of characteristic 2. Thenthe mapping a S a2 is a ring hom*omorphism from R to R.

EXAMPLE 7 Although 2Z, the group of even integers under addi-tion, is group-isomorphic to the group Z under addition, the ring 2Z isnot ring-isomorphic to the ring Z. (Quick! What does Z have that 2Zdoesn’t?)

Our next two examples are applications to number theory of the nat-ural hom*omorphism given in Example 1.

EXAMPLE 8 (Test for Divisibility by 9)

An integer n with decimal representation akak21 ? ? ? a0 is divisible by 9if and only if ak 1 ak21 1 ? ? ? 1 a0 is divisible by 9. To verify this, ob-serve that n 5 ak10k 1 ak2110k21 1 ? ? ? 1 a0. Then, letting a denotethe natural hom*omorphism from Z to Z9 [in particular, a(10) 5 1], wenote that n is divisible by 9 if and only if

0 5 a(n) 5 a(ak)(a(10))k 1 a(ak21)(a(10))k21 1 ? ? ? 1 a(a0)

5 a(ak) 1 a(ak21) 1 ? ? ? 1 a(a0)

5 a(ak 1 ak21 1 ? ? ? 1 a0).

But a(ak 1 ak21 1 ? ? ? 1 a0) 5 0 is equivalent to ak 1 ak21 1 ? ? ? 1a0 being divisible by 9.

EXAMPLE 9 (Theorem of Gersonides)

Among the most important unsolved problems in number theory is theso-called “abc conjecture.” This conjecture is a natural generalizationof a theorem first proved in the fourteenth century by the RabbiGersonides. Gersonides proved that the only pairs of positive integersthat are powers of 2 and powers of 3 which differ by 1 are 1, 2; 2, 3; 3,4; and 8, 9. That is, these four pairs are the only solutions to the equa-tions 2m 5 3n 6 1. To verify that this is so for 2m 5 3n 1 1, observe thatfor all n we have 3n mod 8 5 3 or 1. Thus, 3n 1 1 mod 8 5 4 or 2. Onthe other hand, for m . 3, we have 2m mod 8 5 0. To handle the casewhere 2m 5 3n 2 1, we first note that for all n, 3n mod 16 5 3, 9, 11, or1, depending on the value of n mod 4. Thus, (3n 2 1) mod 16 5 2, 8, 10,or 0. Since 2m mod 16 5 0 for m $ 4, we have ruled out the cases wheren mod 4 5 1, 2, or 3. Because 34k mod 5 5 (34)k mod 5 5 1k mod 5 51, we know that (34k 2 1) mod 5 5 0. But the only values for 2m mod 5are 2, 4, 3, and 1. This contradiction completes the proof.

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15 | Ring hom*omorphisms 283

Properties of Ring hom*omorphismsTheorem 15.1 Properties of Ring hom*omorphisms

PROOF The proofs of these properties are similar to those given inTheorems 10.1 and 10.2 and are left as exercises (Exercise 1).

The student should learn the various properties of Theorem 15.1in words in addition to the symbols. Property 2 says that the hom*omor-phic image of a subring is a subring. Property 4 says that the pullbackof an ideal is an ideal, and so on.

The next three theorems parallel results we had for groups. Theproofs are nearly identical to their group theory counterparts and areleft as exercises (Exercises 2, 3, and 4).

Theorem 15.2 Kernels Are Ideals

Theorem 15.3 First Isomorphism Theorem for Rings

Let f be a ring hom*omorphism from R to S. Then the mapping fromR/Ker f to f(R), given by r 1 Ker f S f(r), is an isomorphism. Insymbols, R/Ker f < f(R).

Let f be a ring hom*omorphism from a ring R to a ring S. Then Ker f5 {r [ R | f(r) 5 0} is an ideal of R.

Let f be a ring hom*omorphism from a ring R to a ring S. Let A be asubring of R and let B be an ideal of S.

1. For any r [ R and any positive integer n, f(nr) 5 nf(r) andf(rn) 5 (f(r))n.

2. f(A) 5 {f(a) | a [ A} is a subring of S.3. If A is an ideal and f is onto S, then f(A) is an ideal.4. f21(B) 5 {r [ R | f(r) [ B} is an ideal of R.5. If R is commutative, then f(R) is commutative.6. If R has a unity 1, S 2 {0}, and f is onto, then f(1) is the unity

of S.7. f is an isomorphism if and only if f is onto and Ker f 5

{r [ R | f(r) 5 0} 5 {0}.8. If f is an isomorphism from R onto S, then f21 is an

isomorphism from S onto R.

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284 Rings

Theorem 15.4 Ideals Are Kernels

The hom*omorphism from R to R/A given in Theorem 15.4 is calledthe natural hom*omorphism from R to R/A. Theorem 15.3 is often re-ferred to as the Fundamental Theorem of Ring hom*omorphisms.

In Example 17 in Chapter 14 we gave a direct proof that �x� is aprime ideal of Z[x] but not a maximal ideal. In the following examplewe illustrate a better way to do this kind of problem.

EXAMPLE 10 Since the mapping f from Z[x] onto Z given by f( f(x)) 5 f(0) is a ring hom*omorphism with Ker f 5 �x� (see Exercise29 in Chapter 14), we have, by Theorem 15.3, Z[x]/�x� < Z. And becauseZ is an integral domain but not a field, we know by Theorems 14.3 and14.4 that the ideal �x� is prime but not maximal in Z[x].

Theorem 15.5 hom*omorphism from Z to a Ring with Unity

PROOF Since the multiplicative group property am+n 5 aman translates to(m 1 n)a 5 ma 1 na when the operation is addition, we have f(m 1 n) 5(m 1 n) ? 1 5 m ? 1 1 n ? 1. So, f preserves addition.

That f also preserves multiplication follows from Exercise 15 inChapter 12, which says that (m ? a)(n ? b) 5 (mn) ? (ab) for all integersm and n. Thus, f(mn) 5 (mn) ? 1 5 (mn) ? ((1)(1)) 5 (m ? 1)(n ? 1) 5f(m)f(n). So, f preserves multiplication as well.

Corollary 1 A Ring with Unity Contains Zn or Z

PROOF Let 1 be the unity of R and let S 5 {k ? 1 | k [ Z}. Theorem 15.5shows that the mapping f from Z to S given by f(k) 5 k ? 1 is a hom*o-morphism, and by the First Isomorphism Theorem for rings, we haveZ/Ker f < S. But, clearly, Ker f 5 �n�, where n is the additive order of 1

If R is a ring with unity and the characteristic of R is n . 0, thenR contains a subring isomorphic to Zn. If the characteristic of R is 0,then R contains a subring isomorphic to Z.

Let R be a ring with unity 1. The mapping f: Z S R given by n S n ? 1is a ring hom*omorphism.

Every ideal of a ring R is the kernel of a ring hom*omorphism of R. In particular, an ideal A is the kernel of the mapping r S r 1 A from R to R/A.

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15 | Ring hom*omorphisms 285

and, by Theorem 13.3, n is also the characteristic of R. So, when Rhas characteristic n, S < Z/�n� < Zn. When R has characteristic 0, S <Z/�0� < Z.

Corollary 2 Zm Is a hom*omorphic Image of Z

PROOF This follows directly from the statement of Theorem 15.5,since in the ring Zm, the integer x mod m is x ? 1. (For example, in Z3, ifx 5 5, we have 5 ? 1 5 1 1 1 1 1 1 1 1 1 5 2.)

Corollary 3 A Field Contains Zp or Q (Steinitz, 1910)

PROOF By Corollary 1, F contains a subring isomorphic to Zp if F hascharacteristic p, and F has a subring S isomorphic to Z if F has charac-teristic 0. In the latter case, let

T 5 {ab21 | a, b [ S, b 2 0}.

Then T is isomorphic to the rationals (Exercise 63).

Since the intersection of all subfields of a field is itself a subfield(Exercise 11), every field has a smallest subfield (that is, a subfieldthat is contained in every subfield). This subfield is called the primesubfield of the field. It follows from Corollary 3 that the primesubfield of a field of characteristic p is isomorphic to Zp, whereas theprime subfield of a field of characteristic 0 is isomorphic to Q. (SeeExercise 67.)

The Field of QuotientsAlthough the integral domain Z is not a field, it is at least contained in afield—the field of rational numbers. And notice that the field of rationalnumbers is nothing more than quotients of integers. Can we mimic theconstruction of the rationals from the integers for other integral do-mains? Yes. The field constructed in Theorem 15.6 is called the field ofquotients of D. Throughout the proof of Theorem 15.6, you should keep

If F is a field of characteristic p, then F contains a subfieldisomorphic to Zp. If F is a field of characteristic 0, then F containsa subfield isomorphic to the rational numbers.

For any positive integer m, the mapping of f: Z S Zm given by x Sx mod m is a ring hom*omorphism.

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286 Rings

in mind that we are using the construction of the rationals from the inte-gers as a model for our construction of the field of quotients of D.

Theorem 15.6 Field of Quotients

PROOF Let S 5 {(a, b) | a, b [ D, b 2 0}. We define an equivalencerelation on S by (a, b) ; (c, d ) if ad 5 bc (compare with Example 14in Chapter 0). Now, let F be the set of equivalence classes of S underthe relation ; and denote the equivalence class that contains (x, y) byx/y. We define addition and multiplication on F by

a/b 1 c/d 5 (ad 1 bc)/(bd ) and a/b ? c/d 5 (ac)/(bd ).

(Notice that here we need the fact that D is an integral domain to ensurethat multiplication is closed; that is, bd 2 0 whenever b 2 0 and d 2 0.)

Since there are many representations of any particular element of F( just as in the rationals, we have 1/2 5 3/6 5 4/8), we must show thatthese two operations are well defined. To do this, suppose that a/b 5 a9/b9and c/d 5 c9/d9, so that ab9 5 a9b and cd9 5 c9d. It then follows that

(ad 1 bc)b9d9 5 adb9d9 1 bcb9d9 5 (ab9)dd9 1 (cd9)bb95 (a9b)dd9 1 (c9d)bb9 5 a9d9bd 1 b9c9bd5 (a9d9 1 b9c9)bd.

Thus, by definition, we have

(ad 1 bc)/(bd) 5 (a9d9 1 b9c9)/(b9d9),

and, therefore, addition is well defined. We leave the verification thatmultiplication is well defined as an exercise (Exercise 55). That F is afield is straightforward. Let 1 denote the unity of D. Then 0/1 is theadditive identity of F. The additive inverse of a/b is 2a/b; the multi-plicative inverse of a nonzero element a/b is b/a. The remaining fieldproperties can be checked easily.

Finally, the mapping f: D S F given by x S x/1 is a ring isomor-phism from D to f(D) (see Exercise 7).

EXAMPLE 11 Let D 5 Z[x]. Then the field of quotients of D is{f(x)/g(x) | f(x), g(x) [ D, where g(x) is not the zero polynomial}.

When F is a field, the field of quotients of F[x] is traditionally de-noted by F(x).

Let D be an integral domain. Then there exists a field F (called thefield of quotients of D) that contains a subring isomorphic to D.

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15 | Ring hom*omorphisms 287

EXAMPLE 12 Let p be a prime. Then Zp(x) 5 {f(x)/g(x) | f(x), g(x) [Zp[x], g(x) 2 0} is an infinite field of characteristic p.

