Abstract Algebra (Herstein 3rd Ed)

Third Edition

lo N„ Hen-stein Late Professor of Mathematics University of Chicago

JOHN W I L E Y & SONS, INC. N E W Y O R K • CHICHESTER • W E I N H E I M • BRISBANE • SINGAPORE • T O R O N T O

Cover Photograph: Charnley Residence, entryway. Photo by © Nick Merrick/Hedrich-Blessing This book was typeset in 10/12 Times Ten Roman by University Graphics, Inc. The paper in this book was manufactured by a mill whose forest management programs include sustained yield harvesting of it timberlands. Sustained yield harvesting principles ensure that the number of trees cut each year does not exceed the amount of new growth. Copyright © 1999 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by means, electronic, mechanical, photocopying recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (508) 750-8400, fax (508) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc. 605 Third Avenue, New York, NY 10158-0012, (212) 850-6008, E-mail: [email protected]. To order books or for customer service call l-800-CALL-WILEY(225-5945). Library of Congress Cataloging in Publication Data Herstein, I. N. Abstract algebra / I.N. Herstein. — 3rd ed. p. cm. Includes index. ISBN 0-471-36879-2 1. Algebra, Abstract. I. Title QA162.H47 1995 95-21470 512'.02^dc20 CIP Printed in the United States of America. 10 9 8 7 6 5. 4 3 2

To Biska

Preface

U

ix

Things Familiar and Less Familiar 1 2 3 4 5 6 7

A Few Preliminary R e m a r k s 1 Set Theory 3 Mappings 8 A(S) (The Set of 1-1 Mappings of S onto Itself) T h e Integers 21 Mathematical Induction 29 Complex N u m b e r s 32

J£M G r o u p s 1 2 3 4 5 6 7 8 9 10 11

1

40

Definitions and E x a m p l e s of G r o u p s 40 Some Simple R e m a r k s 48 Subgroups 51 Lagrange's T h e o r e m 56 66 H o m o m o r p h i s m s and N o r m a l Subgroups Factor G r o u p s 77 The H o m o m o r p h i s m T h e o r e m s 84 Cauchy's T h e o r e m 88 Direct Products 92 Finite Abelian G r o u p s (Optional) 96 101 Coniugacy and Svlow's T h e o r e m (Optional)

16

Contents

The Symmetric G r o u p 1 2 3

Preliminaries 108 Cycle Decomposition 111 O d d and Even Permutations

Ring T h e o r y 1 2 3 4 5 6 7

119

125

Definitions and Examples 125 Some Simple Results 137 Ideals, H o m o m o r p h i s m s , and Quotient Rings 139 Maximal Ideals 148 Polynomial Rings 151 Polynomials over the Rationals 166 Field of Quotients of an Integral D o m a i n 172

Fields

176

1 2 3 4 5 6

Examples of Fields 176 A Brief Excursion into Vector Spaces Field Extensions 191 Finite Extensions 198 Constructibility 201 R o o t s of Polynomials 207

S p e c i a l Topics ( O p t i o n a l ) 1 2 3 4 5 6 7

108

215

T h e Simplicity of A„ 215 Finite Fields I 221 Finite Fields II: Existence 224 Finite Fields III: Uniqueness 227 Cyclotomic Polynomials 229 Liouville's Criterion 236 T h e Irrationality of IT 239

Index

243

180

PREFACE TO THE THIRD EDITION

W h e n we were asked to p r e p a r e the third edition of this book, it was our consensus that it should not be altered in any significant way, and that Herstein's informal style should be preserved. W e feel that one of the b o o k ' s virtues is the fact that it covers a big chunk of abstract algebra in a condensed and interesting way. A t the same time, without trivializing the subject, it remains accessible to most undergraduates. W e have, however, corrected m i n o r errors, straightened out inconsistencies, clarified and expanded some proofs, and added a few examples. To resolve the many typographical problems of the second edition, Prentice Hall has had the b o o k completely retypeset—making it easier a n d m o r e pleasurable to read. It has b e e n pointed out t o us that some instructors would find it useful to have the Symmetric G r o u p S„ and the cycle notation available in C h a p t e r 2, in order to provide m o r e examples of groups. R a t h e r t h a n alter the arrangement of the contents, thereby disturbing the original balance, we suggest an alternate route through the material, which addresses this concern. After Section 2.5, one could spend an hour discussing permutations and their cycle decomposition (Sections 3.1 and 3.2), leaving the proofs until later. T h e students might then go over several past examples of finite groups and explicitly set u p isomorphisms with subgroups of S„. This exercise would be motivated by Cayley's theorem, q u o t e d in Section 2.5. At the same time, it would have the beneficial result of making the students more comfortable with the concept of an isomorphism. T h e instructor could then weave in the various subgroups of the Symmetric G r o u p s S„ as examples throughout the remain-