Exercises

We can work it out.TITLE OF SONG BY JOHN LENNON AND

PAUL MCCARTNEY, December 1965

1. Prove Theorem 15.1.2. Prove Theorem 15.2.3. Prove Theorem 15.3.4. Prove Theorem 15.4.5. Show that the correspondence x S 5x from Z5 to Z10 does not pre-

serve addition.6. Show that the correspondence x S 3x from Z4 to Z12 does not pre-

serve multiplication.7. Show that the mapping f: D S F in the proof of Theorem 15.6 is a

ring hom*omorphism.8. Prove that every ring hom*omorphism f from Zn to itself has the

form f(x) 5 ax, where a2 5 a.9. Suppose that f is a ring hom*omorphism from Zm to Zn. Prove that

if f(1) 5 a, then a2 5 a. Give an example to show that the converseis false.

10. a. Is the ring 2Z isomorphic to the ring 3Z?b. Is the ring 2Z isomorphic to the ring 4Z?

11. Prove that the intersection of any collection of subfields of a fieldF is a subfield of F. (This exercise is referred to in this chapter.)

12. Let Z3[i] 5 {a 1 bi | a, b [ Z3} (see Example 9 in Chapter 13). Showthat the field Z3[i] is ring-isomorphic to the field Z3[x]/�x2 1 1�.

13. Let

S 5 .

Show that f: C S S given by

f(a 1 bi) 5

is a ring isomorphism.

c a b

2b ad

e c a b

2b ad `a, b [ R f

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288 Rings

14. Let Z[ ] 5 {a 1 b | a, b [ Z}. Let

H 5

Show that Z[ ] and H are isomorphic as rings.

15. Consider the mapping from M2(Z ) into Z given by S a.

Prove or disprove that this is a ring hom*omorphism.

16. Let R 5 . Prove or disprove that the map-

ping S a is a ring hom*omorphism.

17. Is the mapping from Z5 to Z30 given by x S 6x a ring hom*omor-phism? Note that the image of the unity is the unity of the imagebut not the unity of Z30.

18. Is the mapping from Z10 to Z10 given by x S 2x a ring hom*omor-phism?

19. Describe the kernel of the hom*omorphism given in Example 3.20. Recall that a ring element a is called an idempotent if a2 5 a. Prove

that a ring hom*omorphism carries an idempotent to an idempotent.21. Determine all ring hom*omorphisms from Z6 to Z6. Determine all

ring hom*omorphisms from Z20 to Z30.22. Determine all ring isomorphisms from Zn to itself.23. Determine all ring hom*omorphisms from Z to Z.24. Suppose f is a ring hom*omorphism from Z % Z into Z % Z. What

are the possibilities for f((1, 0))?25. Determine all ring hom*omorphisms from Z % Z into Z % Z.26. In Z, let A 5 �2� and B 5 �8�. Show that the group A/B is isomor-

phic to the group Z4 but that the ring A/B is not ring-isomorphic tothe ring Z4.

27. Let R be a ring with unity and let f be a ring hom*omorphism from Ronto S where S has more than one element. Prove that S has a unity.

28. Show that (Z % Z )/(�a� % �b�) is ring-isomorphic to Za % Zb.29. Determine all ring hom*omorphisms from Z % Z to Z.30. Prove that the sum of the squares of three consecutive integers can-

not be a square.31. Let m be a positive integer and let n be an integer obtained from m

by rearranging the digits of m in some way. (For example, 72345 isa rearrangement of 35274.) Show that m 2 n is divisible by 9.

ca b

0 cd

e ca b

0 cd ` a, b, c [ Z f

ca b

c dd

"2

e ca 2b

b ad ` a, b [ Z f .

"2"2

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15 | Ring hom*omorphisms 289

32. (Test for divisibility by 11) Let n be an integer with decimal repre-sentation akak21 ? ? ? a1a0. Prove that n is divisible by 11 if and onlyif a0 2 a1 1 a2 2 ? ? ? (21)kak is divisible by 11.

33. Show that the number 7,176,825,942,116,027,211 is divisible by 9but not divisible by 11.

34. Show that the number 9,897,654,527,609,805 is divisible by 99.35. (Test for divisibility by 3) Let n be an integer with decimal repre-

sentation akak21 ? ? ? a1a0. Prove that n is divisible by 3 if and onlyif ak 1 ak21 1 ? ? ? 1 a1 1 a0 is divisible by 3.

36. (Test for divisibility by 4) Let n be an integer with decimal repre-sentation akak21 ? ? ? a1a0. Prove that n is divisible by 4 if and onlyif a1a0 is divisible by 4.

37. Show that no integer of the form 111,111,111, . . . ,111 is prime.38. Consider an integer n of the form a, 111,111,111,111,111,111,

111,111,12b. Find values for a and b such that n is divisible by 99.39. Suppose n is a positive integer written in the form n 5 ak3k 1

ak213k21 1 ? ? ? 1 a13 1 a0, where each of the ai’s is 0, 1, or 2 (thebase 3 representative of n). Show that n is even if and only if ak 1ak21 1 ? ? ? 1 a1 1 a0 is even.

40. Find an analog of the condition given in the previous exercise forcharacterizing divisibility by 4.

41. In your head, determine (2 ? 1075 1 2)100 mod 3 and (10100 1 1)99

mod 3.42. Determine all ring hom*omorphisms from Q to Q.43. Let R and S be commutative rings with unity. If f is a hom*omor-

phism from R onto S and the characteristic of R is nonzero, provethat the characteristic of S divides the characteristic of R.

44. Let R be a commutative ring of prime characteristic p. Show thatthe Frobenius map x S xp is a ring hom*omorphism from R to R.

45. Is there a ring hom*omorphism from the reals to some ring whosekernel is the integers?

46. Show that a hom*omorphism from a field onto a ring with morethan one element must be an isomorphism.

47. Suppose that R and S are commutative rings with unities. Let f be aring hom*omorphism from R onto S and let A be an ideal of S.a. If A is prime in S, show that f21(A) 5 {x [ R | f(x) [ A} is

prime in R.b. If A is maximal in S, show that f21(A) is maximal in R.

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290 Rings

48. A principal ideal ring is a ring with the property that every idealhas the form �a�. Show that the hom*omorphic image of a principalideal ring is a principal ideal ring.

49. Let R and S be rings.a. Show that the mapping from R % S onto R given by (a, b) S a

is a ring hom*omorphism.b. Show that the mapping from R to R % S given by a S (a, 0) is a

one-to-one ring hom*omorphism.c. Show that R % S is ring-isomorphic to S % R.

50. Show that if m and n are distinct positive integers, then mZ is notring-isomorphic to nZ.

51. Prove or disprove that the field of real numbers is ring-isomorphicto the field of complex numbers.

52. Show that the only ring automorphism of the real numbers is theidentity mapping.

53. Determine all ring hom*omorphisms from R to R.54. Suppose that n divides m and that a is an idempotent of Zn (that is,

a2 5 a). Show that the mapping x S ax is a ring hom*omorphismfrom Zm to Zn. Show that the same correspondence need not yield aring hom*omorphism if n does not divide m.

55. Show that the operation of multiplication defined in the proof ofTheorem 15.6 is well defined.

56. Let Q[ ] 5 {a 1 b | a, b [ Q} and Q[ ] 5 {a 1 b |a, b [ Q}. Show that these two rings are not ring-isomorphic.

57. Let Z[i] 5 {a 1 bi | a, b [ Z}. Show that the field of quotients ofZ[i] is ring-isomorphic to Q[i] 5 {r 1 si | r, s [ Q}. (This exerciseis referred to in Chapter 18.)

58. Let F be a field. Show that the field of quotients of F is ring-isomorphic to F.

59. Let D be an integral domain and let F be the field of quotients of D.Show that if E is any field that contains D, then E contains asubfield that is ring-isomorphic to F. (Thus, the field of quotientsof an integral domain D is the smallest field containing D.)

60. Explain why a commutative ring with unity that is not an integral do-main cannot be contained in a field. (Compare with Theorem 15.6.)

61. Show that the relation ; defined in the proof of Theorem 15.6 is anequivalence relation.

62. Give an example of a ring without unity that is contained in a field.63. Prove that the set T in the proof of Corollary 3 to Theorem 15.5 is

ring-isomorphic to the field of rational numbers.

"5"5"2"2

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15 | Ring hom*omorphisms 291

64. Suppose that f: R S S is a ring hom*omorphism and that theimage of f is not {0}. If R has a unity and S is an integral domain,show that f carries the unity of R to the unity of S. Give an ex-ample to show that the preceding statement need not be true if Sis not an integral domain.

65. Let f(x) [ R[x]. If a 1 bi is a complex zero of f(x) (here i 5 ),show that a 2 bi is a zero of f(x). (This exercise is referred to inChapter 32.)

66. Let R 5 , and let f be the mapping that

takes to a 2 b.

a. Show that f is a hom*omorphism.b. Determine the kernel of f.c. Show that R/Ker f is isomorphic to Z.d. Is Ker f a prime ideal?e. Is Ker f a maximal ideal?

67. Show that the prime subfield of a field of characteristic p is ring-isomorphic to Zp and that the prime subfield of a field of charac-teristic 0 is ring-isomorphic to Q. (This exercise is referred to inthis chapter.)

68. Let n be a positive integer. Show that there is a ring isomorphismfrom Z2 to a subring of Z2n if and only if n is odd.

69. Show that Zmn is ring-isomorphic to Zm % Zn when m and n are rel-atively prime.

Suggested Readings

J. A. Gallian and J. Van Buskirk, “The Number of hom*omorphisms fromZm into Zn,” American Mathematical Monthly 91 (1984): 196–197.

In this article, formulas are given for the number of group hom*omor-phisms from Zm into Zn and the number of ring hom*omorphisms fromZm into Zn. This article can be downloaded at http://www.d.umn.edu/~jgallian/homs.pdf

Lillian Kinkade and Joyce Wagner, “When Polynomial Rings ArePrincipal Ideal Rings,” Journal of Undergraduate Mathematics 23(1991): 59–62.

In this article written by undergraduates, it is shown that R[x] is aprincipal ideal ring if and only if R < R1 % R2 % ? ? ? % Rn, whereeach Ri is a field.

ca b

b ad

e ca b

b ad ` a, b [ Z f

"21

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292 Rings

Mohammad Saleh and Hasan Yousef, “The Number of Ring hom*omor-phisms from Zm1

% ? ? ? % Zmr into Zk1% ? ? ? % Zk s,” American Mathe-

matical Monthly 105 (1998): 259–260.

This article gives a formula for the number described in the title.

Suggested Website

http://www.d.umn.edu/~jgallian/puzzle

This site has a math puzzle that is based on the ideas presented in thischapter. The user selects an integer and then proceeds through a series ofsteps to produce a new integer. Finally, another integer is created by usingall but one of the digits of the previous integer in any order. The softwarethen determines the digit not used.

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Wit lies in recognizing the resemblance among things which differ and thedifference between things which are alike.

MADAME DE STAEL

293

16 Polynomial Rings

293

Notation and TerminologyOne of the mathematical concepts that students are most familiar withand most comfortable with is that of a polynomial. In high school,students study polynomials with integer coefficients, rational coeffi-cients, real coefficients, and perhaps even complex coefficients. In ear-lier chapters of this book, we introduced something that was probablynew—polynomials with coefficients from Zn. Notice that all of thesesets of polynomials are rings, and, in each case, the set of coefficients isalso a ring. In this chapter, we abstract all of these examples into one.

Definition Ring of Polynomials over R

Let R be a commutative ring. The set of formal symbols

R[x] 5 {anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 | ai [ R,n is a nonnegative integer}

is called the ring of polynomials over R in the indeterminate x.

Two elements

anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0

and

bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0

of R[x] are considered equal if and only if ai 5 bi for all nonnegativeintegers i. (Define ai 5 0 when i . n and bi 5 0 when i . m.)

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294 Rings

In this definition, the symbols x, x2, . . . , xn do not represent“unknown” elements or variables from the ring R. Rather, their purposeis to serve as convenient placeholders that separate the ring elementsan, an21, . . . , a0. We could have avoided the x’s by defining a polyno-mial as an infinite sequence a0, a1, a2, . . . , an, 0, 0, 0, . . . , but ourmethod takes advantage of the student’s experience in manipulatingpolynomials where x does represent a variable. The disadvantage of ourmethod is that one must be careful not to confuse a polynomial with thefunction determined by a polynomial. For example, in Z3[x], the poly-nomials f (x) 5 x3 and g(x) 5 x5 determine the same function from Z3

to Z3, since f(a) 5 g(a) for all a in Z3.† But f(x) and g(x) are differentelements of Z3[x]. Also, in the ring Zn[x], be careful to reduce only thecoefficients and not the exponents modulo n. For example, in Z3[x],5x 5 2x, but x5 2 x2.

To make R[x] into a ring, we define addition and multiplication inthe usual way.

Definition Addition and Multiplication in R[x]

Let R be a commutative ring and let

f (x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0

and

g(x) 5 bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0

belong to R[x]. Then

f (x) 1 g(x) 5 (as 1 bs)xs 1 (as21 1 bs21)xs21

1 ? ? ? 1 (a1 1 b1)x 1 a0 1 b0,

where s is the maximum of m and n, ai 5 0 for i . n, and bi 5 0 for i . m. Also,

f (x)g(x) 5 cm1nxm1n 1 cm1n21xm1n21 1 ? ? ? 1 c1x 1 c0,

where

ck 5 akb0 1 ak21b1 1 ? ? ? 1 a1bk21 1 a0bk

for k 5 0, . . . , m 1 n.

Although the definition of multiplication might appear complicated,it is just a formalization of the familiar process of using the distributive

†In general, given f(x) in R[x] and a in R, f(a) means substitute a for x in the formulafor f(x). This substitution is a hom*omorphism from R[x] to R.

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16 | Polynomial Rings 295

property and collecting like terms. So, just multiply polynomials over acommutative ring R in the same way that polynomials are always mul-tiplied. Here is an example.

Consider f (x) 5 2x3 1 x2 1 2x 1 2 and g(x) 5 2x2 1 2x 1 1 in Z3[x].Then, in our preceding notation, a5 5 0, a4 5 0, a3 5 2, a2 5 1, a1 5 2,a0 5 2, and b5 5 0, b4 5 0, b3 5 0, b2 5 2, b1 5 2, b0 5 1. Now, usingthe definitions and remembering that addition and multiplication of thecoefficients are done modulo 3, we have

f(x) 1 g(x) 5 (2 1 0)x3 1 (1 1 2)x2 1 (2 1 2)x 1 (2 1 1)

5 2x3 1 0x2 1 1x 1 0

5 2x3 1 x

and

f(x) ? g(x) 5 (0 ? 1 1 0 ? 2 1 2 ? 2 1 1 ? 0 1 2 ? 0 1 2 ? 0)x5

1 (0 ? 1 1 2 ? 2 1 1 ? 2 1 2 ? 0 1 2 ? 0)x4

1 (2 ? 1 1 1 ? 2 1 2 ? 2 1 2 ? 0)x3

1 (1 ? 1 1 2 ? 2 1 2 ? 2)x2 1 (2 ? 1 1 2 ? 2)x 1 2 ? 15 x5 1 0x4 1 2x3 1 0x2 1 0x 1 25 x5 1 2x3 1 2

Our definitions for addition and multiplication of polynomials wereformulated so that they are commutative and associative, and so thatmultiplication is distributive over addition. We leave the verificationthat R[x] is a ring to the reader.

It is time to introduce some terminology for polynomials. If

f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0,

where an 2 0, we say that f(x) has degree n; the term an is called theleading coefficient of f(x), and if the leading coefficient is the multi-plicative identity element of R, we say that f(x) is a monic polynomial.The polynomial f(x) 5 0 has no degree. Polynomials of the formf(x) 5 a0 are called constant. We often write deg f(x) 5 n to indicatethat f(x) has degree n. In keeping with our experience with polynomialswith real coefficients, we adopt the following notational conventions:We may insert or delete terms of the form 0xk; 1xk will be denoted byxk; 1 (2ak)xk will be denoted by 2akxk.

Very often properties of R carry over to R[x]. Our first theorem is acase in point.

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296 Rings

Theorem 16.1 D an Integral Domain Implies D[x] an Integral Domain

PROOF Since we already know that D[x] is a ring, all we need to show is that D[x] is commutative with a unity and has no zero-divisors.Clearly, D[x] is commutative whenever D is. If 1 is the unity element ofD, it is obvious that f(x) 5 1 is the unity element of D[x]. Finally, sup-pose that

f (x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0

and

g(x) 5 bmxm 1 bm21xm21 1 ? ? ? 1 b0,

where an 2 0 and bm 2 0. Then, by definition, f(x)g(x) has leading co-efficient anbm and, since D is an integral domain, anbm 2 0.

The Division Algorithm and Consequences

One of the properties of integers that we have used repeatedly is thedivision algorithm: If a and b are integers and b 2 0, then there existunique integers q and r such that a 5 bq 1 r, where 0 # r , |b|. Thenext theorem is the analogous statement for polynomials over a field.

Theorem 16.2 Division Algorithm for F[x]

PROOF We begin by showing the existence of q(x) and r(x). Iff(x) 5 0 or deg f(x) , deg g(x), we simply set q(x) 5 0 and r(x) 5 f(x).So, we may assume that n 5 deg f(x) $ deg g(x) 5 m and let f(x) 5anxn 1 ? ? ? 1 a0 and g(x) 5 bmxm 1 ? ? ? 1 b0. The idea behind thisproof is to begin just as if you were going to “long divide” g(x) intof(x), then use the Second Principle of Mathematical Induction ondeg f(x) to finish up. Thus, resorting to long division, we let f1(x) 5

Let F be a field and let f(x) and g(x) [ F[x] with g(x) 2 0. Then there exist unique polynomials q(x) and r(x) in F[x] such that f(x) 5

g(x)q(x) 1 r(x) and either r(x) 5 0 or deg r(x) , deg g(x).

If D is an integral domain, then D [x] is an integral domain.

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16 | Polynomial Rings 297

f(x) 2 anbm21xn2mg(x).† Then, f1(x) 5 0 or deg f1(x) , deg f(x); so, by

our induction hypothesis, there exist q1(x) and r1(x) in F[x] such that f1(x) 5 g(x)q1(x) 1 r1(x), where r1(x) 5 0 or deg r1(x) , deg g(x).[Technically, we should get the induction started by proving the casein which deg f(x) 5 0, but this is trivial.] Thus,

f(x) 5 anbm21xn2mg(x) 1 f1(x)

5 anbm21xn2mg(x) 1 q1(x)g(x) 1 r1(x)

5 [anbm21xn2m 1 q1(x)]g(x) 1 r1(x).

So, the polynomials q(x) 5 anbm21xn2m 1 q1(x) and r(x) 5 r1(x) have

the desired properties.To prove uniqueness, suppose that f(x) 5 g(x)q(x) 1 r(x) and f(x) 5

g(x) (x) 1 (x), where r(x) 5 0 or deg r(x) , deg g(x) and (x) 5 0or deg (x) , deg g(x). Then, subtracting these two equations, we obtain

0 5 g(x)[q(x) 2 (x)] 1 [r(x) 2 (x)]

or

(x) 2 r(x) 5 g(x)[q(x) 2 (x)].

Thus, (x) 2 r(x) is 0, or the degree of (x) 2 r(x) is at least that of g(x). Since the latter is clearly impossible, we have (x) 5 r(x) and q(x) 5 (x) as well.

The polynomials q(x) and r(x) in the division algorithm are calledthe quotient and remainder in the division of f(x) by g(x). When thering of coefficients of a polynomial ring is a field, we can use the longdivision process to determine the quotient and remainder.

qr

rr

qr

rq

rrrq

†For example,

So,23x2 1 x 1 1 5 3x4 1 x 1 1 2 (3/2)x2(2x2 1 2)

In general,

So,f1(x) 5 (anxn 1 ? ? ?) 2 anbm

21xn2m(bmxm 1 ? ? ?)

anbm21xn2m

qan xn 1 . . .

an xn 1 . . .

f1(x)

bm xm 1 . . .

(3>2)x2

q3x4 1 x 1 1

3x4 1 3x2

2 3x2 1 x 1 1

2x2 1 2

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298 Rings

EXAMPLE 1 To find the quotient and remainder upon dividingf(x) 5 3x4 1 x3 1 2x2 1 1 by g(x) 5 x2 1 4x 1 2, where f(x) and g(x)belong to Z5[x], we may proceed by long division, provided we keep inmind that addition and multiplication are done modulo 5. Thus,

So, 3x2 1 4x is the quotient and 2x 1 1 is the remainder. Therefore,

3x4 1 x3 1 2x2 1 1 5 (x2 1 4x 1 2)(3x2 1 4x) 1 2x 1 1.

Let D be an integral domain. If f(x) and g(x) [ D[x], we say that g(x)divides f(x) in D[x] [and write g(x) | f(x)] if there exists an h(x) [ D[x]such that f(x) 5 g(x)h(x). In this case, we also call g(x) a factor of f(x).An element a is a zero (or a root) of a polynomial f(x) if f(a) 5 0.[Recall that f(a) means substitute a for x in the expression for f(x).]When F is a field, a [ F, and f(x) [ F[x], we say that a is a zero ofmultiplicity k (k $ 1) if (x 2 a)k is a factor of f(x) but (x 2 a)k11 is nota factor of f(x). With these definitions, we may now give several impor-tant corollaries of the division algorithm. No doubt you have seen thesefor the special case where F is the field of real numbers.

Corollary 1 The Remainder Theorem

PROOF The proof of Corollary 1 is left as an exercise (Exercise 5).

Corollary 2 The Factor Theorem

PROOF The proof of Corollary 2 is left as an exercise (Exercise 7).

3x2 1 4x

q3x4 1 x3 1 2x2 1 1

3x4 1 2x3 1 x2

4x3 1 x2 1 1

4x3 1 x2 1 3x

2x 1 1

x2 1 4x 1 2

Let F be a field, a [ F, and f(x) [ F [x]. Then f(a) is the remainder inthe division of f(x) by x 2 a.

Let F be a field, a [ F, and f(x) [ F[x]. Then a is a zero of f(x) ifand only if x 2 a is a factor of f(x).

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16 | Polynomial Rings 299

Corollary 3 Polynomials of Degree n Have at Most n Zeros

PROOF We proceed by induction on n. Clearly, a polynomial ofdegree 0 over a field has no zeros. Now suppose that f(x) is a polyno-mial of degree n over a field and a is a zero of f(x) of multiplicity k.Then, f(x) 5 (x 2 a)kq(x) and q(a) 2 0; and, since n 5 deg f(x) 5deg (x 2 a)k q(x) 5 k 1 deg q(x), we have k # n (see Exercise 17). Iff(x) has no zeros other than a, we are done. On the other hand, if b 2 aand b is a zero of f(x), then 0 5 f(b) 5 (b 2 a)kq(b), so that b is also azero of q(x) with the same multiplicity as it has for f (x) (see Exercise19). By the Second Principle of Mathematical Induction, we knowthat q(x) has at most deg q(x) 5 n 2 k zeros, counting multiplicity. Thus,f(x) has at most k 1 n 2 k 5 n zeros, counting multiplicity.

We remark that Corollary 3 is not true for arbitrary polynomial rings.For example, the polynomial x2 1 3x 1 2 has four zeros in Z6. (SeeExercise 3.) Lagrange was the first to prove Corollary 3 for polynomi-als in Zp[x].

EXAMPLE 2 The Complex Zeros of xn 2 1

We find all complex zeros of xn 2 1. Let v 5 cos(360°/n) 1i sin(360°/n). It follows from DeMoivre’s Theorem (see Example 7in Chapter 0) that vn 5 1 and vk 2 1 for 1 # k , n. Thus, each of 1,v, v2, . . . , vn21 is a zero of xn 2 1 and, by Corollary 3, there are noothers.

The complex number v in Example 2 is called the primitive nth rootof unity.

We conclude this chapter with an important theoretical applicationof the division algorithm, but first an important definition.

Definition Principal Ideal Domain (PID)

A principal ideal domain is an integral domain R in which every idealhas the form �a� 5 {ra | r [ R} for some a in R.

Theorem 16.3 F[x] Is a PID

A polynomial of degree n over a field has at most n zeros, countingmultiplicity.

Let F be a field. Then F[x] is a principal ideal domain.

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300 Rings

PROOF By Theorem 16.1, we know that F[x] is an integral domain.Now, let I be an ideal in F[x]. If I 5 {0}, then I 5 �0�. If I 2 {0}, thenamong all the elements of I, let g(x) be one of minimum degree. We willshow that I 5 �g(x)�. Since g(x) [ I, we have �g(x)� # I. Nowlet f(x) [ I. Then, by the division algorithm, we may write f(x) 5g(x)q(x) 1 r(x), where r(x) 5 0 or deg r(x) , deg g(x). Since r(x) 5 f(x) 2g(x)q(x) [ I, the minimality of deg g(x) implies that the latter conditioncannot hold. So, r(x) 5 0 and, therefore, f(x) [ �g(x)�. This shows thatI # �g(x)�.

The proof of Theorem 16.3 also establishes the following.

Theorem 16.4 Criterion for I 5 �g(x)�

As an application of the First Isomorphism Theorem for Rings(Theorem 15.3) and Theorem 16.4, we verify the remark we made inExample 12 in Chapter 14 that the ring R[x]/�x2 1 1� is isomorphic tothe ring of complex numbers.

EXAMPLE 3 Consider the hom*omorphism f from R[x] onto C givenby f(x) → f(i) (that is, evaluate a polynomial in R[x] at i). Then x2 1 1 [ Ker f and is clearly a polynomial of minimum degree in Ker f.Thus, Ker f 5 �x2 1 1� and R[x]/�x2 1 1� is isomorphic to C.

Exercises

If I feel unhappy, I do mathematics to become happy. If I am happy, I domathematics to keep happy.

PAUL TURÁN

1. Let f(x) 5 4x3 1 2x2 1 x 1 3 and g(x) 5 3x4 1 3x3 1 3x2 1 x 1 4,where f(x), g(x) [ Z5[x]. Compute f(x) 1 g(x) and f(x) ? g(x).

2. In Z3[x], show that the distinct polynomials x4 1 x and x2 1 xdetermine the same function from Z3 to Z3.

3. Show that x2 1 3x 1 2 has four zeros in Z6. (This exercise isreferred to in this chapter.)

Let F be a field, I a nonzero ideal in F [x], and g(x) an element ofF [x]. Then, I 5 8g(x)9 if and only if g(x) is a nonzero polynomial ofminimum degree in I.

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16 | Polynomial Rings 301

4. If R is a commutative ring, show that the characteristic of R[x] isthe same as the characteristic of R.

5. Prove Corollary 1 of Theorem 16.2.6. List all the polynomials of degree 2 in Z2[x].7. Prove Corollary 2 of Theorem 16.2.8. Let R be a commutative ring. Show that R[x] has a subring isomor-

phic to R.9. If f: R → S is a ring hom*omorphism, define :R[x] → S[x] by

(anxn 1 ? ? ? 1 a0) → f(an)xn 1 ? ? ? 1 f(a0). Show that is a ringhom*omorphism. (This exercise is referred to in Chapter 33.)

10. If the rings R and S are isomorphic, show that R[x] and S[x] are isomorphic.

11. Let f(x) 5 x3 1 2x 1 4 and g(x) 5 3x 1 2 in Z5[x]. Determine the quotient and remainder upon dividing f(x) by g(x).

12. Let f(x) 5 5x4 1 3x3 1 1 and g(x) 5 3x2 1 2x 1 1 in Z7[x].Determine the quotient and remainder upon dividing f(x) by g(x).

13. Show that the polynomial 2x 1 1 in Z4[x] has a multiplicative in-verse in Z4[x].

14. Are there any nonconstant polynomials in Z[x] that have multi-plicative inverses? Explain your answer.

15. Let p be a prime. Are there any nonconstant polynomials in Zp[x]that have multiplicative inverses? Explain your answer.

16. Show that Corollary 3 of Theorem 16.2 is false for any commuta-tive ring that has a zero divisor.

17. (Degree Rule) Let D be an integral domain and f(x), g(x) [ D[x].Prove that deg ( f(x) ? g(x)) 5 deg f(x) 1 deg g(x). Show, by exam-ple, that for commutative ring R it is possible that deg f(x)g(x) ,deg f(x) 1 deg g(x) where f(x) and g(x) are nonzero elements inR[x]. (This exercise is referred to in this chapter, Chapter 17, andChapter 18.)

18. Prove that the ideal �x� in Q[x] is maximal.19. Let f(x) belong to F[x], where F is a field. Let a be a zero of f(x) of

multiplicity n, and write f(x) 5 (x 2 a)nq(x). If b a is a zero ofq(x), show that b has the same multiplicity as a zero of q(x) as itdoes for f(x). (This exercise is referred to in this chapter.)

20. Prove that for any positive integer n, a field F can have at most afinite number of elements of order at most n.

21. Let F be an infinite field and let f(x) [ F[x]. If f(a) 5 0 for infi-nitely many elements a of F, show that f(x) 5 0.

?

ff

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302 Rings

22. Let F be an infinite field and let f(x), g(x) [ F[x]. If f(a) 5 g(a) forinfinitely many elements a of F, show that f(x) 5 g(x).

23. Let F be a field and let p(x) [ F[x]. If f(x), g(x) [ F[x] and deg f(x) , deg p(x) and deg g(x) , deg p(x), show that f(x) 1�p(x)� 5 g(x) 1 �p(x)� implies f(x) 5 g(x). (This exercise isreferred to in Chapter 20.)

24. Prove that Z[x] is not a principal ideal domain. (Compare this withTheorem 16.3.)

25. Find a polynomial with integer coefficients that has 1/2 and 21/3as zeros.

26. Let f(x) [ R[x]. Suppose that f(a) 5 0 but f9(a) 2 0, where f9(x) isthe derivative of f(x). Show that a is a zero of f(x) of multiplicity 1.

27. Show that Corollary 2 of Theorem 16.2 is true over any commuta-tive ring with unity.

28. Show that Corollary 3 of Theorem 16.2 is true for polynomialsover integral domains.

29. Let F be a field and let

I 5 {anxn 1 an21xn21 1 ? ? ? 1 a0 | an, an21, . . . , a0 [ F andan 1 an21 1 ? ? ? 1 a0 5 0}.

Show that I is an ideal of F[x] and find a generator for I.30. Let F be a field and let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ F[x].

Prove that x 2 1 is a factor of f(x) if and only if an 1 an21 1 ? ? ? 1a0 5 0.

31. Let m be a fixed positive integer. For any integer a, let denotea mod m. Show that the mapping of f: Z[x] → Zm[x] given by

f(anxn 1 an21xn21 1 ? ? ? 1 a0) 5 nxn 1 n21xn21 1 ? ? ? 1 0

is a ring hom*omorphism. (This exercise is referred to in Chapters17 and 33.)

32. Find infinitely many polynomials f(x) in Z3[x] such that f(a) 5 0 forall a in Z3.

33. For every prime p, show that

xp21 2 1 5 (x 2 1)(x 2 2) ? ? ? [x 2 (p 2 1)]in Zp[x].

34. (Wilson’s Theorem) For every integer n . 1, prove that (n 2 1)!mod n 5 n 2 1 if and only if n is prime.

35. For every prime p, show that ( p 2 2)! mod p 5 1.36. Find the remainder upon dividing 98! by 101.37. Prove that (50!)2 mod 101 5 21 mod 101.

aaa

a

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16 | Polynomial Rings 303

38. If I is an ideal of a ring R, prove that I[x] is an ideal of R[x].39. Give an example of a commutative ring R with unity and a

maximal ideal I of R such that I[x] is not a maximal ideal of R[x].40. Let R be a commutative ring with unity. If I is a prime ideal of R,

prove that I[x] is a prime ideal of R[x].41. Let F be a field, and let f(x) and g(x) belong to F[x]. If there is no

polynomial of positive degree in F[x] that divides both f(x) and g(x)[in this case, f(x) and g(x) are said to be relatively prime], prove thatthere exist polynomials h(x) and k(x) in F[x] with the property thatf(x)h(x) 1 g(x)k(x) 5 1. (This exercise is referred to in Chapter 20.)

42. Prove that Q[x]/�x2 2 2� is ring-isomorphic to Q[ ] 5 {a 1b | a, b [ Q}.

43. Let f(x) [ R[x]. If f(a) 5 0 and f 9(a) 5 0 [f 9(a) is the derivative off(x) at a], show that (x 2 a)2 divides f(x).

44. Let F be a field and let I 5 {f(x) [ F[x] | f(a) 5 0 for all a in F}.Prove that I is an ideal in F[x]. Prove that I is infinite when F is fi-nite and I 5 {0} when F is infinite. When F is finite, find a monicpolynomial g(x) such that I 5 �g(x)�.

45. Let g(x) and h(x) belong to Z[x] and let h(x) be monic. If h(x) di-vides g(x) in Q[x], show that h(x) divides g(x) in Z[x]. (This exer-cise is referred to in Chapter 33.)

46. For any field F, recall that F(x) denotes the field of quotients of thering F[x]. Prove that there is no element in F(x) whose square is x.

47. Let F be a field. Show that there exist a, b [ F with the propertythat x2 1 x 1 1 divides x43 1 ax 1 b.

48. Let f(x) 5 amxm 1 am21xm21 1 ? ? ? 1 a0 and g(x) 5 bnxn 1 bn21xn21 1? ? ? 1 b0 belong to Q[x] and suppose that f 8 g belongs to Z[x]. Provethat aibj is an integer for every i and j.

49. Let f (x) belong to Z[x]. If a mod m 5 b mod m, prove that f (a)mod m 5 f(b) mod m. Prove that if both f(0) and f(1) are odd then f has no zero in Z.

50. Find the remainder when x51 is divided by x 1 4 in Z7[x].51. Show that 1 is the only solution of x25 2 1 5 0 in Z37.

"2"2

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Saunders Mac Lane

SAUNDERS MAC LANE ranks among the mostinfluential mathematicians in the twentiethcentury. He was born on August 4, 1909, inNorwich, Connecticut. In 1933, at the heightof the Depression, he was newly married; de-spite having degrees from Yale, the Universityof Chicago, and the University of Göttingen,he had no prospects for a position at a collegeor university. After applying for employmentas a master at a private preparatory school forboys, Mac Lane received a two-year instruc-torship at Harvard in 1934. He then spent ayear at Cornell and a year at the Universityof Chicago before returning to Harvard in1938. In 1947, he went back to Chicago per-manently.

Much of Mac Lane’s work focuses on theinterconnections among algebra, topology,

The 1986 Steele Prize for cumulativeinfluence is awarded to Saunders Mac Lanefor his many contributions to algebra andalgebraic topology, and in particular for hispioneering work in hom*ological andcategorical algebra.

Citation for the Steele Prize

304

and geometry. His book, Survey of ModernAlgebra, coauthored with Garrett Birkhoff,influenced generations of mathematiciansand is now a classic. Mac Lane has served aspresident of the Mathematical Association ofAmerica and the American MathematicalSociety. He was elected to the NationalAcademy of Sciences, received the NationalMedal of Science and supervised 41 Ph.D.theses. Mac Lane died April 14, 2005, at ageof 95.

To find more information about MacLane, visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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Factorization of Polynomials

The value of a principle is the number of things it will explain.RALPH WALDO EMERSON

305

17

Reducibility TestsIn high school, students spend much time factoring polynomials andfinding their zeros. In this chapter, we consider the same problems in amore abstract setting.

To discuss factorization of polynomials, we must first introduce thepolynomial analog of a prime integer.

Definition Irreducible Polynomial, Reducible Polynomial

Let D be an integral domain. A polynomial f(x) from D[x] that isneither the zero polynomial nor a unit in D[x] is said to be irreducibleover D if, whenever f(x) is expressed as a product f(x) 5 g(x)h(x), withg(x) and h(x) from D[x], then g(x) or h(x) is a unit in D[x]. A nonzero,nonunit element of D[x] that is not irreducible over D is calledreducible over D.

In the case that an integral domain is a field F, it is equivalent and moreconvenient to define a nonconstant f(x) [ F[x] to be irreducible if f(x) can-not be expressed as a product of two polynomials of lower degree.

EXAMPLE 1 The polynomial f(x) 5 2x2 1 4 is irreducible over Qbut reducible over Z, since 2x2 1 4 5 2(x2 1 2) and neither 2 nor x2 1 2is a unit in Z[x].

EXAMPLE 2 The polynomial f(x) 5 2x2 1 4 is irreducible over Rbut reducible over C.

EXAMPLE 3 The polynomial x2 2 2 is irreducible over Q but re-ducible over R.

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EXAMPLE 4 The polynomial x2 1 1 is irreducible over Z3 but re-ducible over Z5.

In general, it is a difficult problem to decide whether or not a partic-ular polynomial is reducible over an integral domain, but there are spe-cial cases when it is easy. Our first theorem is a case in point. It appliesto the three preceding examples.

Theorem 17.1 Reducibility Test for Degrees 2 and 3

PROOF Suppose that f(x) 5 g(x)h(x), where both g(x) and h(x) belongto F[x] and have degrees less than that of f(x). Since deg f(x) 5 deg g(x) 1deg h(x) (Exercise 17 in Chapter 16) and deg f(x) is 2 or 3, at least oneof g(x) and h(x) has degree 1. Say g(x) 5 ax 1 b. Then, clearly, 2a21bis a zero of g(x) and therefore a zero of f(x) as well.

Conversely, suppose that f(a) 5 0, where a [ F. Then, by the FactorTheorem, we know that x 2 a is a factor of f(x) and, therefore, f(x) isreducible over F.

Theorem 17.1 is particularly easy to use when the field is Zp, be-cause, in this case, we can check for reducibility of f(x) by simply test-ing to see if f(a) 5 0 for a 5 0, 1, . . . , p 2 1. For example, since 2 is azero of x2 1 1 over Z5, x2 1 1 is reducible over Z5. On the other hand,because neither 0, 1, nor 2 is a zero of x2 1 1 over Z3, x2 1 1 is irre-ducible over Z3.

Note that polynomials of degree larger than 3 may be reducible overa field, even though they do not have zeros in the field. For example, inQ[x], the polynomial x4 1 2x2 1 1 is equal to (x2 1 1)2, but has nozeros in Q.

Our next three tests deal with polynomials with integer coefficients.To simplify the proof of the first of these, we introduce some terminol-ogy and isolate a portion of the argument in the form of a lemma.

Definition Content of Polynomial, Primitive Polynomial

The content of a nonzero polynomial anxn 1 an21xn21 1 ? ? ? 1 a0,

where the a’s are integers, is the greatest common divisor of theintegers an, an21, . . . , a0. A primitive polynomial is an element of Z[x]with content 1.

Let F be a field. If f(x) [ F[x] and deg f(x) is 2 or 3, then f(x) isreducible over F if and only if f(x) has a zero in F.

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Gauss’s Lemma

PROOF Let f(x) and g(x) be primitive polynomials, and suppose thatf(x)g(x) is not primitive. Let p be a prime divisor of the content off(x)g(x), and let , , and be the polynomials obtainedfrom f(x), g(x), and f(x)g(x) by reducing the coefficients modulo p.Then, and belong to the integral domain Zp[x] and 5

5 0, the zero element of Zp[x] (see Exercise 31 in Chapter 16).Thus, 5 0 or 5 0. This means that either p divides every co-efficient of f(x) or p divides every coefficient of g(x). Hence, either f(x)is not primitive or g(x) is not primitive. This contradiction completesthe proof.

Remember that the question of reducibility depends on which ring ofcoefficients one permits. Thus, x2 2 2 is irreducible over Z but reducible over Q[ ]. In Chapter 20, we will prove that every poly-nomial of degree greater than 1 with coefficients from an integral domain is reducible over some field. Theorem 17.2 shows that in thecase of polynomials irreducible over Z, this field must be larger thanthe field of rational numbers.

Theorem 17.2 Reducibility over Q Implies Reducibility Over Z

PROOF Suppose that f (x) 5 g(x)h(x), where g(x) and h(x) [ Q[x].Clearly, we may assume that f(x) is primitive because we can divideboth f (x) and g(x) by the content of f(x). Let a be the least commonmultiple of the denominators of the coefficients of g(x), and b the leastcommon multiple of the denominators of the coefficients of h(x). Thenabf(x) 5 ag(x) ? bh(x), where ag(x) and bh(x) [ Z[x]. Let c1 be the con-tent of ag(x) and let c2 be the content of bh(x). Then ag(x) 5 c1g1(x) andbh(x) 5 c2h1(x), where both g1(x) and h1(x) are primitive and abf(x) 5c1c2g1(x)h1(x). Since f(x) is primitive, the content of abf(x) is ab. Also,since the product of two primitive polynomials is primitive, it followsthat the content of c1c2g1(x)h1(x) is c1c2. Thus, ab 5 c1c2 and f(x) 5g1(x)h1(x), where g1(x) and h1(x) [ Z[x] and deg g1(x) 5 deg g(x) anddeg h1(x) 5 deg h(x).

Let f(x) [ Z[x]. If f(x) is reducible over Q, then it is reducible over Z.

"2

g(x)f (x)f(x)g(x)

g(x)f (x)g(x)f (x)

f(x)g(x)g(x)f (x)

The product of two primitive polynomials is primitive.

17 | Factorization of Polynomials 307

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EXAMPLE 5 We illustrate the proof of Theorem 17.2 by tracingthrough it for the polynomial f (x) 5 6x2 1 x 2 2 5 (3x 2 3/2)(2x 14/3) 5 g(x)h(x). In this case we have a 5 2, b 5 3, c1 5 3, c2 5 2, g1(x) 52x 2 1, and h1(x) 5 3x 1 2, so that 2 ? 3(6x2 1 x 2 2) 5 3 ? 2(2x 21)(3x 1 2) or 6x2 1 x 2 2 5 (2x 2 1)(3x 1 2).

Irreducibility TestsTheorem 17.1 reduces the question of irreducibility of a polynomial ofdegree 2 or 3 to one of finding a zero. The next theorem often allows usto simplify the problem even further.

Theorem 17.3 Mod p Irreducibility Test

PROOF It follows from the proof of Theorem 17.2 that if f(x) is re-ducible over Q, then f(x) 5 g(x)h(x) with g(x), h(x) [ Z[x], and bothg(x) and h(x) have degree less than that of f(x). Let , , and be the polynomials obtained from f(x), g(x), and h(x) by reducing allthe coefficients modulo p. Since deg f(x) 5 deg , we have deg

# deg g(x) , deg and deg # deg h(x) , deg . But,5 , and this contradicts our assumption that is irre-

ducible over Zp.

EXAMPLE 6 Let f(x) 5 21x3 2 3x2 1 2x 1 9. Then, over Z2, wehave 5 x3 1 x2 1 1 and, since 5 1 and 5 1, we see that

is irreducible over Z2. Thus, f (x) is irreducible over Q. Notice that,over Z3, 5 2x is irreducible, but we may not apply Theorem 17.3to conclude that f(x) is irreducible over Q.

Be careful not to use the converse of Theorem 17.3. If f(x) [ Z[x]and is reducible over Zp for some p, f(x) may still be irreducibleover Q. For example, consider f(x) 5 21x3 2 3x2 1 2x 1 8. Then, overZ2, 5 x3 1 x2 5 x2(x 1 1). But over Z5, has no zeros andtherefore is irreducible over Z5. So, f(x) is irreducible over Q. Note thatthis example shows that the Mod p Irreducibility Test may fail forsome p and work for others. To conclude that a particular f(x) in Z[x] is

f (x)f (x)

f (x)

f (x)f (x)

f (1)f (0)f (x)

f (x)g(x)h(x)f (x)f (x)h(x)f (x)g(x)

f (x)

h(x)g(x)f (x)

308 Rings

Let p be a prime and suppose that f(x) [ Z[x] with deg f(x) $ 1.Let (x) be the polynomial in Zp[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If (x) is irreducible over Zp anddeg (x) 5 deg f(x), then f(x) is irreducible over Q.f

ff

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irreducible over Q, all we need to do is find a single p for which the cor-responding polynomial in Zp is irreducible. However, this is not al-ways possible, since f(x) 5 x4 1 1 is irreducible over Q but reducibleover Zp for every prime p. (See Exercise 29.)

The Mod p Irreducibility Test can also be helpful in checking forirreducibility of polynomials of degree greater than 3 and polynomialswith rational coefficients.

EXAMPLE 7 Let f(x) 5 (3/7)x4 2 (2/7)x2 1 (9/35)x 1 3/5. We willshow that f(x) is irreducible over Q. First, let h(x) 5 35f(x) 5 15x4 210x2 1 9x 1 21. Then f(x) is irreducible over Q if h(x) is irreducibleover Z. Next, applying the Mod 2 Irreducibility Test to h(x), we get

5 x4 1 x 1 1. Clearly, has no zeros in Z2. Furthermore,has no quadratic factor in Z2[x] either. [For if so, the factor would haveto be either x2 1 x 1 1 or x2 1 1. Long division shows that x2 1 x 1 1is not a factor, and x2 1 1 cannot be a factor because it has a zerowhereas does not.] Thus is irreducible over Z2[x]. This guaran-tees that h(x) is irreducible over Q.

EXAMPLE 8 Let f(x) 5 x5 1 2x 1 4. Obviously, neither Theorem17.1 nor the Mod 2 Irreducibility Test helps here. Let’s try mod 3.Substitution of 0, 1, and 2 into does not yield 0, so there are no linearfactors. But may have a quadratic factor. If so, we may assume it hasthe form x2 1 ax 1 b (see Exercise 5). This gives nine possibilities tocheck. We can immediately rule out each of the nine that has a zero overZ3, since does not have one. This leaves only x2 1 1, x2 1 x 1 2, andx2 1 2x 1 2 to check. These are eliminated by long division. So, since

is irreducible over Z3, f(x) is irreducible over Q. (Why is it unnec-essary to check for cubic or fourth-degree factors?)

Another important irreducibility test is the following one, credited toFerdinand Eisenstein (1823–1852), a student of Gauss. The corollarywas first proved by Gauss by a different method.

Theorem 17.4 Eisenstein’s Criterion (1850)

Let

f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x].

If there is a prime p such that p B an, p | an21, . . . , p | a0 and p2 B a0,then f(x) is irreducible over Q.

f (x)

f (x)

f (x)f (x)

h(x)h(x)

h(x)h(x)h(x)

f (x)

17 | Factorization of Polynomials 309

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PROOF If f (x) is reducible over Q, we know by Theorem 17.2 thatthere exist elements g(x) and h(x) in Z[x] such that f(x) 5 g(x)h(x) and1 # deg g(x), and 1 # deg h(x) , n. Say g(x) 5 br x

r 1 ? ? ? 1 b0 andh(x) 5 cs x

s 1 ? ? ? 1 c0. Then, since p | a0, p2 B a0, and a0 5 b0c0, it fol-lows that p divides one of b0 and c0 but not the other. Let us say p | b0and p B c0. Also, since p B an 5 brcs, we know that p B br. So, there is aleast integer t such that p B bt. Now, consider at 5 btc0 1 bt21c1 1 ? ? ?1 b0ct. By assumption, p divides at and, by choice of t, every summandon the right after the first one is divisible by p. Clearly, this forces p todivide btc0 as well. This is impossible, however, since p is prime and pdivides neither bt nor c0.

Corollary Irreducibility of pth Cyclotomic Polynomial

PROOF Let

Then, since every coefficient except that of xp21 is divisible by p andthe constant term is not divisible by p2, by Eisenstein’s Criterion, f(x) isirreducible over Q. So, if Fp(x) 5 g(x)h(x) were a nontrivial factoriza-tion of Fp(x) over Q, then f(x) 5 Fp(x 1 1) 5 g(x 1 1) ? h(x 1 1)would be a nontrivial factorization of f(x) over Q. Since this is impossi-ble, we conclude that Fp(x) is irreducible over Q.

EXAMPLE 9 The polynomial 3x5 1 15x4 2 20x3 1 10x 1 20 isirreducible over Q because 5 B 3 and 25 B 20 but 5 does divide 15,220, 10, and 20.

The principal reason for our interest in irreducible polynomialsstems from the fact that there is an intimate connection among them,maximal ideals, and fields. This connection is revealed in the next the-orem and its first corollary.

f(x)5Fp(x11)5(x11)p21

(x11)215x p211ap

1b xp221ap

2b xp231.

. .1ap

1b.

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For any prime p, the pth cyclotomic polynomial

Fp(x) 5 5 xp21 1 xp22 1 ? ? ? 1 x 1 1

is irreducible over Q.

xp 2 1x 2 1

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Theorem 17.5 � p(x)� Is Maximal If and Only If p(x) Is Irreducible

PROOF Suppose first that �p(x)� is a maximal ideal in F[x]. Clearly,p(x) is neither the zero polynomial nor a unit in F[x], because neither{0} nor F[x] is a maximal ideal in F[x]. If p(x) 5 g(x)h(x) is a factor-ization of p(x) over F, then �p(x)� # �g(x)� # F[x]. Thus, �p(x)� 5 �g(x)�or F[x] 5 �g(x)�. In the first case, we must have deg p(x) 5 deg g(x). Inthe second case, it follows that deg g(x) 5 0 and, consequently, deg h(x) 5deg p(x). Thus, p(x) cannot be written as a product of two polynomialsin F[x] of lower degree.

Now, suppose that p(x) is irreducible over F. Let I be any ideal ofF[x] such that �p(x)� # I # F[x]. Because F[x] is a principal ideal do-main, we know that I 5 �g(x)� for some g(x) in F[x]. So, p(x) [ �g(x)�and, therefore, p(x) 5 g(x)h(x), where h(x) [ F[x]. Since p(x) is irre-ducible over F, it follows that either g(x) is a constant or h(x) is a con-stant. In the first case, we have I 5 F[x]; in the second case, we have �p(x)� 5 �g(x)� 5 I. So, �p(x)� is maximal in F[x].

Corollary 1 F[x]/� p(x)� Is a Field

PROOF This follows directly from Theorems 17.5 and 14.4.

The next corollary is a polynomial analog of Euclid’s Lemma forprimes (see Chapter 0).

Corollary 2 p(x) | a(x)b(x) Implies p(x) | a(x) or p(x) | b(x)

PROOF Since p(x) is irreducible, F[x]/� p(x)� is a field and, therefore, anintegral domain. From Theorem 14.3, we know that �p(x)� is a primeideal, and since p(x) divides a(x)b(x), we have a(x)b(x) [ �p(x)�. Thus,a(x) [ �p(x)� or b(x) [ �p(x)�. This means that p(x) | a(x) or p(x) | b(x).

The next two examples put the theory to work.

Let F be a field and let p(x), a(x), b(x) [ F[x]. If p(x) is irreducibleover F and p(x) | a(x)b(x), then p(x) | a(x) or p(x) | b(x).

Let F be a field and p(x) an irreducible polynomial over F. ThenF[x]/�p(x)� is a field.

Let F be a field and let p(x) [ F[x]. Then �p(x)� is a maximal ideal in F[x] if and only if p(x) is irreducible over F.

17 | Factorization of Polynomials 311

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EXAMPLE 10 We construct a field with eight elements. ByTheorem 17.1 and Corollary 1 of Theorem 17.5, it suffices to find acubic polynomial over Z2 that has no zero in Z2. By inspection, x3 1x 1 1 fills the bill. Thus, Z2[x]/�x3 1 x 1 1� 5 {ax2 1 bx 1 c 1 �x3 1x 1 1� | a, b, c [ Z2} is a field with eight elements. For practice, let usdo a few calculations in this field. Since the sum of two polynomials ofthe form ax2 1 bx 1 c is another one of the same form, addition is easy.For example,

(x2 1 x 1 1 1 �x3 1 x 1 1�) 1 (x2 1 1 1 �x3 1 x 1 1�)5 x 1 �x3 1 x 1 1�.

On the other hand, multiplication of two coset representatives need notyield one of the original eight coset representatives:

(x2 1 x 1 1 1 �x3 1 x 1 1�) ? (x2 1 1 1 �x3 1 x 1 1�)5 x4 1 x3 1 x 1 1 1 �x3 1 x 1 1� 5 x4 1 �x3 1 x 1 1�

(since the ideal absorbs the last three terms). How do we express this inthe form ax2 1 bx 1 c 1 �x3 1 x 1 1�? One way is to long divide x4 byx3 1 x 1 1 to obtain the remainder of x2 1 x (just as one reduces 12 1 �5� to 2 1 �5� by dividing 12 by 5 to obtain the remainder 2).Another way is to observe that x3 1 x 1 1 1 �x3 1 x 1 1� 5 0 1�x3 1 x 1 1� implies x3 1 �x3 1 x 1 1� 5 x 1 1 1 �x3 1 x 1 1�. Thus,we may multiply both sides by x to obtain

x4 1 �x3 1 x 1 1� 5 x2 1 x 1 �x3 1 x 1 1�.

Similarly,

(x2 1 x 1 kx3 1 x 1 1l) ? (x 1 kx3 1 x 1 1l)5 x3 1 x2 1 kx3 1 x 1 1l5 x2 1 x 1 1 1 kx3 1 x 1 1l.

A partial multiplication table for this field is given in Table 17.1. Tosimplify the notation, we indicate a coset by its representative only.

Table 17.1 A Partial Multiplication Table for Example 10

1 x x 1 1 x2 x2 1 1 x2 1 x x2 1 x 1 1

1 1 x x 1 1 x2 x2 1 1 x2 1 x x2 1 x 1 1x x x2 x2 1 x x 1 1 1 x2 1 x 1 1 x2 1 1x 1 1 x 1 1 x2 1 x x2 1 1 x2 1 x 1 1 x2 1 xx2 x2 x 1 1 x2 1 x 1 1 x2 1 x x x2 1 1 1x2 1 1 x2 1 1 1 x2 x x2 1 x 1 1 x 1 1 x2 1 x

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(Complete the table yourself. Keep in mind that x3 can be replaced by x 1 1 and x4 by x2 1 x.)

EXAMPLE 11 Since x2 1 1 has no zero in Z3, it is irreducible overZ3. Thus, Z3[x]/�x2 1 1� is a field. Analogous to Example 12 in Chapter 14,Z3[x]/�x2 1 1� 5 {ax 1 b 1 �x2 1 1� | a, b [ Z3}. Thus, this field hasnine elements. A multiplication table for this field can be obtained fromTable 13.1 by replacing i by x. (Why does this work?)

Unique Factorization in Z[x]As a further application of the ideas presented in this chapter, we nextprove that Z[x] has an important factorization property. In Chapter 18,we will study this property in greater depth. The first proof of Theorem17.6 was given by Gauss. In reading this theorem and its proof, keep inmind that the units in Z[x] are precisely f(x) 5 1 and f(x) 5 21 (seeExercise 25 in Chapter 12), the irreducible polynomials of degree 0over Z are precisely those of the form f(x) 5 p and f(x) 5 2p where p isa prime, and every nonconstant polynomial from Z[x] that is irreducibleover Z is primitive (see Exercise 3).

Theorem 17.6 Unique Factorization in Z[x]

PROOF Let f(x) be a nonzero, nonunit polynomial from Z[x]. Ifdeg f(x) 5 0, then f(x) is constant and the result follows from theFundamental Theorem of Arithmetic. If deg f(x) . 0, let b denote thecontent of f(x), and let b1b2 ? ? ? bs be the factorization of b as a productof primes. Then, f(x) 5 b1b2 ? ? ? bs f1(x), where f1(x) belongs to Z[x], is

Every polynomial in Z[x] that is not the zero polynomial or a unitin Z[x] can be written in the form b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x),where the bi’s are irreducible polynomials of degree 0, and the pi(x)’sare irreducible polynomials of positive degree. Furthermore, if

b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ? qn(x),

where the b’s and c’s are irreducible polynomials of degree 0, and thep(x)’s and q(x)’s are irreducible polynomials of positive degree, thens 5 t, m 5 n, and, after renumbering the c’s and q(x)’s, we have bi 56ci for i 5 1, . . . , s; and pi(x) 5 6qi(x) for i 5 1, . . . , m.

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primitive and deg f1(x) 5 deg f(x). Thus, to prove the existence portionof the theorem, it suffices to show that a primitive polynomial f(x) ofpositive degree can be written as a product of irreducible polynomialsof positive degree. We proceed by induction on deg f(x). If deg f(x) 5 1,then f(x) is already irreducible and we are done. Now suppose thatevery primitive polynomial of degree less than deg f(x) can be writtenas a product of irreducibles of positive degree. If f(x) is irreducible,there is nothing to prove. Otherwise, f(x) 5 g(x)h(x), where both g(x)and h(x) are primitive and have degree less than that of f(x). Thus, by in-duction, both g(x) and h(x) can be written as a product of irreducibles ofpositive degree. Clearly, then, f(x) is also such a product.

To prove the uniqueness portion of the theorem, suppose thatf(x) 5 b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ?qn(x), where the b’s and c’s are irreducible polynomials of degree 0, andthe p(x)’s and q(x)’s are irreducible polynomials of positive degree. Letb 5 b1b2 ? ? ? bs and c 5 c1c2 ? ? ? ct. Since the p(x)’s and q(x)’s areprimitive, it follows from Gauss’s Lemma that p1(x)p2(x) ? ? ? pm(x) andq1(x)q2(x) ? ? ? qn(x) are primitive. Hence, both b and c must equal plusor minus the content of f(x) and, therefore, are equal in absolute value.It then follows from the Fundamental Theorem of Arithmetic that s 5 tand, after renumbering, bi 5 6ci for i 5 1, 2, . . . , s. Thus, by cancel-ing the constant terms in the two factorizations for f(x), we havep1(x)p2(x) ? ? ? pm(x) 5 6q1(x) q2(x) ? ? ? qn(x). Now, viewing the p(x)’sand q(x)’s as elements of Q[x] and noting that p1(x) divides q1(x) ? ? ?qn(x), it follows from Corollary 2 of Theorem 17.5 and induction (seeExercise 27) that p1(x) | qi(x) for some i. By renumbering, we may as-sume i 5 1. Then, since q1(x) is irreducible, we have q1(x) 5 (r/s)p1(x),where r, s [ Z. However, because both q1(x) and p1(x) are primitive, wemust have r/s 5 61. So, q1(x) 5 6p1(x). Also, after canceling, we havep2(x) ? ? ? pm(x) 5 6q2(x) ? ? ? qn(x). Now, we may repeat the argumentabove with p2(x) in place of p1(x). If m , n, after m such steps wewould have 1 on the left and a nonconstant polynomial on the right.Clearly, this is impossible. On the other hand, if m . n, after n steps wewould have 61 on the right and a nonconstant polynomial on the left—another impossibility. So, m 5 n and pi(x) 5 6qi(x) after suitablerenumbering of the q(x)’s.

Weird Dice: An Applicationof Unique Factorization

EXAMPLE 12 Consider an ordinary pair of dice whose faces arelabeled 1 through 6. The probability of rolling a sum of 2 is 1/36, theprobability of rolling a sum of 3 is 2/36, and so on. In a 1978 issue of

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Scientific American [1], Martin Gardner remarked that if one were tolabel the six faces of one cube with the integers 1, 2, 2, 3, 3, 4 and the sixfaces of another cube with the integers 1, 3, 4, 5, 6, 8, then the probabil-ity of obtaining any particular sum with these dice (called Sichermandice) would be the same as the probability of rolling that sum with ordi-nary dice (that is, 1/36 for a 2, 2/36 for a 3, and so on). See Figure 17.1.In this example, we show how the Sicherman labels can be derived, andthat they are the only possible such labels besides 1 through 6. To do so,we utilize the fact that Z[x] has the unique factorization property.

Figure 17.1

To begin, let us ask ourselves how we may obtain a sum of 6, say, withan ordinary pair of dice. Well, there are five possibilities for the two faces:(5, 1), (4, 2), (3, 3), (2, 4), and (1, 5). Next we consider the product of thetwo polynomials created by using the ordinary dice labels as exponents:

(x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x).

Observe that we pick up the term x6 in this product in precisely the fol-lowing ways: x5 ? x1, x4 ? x2 , x3 ? x3, x2 ? x4, x1 ? x5. Notice the correspon-dence between pairs of labels whose sums are 6 and pairs of termswhose products are x6. This correspondence is one-to-one, and it is validfor all sums and all dice—including the Sicherman dice and any otherdice that yield the desired probabilities. So, let a1, a2, a3, a4, a5, a6 andb1, b2, b3, b4, b5, b6 be any two lists of positive integer labels for the facesof a pair of cubes with the property that the probability of rolling anyparticular sum with these dice (let us call them weird dice) is the same asthe probability of rolling that sum with ordinary dice labeled 1 through6. Using our observation about products of polynomials, this means that

(x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x)5 (xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6) ?

(xb1 1 xb2 1 xb3 1 xb4 1 xb5 1 xb6). (1)

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Now all we have to do is solve this equation for the a’s and b’s. Here iswhere unique factorization in Z[x] comes in. The polynomial x6 1 x5 1x4 1 x3 1 x2 1 x factors uniquely into irreducibles as

x(x 1 1)(x2 1 x 1 1)(x2 2 x 1 1)

so that the left-hand side of Equation (1) has the irreducible factorization

x2(x 1 1)2(x2 1 x 1 1)2(x2 2 x 1 1)2.

So, by Theorem 17.6, this means that these factors are the only possibleirreducible factors of P(x) 5 xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6. Thus,P(x) has the form

xq(x 1 1)r(x2 1 x 1 1)t(x2 2 x 1 1)u,

where 0 # q, r, t, u # 2.To restrict further the possibilities for these four parameters, we eval-

uate P(1) in two ways. P(1) 5 1a1 1 1a2 1 ? ? ? 1 1a6 5 6 and P(1) 5 1q2r3t1u. Clearly, this means that r 5 1 and t 5 1. What about q?Evaluating P(0) in two ways shows that q 2 0. On the other hand, if q 5 2, the smallest possible sum one could roll with the correspondinglabels for dice would be 3. Since this violates our assumption, we havenow reduced our list of possibilities for q, r, t, and u to q 5 1, r 5 1,t 5 1, and u 5 0, 1, 2. Let’s consider each of these possibilities in turn.

When u 5 0, P(x) 5 x4 1 x3 1 x3 1 x2 1 x2 1 x, so the die labelsare 4, 3, 3, 2, 2, 1—a Sicherman die.

When u 5 1, P(x) 5 x6 1 x5 1 x4 1 x3 1 x2 1 x, so the die labelsare 6, 5, 4, 3, 2, 1—an ordinary die.

When u 5 2, P(x) 5 x8 1 x6 1 x5 1 x4 1 x3 1 x, so the die labelsare 8, 6, 5, 4, 3, 1—the other Sicherman die.

This proves that the Sicherman dice do give the same probabilitiesas ordinary dice and that they are the only other pair of dice that havethis property.

Exercises

No matter how good you are at something, there’s always about a millionpeople better than you.

HOMER SIMPSON

1. Verify the assertion made in Example 2.2. Suppose that D is an integral domain and F is a field containing D.

If f(x) [ D[x] and f(x) is irreducible over F but reducible over D,what can you say about the factorization of f(x) over D?

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3. Show that a nonconstant polynomial from Z[x] that is irreducibleover Z is primitive. (This exercise is referred to in this chapter.)

4. Suppose that f(x) 5 xn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]. If r is ra-

tional and x 2 r divides f(x), show that r is an integer.5. Let F be a field and let a be a nonzero element of F.

a. If af(x) is irreducible over F, prove that f(x) is irreducible over F.b. If f(ax) is irreducible over F, prove that f(x) is irreducible over F.c. If f(x 1 a) is irreducible over F, prove that f(x) is irreducible

over F.d. Use part c to prove that 8x3 2 6x 1 1 is irreducible over Q.(This exercise is referred to in this chapter.)

6. Suppose that f(x) [ Zp[x] and is irreducible over Zp, where p is aprime. If deg f(x) 5 n, prove that Zp[x]/� f(x)� is a field with pn ele-ments.

7. Construct a field of order 25.8. Construct a field of order 27.9. Show that x3 1 x2 1 x 1 1 is reducible over Q. Does this fact con-

tradict the corollary to Theorem 17.4?10. Determine which of the polynomials below is (are) irreducible

over Q.a. x5 1 9x4 1 12x2 1 6b. x4 1 x 1 1c. x4 1 3x2 1 3d. x5 1 5x2 1 1e. (5/2)x5 1 (9/2)x4 1 15x3 1 (3/7)x2 1 6x 1 3/14

11. Show that x4 1 1 is irreducible over Q but reducible over R. (Thisexercise is referred to in this chapter.)

12. Show that x2 1 x 1 4 is irreducible over Z11.13. Let f(x) 5 x3 1 6 [ Z7[x]. Write f(x) as a product of irreducible

polynomials over Z7.14. Let f(x) 5 x3 1 x2 1 x 1 1 [ Z2[x]. Write f(x) as a product of irre-

ducible polynomials over Z2.15. Let p be a prime.

a. Show that the number of reducible polynomials over Zp of theform x2 1 ax 1 b is p( p 1 1)/2.

b. Determine the number of reducible quadratic polynomials over Zp.16. Let p be a prime.

a. Determine the number of irreducible polynomials over Zp of theform x2 1 ax 1 b.

b. Determine the number of irreducible quadratic polynomialsover Zp.

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17. Show that for every prime p there exists a field of order p2.18. Prove that, for every positive integer n, there are infinitely many

polynomials of degree n in Z[x] that are irreducible over Q.19. Show that the field given in Example 11 in this chapter is isomor-

phic to the field given in Example 9 in Chapter 13.20. Let f(x) [ Zp[x]. Prove that if f(x) has no factor of the form x2 1

ax 1 b, then it has no quadratic factor over Zp.21. Find all monic irreducible polynomials of degree 2 over Z3.22. Given that p is not the zero of a nonzero polynomial with rational

coefficients, prove that p 2 cannot be written in the form ap 1 b,where a and b are rational.

23. Find all the zeros and their multiplicities of x5 1 4x4 1 4x3 2 x2 24x 1 1 over Z5.

24. Find all zeros of f(x) 5 3x2 1 x 1 4 over Z7 by substitution. Findall zeros of f(x) by using the Quadratic Formula (2b 6 ) ?(2a)21 (all calculations are done in Z7). Do your answers agree?Should they? Find all zeros of g(x) 5 2x2 1 x 1 3 over Z5 by sub-stitution. Try the Quadratic Formula on g(x). Do your answersagree? State necessary and sufficient conditions for the QuadraticFormula to yield the zeros of a quadratic from Zp[x], where p is aprime greater than 2.

25. (Rational Root Theorem) Let

f (x) 5 anxn 1 an21x

n21 1 ? ? ? 1 a0 [ Z[x]

and an 2 0. Prove that if r and s are relatively prime integers andf (r/s) 5 0, then r | a0 and s | an.

26. Let F be a field and f(x) [ F[x]. Show that, as far as deciding uponthe irreducibility of f(x) over F is concerned, we may assume thatf(x) is monic. (This assumption is useful when one uses a computerto check for irreducibility.)

27. Let F be a field and let p(x), a1(x), a2(x), . . . , ak(x) [ F[x], wherep(x) is irreducible over F. If p(x) | a1(x)a2(x) ? ? ? ak(x), show thatp(x) divides some ai(x). (This exercise is referred to in the proof ofTheorem 17.6.)

28. Explain how the Mod p Irreducibility Test (Theorem 17.3) can beused to test members of Q[x] for irreducibility.

29. Show that x4 1 1 is reducible over Zp for every prime p. (This ex-ercise is referred to in this chapter.)

30. If p is a prime, prove that xp21 2 xp22 1 xp23 2 ? ? ? 2 x 1 1 isirreducible over Q.

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31. Let F be a field and let p(x) be irreducible over F. If E is a fieldthat contains F and there is an element a in E such that p(a) 5 0,show that the mapping f: F[x] → E given by f(x) → f(a) is a ringhom*omorphism with kernel �p(x)�. (This exercise is referred to inChapter 20.)

32. Prove that the ideal �x2 1 1� is prime in Z[x] but not maximal in Z[x].33. Let F be a field and let p(x) be irreducible over F. Show that {a 1

� p(x)� | a [ F} is a subfield of F[x]/�p(x)� isomorphic to F. (Thisexercise is referred to in Chapter 20.)

34. Suppose there is a real number r with the property that r 1 1/r isan odd integer. Prove that r is irrational.

35. In the game of Monopoly, would the probabilities of landing onvarious properties be different if the game were played withSicherman dice instead of ordinary dice? Why?

36. Carry out the analysis given in Example 12 for a pair of tetrahe-drons instead of a pair of cubes. (Define ordinary tetrahedral diceas the ones labeled 1 through 4.)

37. Suppose in Example 12 that we begin with n (n . 2) ordinary diceeach labeled 1 through 6, instead of just two. Show that the onlypossible labels that produce the same probabilities as n ordinarydice are the labels 1 through 6 and the Sicherman labels.

38. Show that one two-sided die labeled with 1 and 4 and another 18-sided die labeled with 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8yield the same probabilities as an ordinary pair of cubes labeled1 through 6. Carry out an analysis similar to that given in Example12 to derive these labels.

Computer Exercises

The experiment serves two purposes, often independent one from theother: it allows the observation of new facts, hitherto either unsuspected,or not yet well defined; and it determines whether a working hypothesisfits the world of observable facts.

RENÉ J. DUBOS

Software for the computer exercises in this chapter is available at thewebsite:

http://www.d.umn.edu/~jgallian

1. This software implements the Mod p Irreducibility Test. Use it totest the polynomials in the examples given in this chapter and thepolynomials given in Exercise 10 for irreducibility.

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2. Use software such as Mathematica, Maple, or GAP to express xn 2 1 as a product of irreducible polynomials with integer coeffi-cients for n 5 4, 8, 12, and 20. On the basis of these data, make aconjecture about the coefficients of the irreducible factors of xn 2 1.Test your conjecture for n 5 105. Does your conjecture hold up?

3. Use software such as Mathematica, Maple, or GAP to express xpn2 x

as a product of irreducibles over Zp for several choices of the primep and n. On the basis of these data, make a conjecture relating thedegrees of the irreducible factors of xpn

2 x and n.

Reference

1. Martin Gardner, “Mathematical Games,” Scientific American 238/2(1978): 19–32.

Suggested Readings

Duane Broline, “Renumbering the Faces of Dice,” Mathematics Magazine52 (1979): 312–315.

In this article, the author extends the analysis we carried out inExample 12 to dice in the shape of Platonic solids.

J. A. Gallian and D. J. Rusin, “Cyclotomic Polynomials and NonstandardDice,” Discrete Mathematics 27 (1979): 245–259.

Here Example 12 is generalized to the case of n dice each with mlabels for all n and m greater than 1.

Randall Swift and Brian Fowler, “Relabeling Dice,” MathematicsMagazine 72 (1999): 204–208.

The authors use the method presented in this chapter to derive positiveinteger labels for a pair of dice that are not six-sided but give thesame probabilities for the sum of the faces as a pair of cubes labeled1 through 6.

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Serge Lang

Lang’s Algebra changed the way graduatealgebra is taught . . . . It has affected allsubsequent graduate-level algebra books.

Citation for the Steele Prize

SERGE LANG was a prolific mathematician,inspiring teacher, and political activist. Hewas born near Paris on May 19, 1927. Hisfamily moved to Los Angeles when he was ateenager. Lang received a B.A. in physicsfrom Caltech in 1946 and a Ph.D. in mathe-matics from Princeton in 1951 under EmilArtin (see the biography in Chapter 19). Hisfirst permanent position was at ColumbiaUniversity in 1955, but in 1971 Lang re-signed his position at Columbia as a protestagainst Columbia’s handling of Vietnam an-tiwar protesters. He joined Yale University in1972 and remained there until his retirement.

Lang made significant contributions tonumber theory, algebraic geometry, differ-ential geometry, and analysis. He wrote morethan 120 research articles and 60 books. Hismost famous and influential book was hisgraduate-level Algebra. Lang was a prize-

winning teacher known for his extraordinarydevotion to students. Lang often got intoheated discussions about mathematics, thearts, and politics. In one incident, he threat-ened to hit a fellow mathematician with abronze bust for not conceding it was self-evident that the Beatles were greater musi-cians than Beethoven.

Among Lang’s honors were the SteelePrize for Mathematical Exposition from theAmerican Mathematcial Society, the ColePrize in Algebra (see Chapter 25), and elec-tion to the National Academy of Sciences.Lang died on September 25, 2005, at theage of 78.

For more information about Lang, visit:

http://wikipedia.org/wiki/Serge_Lang

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322

Divisibility in Integral Domains

Give me a fruitful error anytime, full of seeds, bursting with its owncorrections. You can keep your sterile truth for yourself.

VILFREDO PARETO

Irreducibles, PrimesIn the preceding two chapters, we focused on factoring polynomialsover the integers or a field. Several of those results—unique factoriza-tion in Z[x] and the division algorithm for F[x], for instance—are nat-ural counterparts to theorems about the integers. In this chapter and thenext, we examine factoring in a more abstract setting.

Definition Associates, Irreducibles, Primes

Elements a and b of an integral domain D are called associates ifa 5 ub, where u is a unit of D. A nonzero element a of an integraldomain D is called an irreducible if a is not a unit and, whenever b,c [ D with a 5 bc, then b or c is a unit. A nonzero element a of anintegral domain D is called a prime if a is not a unit and a | bc impliesa | b or a | c.

Roughly speaking, an irreducible is an element that can be factoredonly in a trivial way. Notice that an element a is a prime if and only if�a� is a prime ideal.

Relating the definitions above to the integers may seem a bit confus-ing, since in Chapter 0 we defined a positive integer to be a prime if itsatisfies our definition of an irreducible, and we proved that a prime in-teger satisfies the definition of a prime in an integral domain (Euclid’sLemma). The source of the confusion is that in the case of the integers,the concepts of irreducibles and primes are equivalent, but in general, aswe will soon see, they are not.

The distinction between primes and irreducibles is best illustrated byintegral domains of the form Z[ ] 5 {a 1 b | a, b [ Z}, where d isnot 1 and is not divisible by the square of a prime. (These rings are offundamental importance in number theory.) To analyze these rings, weneed a convenient method of determining their units, irreducibles, and

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primes. To do this, we define a function N, called the norm, from Z[ ]into the nonnegative integers by N(a 1 b ) 5 |a2 2 db2|. We leave itto the reader (Exercise 1) to verify the following four properties: N(x) 5 0if and only if x 5 0; N(xy) 5 N(x)N(y) for all x and y; x is a unit if andonly if N(x) 5 1; and, if N(x) is prime, then x is irreducible in Z[ ].

EXAMPLE 1 We exhibit an irreducible in Z[ ] that is not prime.Here, N(a 1 b ) 5 a2 1 3b2. Consider 1 1 . Suppose that wecan factor this as xy, where neither x nor y is a unit. Then N(xy) 5N(x)N(y) 5 N(1 1 ) 5 4, and it follows that N(x) 5 2. But there areno integers a and b that satisfy a2 1 3b2 5 2. Thus, x or y is a unit and1 1 is an irreducible. To verify that it is not prime, we observe that(1 1 )(1 2 ) 5 4 5 2 ? 2, so that 1 1 divides 2 ? 2. On theother hand, for integers a and b to exist so that 2 5 (1 1 )(a 1b ) 5 (a 2 3b) 1 (a 1 b) , we must have a 2 3b 5 2 and a 1b 5 0, which is impossible.

Showing that an element of a ring of the form Z[ ] is irreducible ismore difficult when d . 1. The next example illustrates one method ofdoing this. The example also shows that the converse of the fourthproperty above for the norm is not true. That is, it shows that x may beirreducible even if N(x) is not prime.

EXAMPLE 2 The element 7 is irreducible in the ring Z[ ]. To verifythis assertion, suppose that 7 5 xy, where neither x nor y is a unit. Then49 5 N(7) 5 N(x) N(y), and since x is not a unit, we cannot have N(x) 51. This leaves only the case N(x) 5 7. Let x 5 a 1 b . Then there areintegers a and b satisfying |a2 2 5b2| 5 7. This means that a2 2 5b2 567. Viewing this equation modulo 7 and trying all possible cases for aand b reveals that the only solutions are a 5 0 5 b. But this means thatboth a and b are divisible by 7, and this implies that |a2 2 5b2| 5 7 isdivisible by 49, which is false.

Example 1 raises the question of whether or not there is an integraldomain containing a prime that is not an irreducible. The answer: no.

Theorem 18.1 Prime Implies Irreducible

In an integral domain, every prime is an irreducible.

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PROOF Suppose that a is a prime in an integral domain and a 5 bc.We must show that b or c is a unit. By the definition of prime, we knowthat a | b or a | c. Say at 5 b. Then 1b 5 b 5 at 5 (bc)t 5 b(ct) and,by cancellation, 1 5 ct. Thus, c is a unit.

Recall that a principal ideal domain is an integral domain in whichevery ideal has the form �a�. The next theorem reveals a circ*mstancein which primes and irreducibles are equivalent.

Theorem 18.2 PID Implies Irreducible Equals Prime

PROOF Theorem 18.1 shows that primes are irreducibles. To prove theconverse, let a be an irreducible element of a principal ideal domain Dand suppose that a | bc. We must show that a | b or a | c. Consider theideal I 5 {ax 1 by | x, y [ D} and let �d� 5 I. Since a [ I, we can writea 5 dr, and because a is irreducible, d is a unit or r is a unit. If d is aunit, then I 5 D and we may write 1 5 ax 1 by. Then c 5 acx 1 bcy,and since a divides both terms on the right, a also divides c.

On the other hand, if r is a unit, then �a� 5 �d� 5 I, and, because b [ I,there is an element t in D such that at 5 b. Thus, a divides b.

It is an easy consequence of the respective division algorithms for Zand F[x], where F is a field, that Z and F[x] are principal ideal domains(see Exercise 41 in Chapter 14 and Theorem 16.3). Our next exampleshows, however, that one of the most familiar rings is not a principalideal domain.

EXAMPLE 3 We show that Z[x] is not a principal ideal domain.Consider the ideal I 5 { f(x) [ Z[x] | f(0) is even}. We claim that I is notof the form �h(x)�. If this were so, there would be f(x) and g(x) in Z[x]such that 2 5 h(x)f(x) and x 5 h(x)g(x), since both 2 and x belong to I.By the degree rule (Exercise 17 in Chapter 16), 0 5 deg 2 5 deg h(x) 1deg f(x), so that h(x) is a constant polynomial. To determine whichconstant, we observe that 2 5 h(1)f(1). Thus, h(1) 5 61 or 62. Since1 is not in I, we must have h(x) 5 62. But then x 5 62g(x), which isnonsense.

We have previously proved that the integral domains Z and Z[x] haveimportant factorization properties: Every integer greater than 1 can be

In a principal ideal domain, an element is an irreducible if and onlyif it is a prime.

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