x

P r e f a c e to T h i r d E d i t i o n

Ch. 6

der of Chapter 2. If desired, one could even introduce Sections 3.1 and 3.2 after Section 2.3 or 2.4. T w o changes in the format have been m a d e since the first edition. First, a Symbol List has b e e n included to facilitate keeping track of terminology. Second, a few problems have b e e n m a r k e d with an asterisk (*). T h e s e serve as a vehicle to introduce concepts and simple arguments that relate in some important way to the discussion. As such, they should be read carefully. Finally, we take this opportunity to thank the m a n y individuals whose collective efforts have helped to improve this edition. W e t h a n k the reviewers: Kwangil K o h from N o r t h Carolina State University, D o n a l d P a s s m a n from the University of Wisconsin, and R o b e r t Zinc from P u r d u e University. And, of course, we thank George Lobell and Elaine W e t t e r a u , and others at Prentice Hall who have been most helpful. B a r b a r a Cortzen David J. Winter

In the last half-century or so abstract algebra has b e c o m e increasingly important not only in mathematics itself, but also in a variety of other disciplines. For instance, the importance of the results and concepts of abstract algebra play an ever m o r e important role in physics, chemistry, and computer science, to cite a few such outside fields. In mathematics itself abstract algebra plays a dual role: that of a unifying link between disparate parts of mathematics and that of a research subject with a highly active life of its own. It has been a fertile and rewarding research area both in the last 100 years a n d at t h e present m o m e n t . Some of the great accomplishments of our twentieth-century mathematics have b e e n precisely in this area. Exciting results have been proved in group theory, commutative and noncommutative ring theory, Lie algebras, Jordan algebras, combinatorics, and a host of other parts of what is k n o w n as abstract algebra. A subject that was once regarded as esoteric has become considered as fairly downto-earth for a large cross section of scholars. T h e p u r p o s e of this b o o k is twofold. F o r those readers w h o either want to go o n to do research in mathematics or in some allied fields that use algebraic notions and methods, this b o o k should serve as an introduction—and, we stress, only as an introduction—to this fascinating subject. F o r interested readers who want to learn what is going on in an engaging part of m o d e r n mathematics, this book could serve that purpose, as well as provide t h e m with some highly usable tools to apply in the areas in which they are interested. T h e choice of subject matter has b e e n m a d e with the objective of introducing readers to some of the fundamental algebraic systems that are b o t h in-

xii

P r e f a c e t o First E d i t i o n

teresting and of wide use. Moreover, in each of these systems the aim has b e e n to arrive at some significant results. T h e r e is little purpose served in studying some abstract object without seeing some nontrivial consequences of the study. W e h o p e that we have achieved the goal of presenting interesting, applicable, and significant results in each of the systems we have chosen to discuss. A s the r e a d e r will soon see, there are many exercises in the book. T h e y are often divided into three categories: easier, middle-level, and h a r d e r (with an occasional very h a r d ) . T h e purpose of these problems is to allow students to test their assimilation of the material, to challenge their mathematical ingenuity, to p r e p a r e the ground for material that is yet to come, and to be a m e a n s of developing mathematical insight, intuition, and techniques. R e a d e r s should not b e c o m e discouraged if they do not manage to solve all the p r o b lems. T h e intent of many of the problems is that they b e tried—even if not solved—for the pleasure (and frustration) of the reader. Some of the p r o b lems a p p e a r several times in the book. Trying to do the problems is undoubtedly the best way of going about learning the subject. We have strived to present the material in the language a n d tone of a classroom lecture. Thus the presentation is somewhat chatty; we h o p e that this will put the readers at their ease. A n attempt is m a d e to give many and revealing examples of the various concepts discussed. Some of these examples are carried forward to be examples of other p h e n o m e n a that come u p . T h e y are often referred to as the discussion progresses. W e feel that the book is self-contained, except in o n e section—the second last one of the b o o k — w h e r e we m a k e implicit use of the fact that a polynomial over the complex field has complex roots (that is the celebrated Fundamental Theorem of Algebra due to Gauss), and in the last section where we m a k e use of a little of the calculus. W e are grateful to many people for their comments and suggestions on earlier drafts of the book. M a n y of the changes they suggested have b e e n incorporated and should improve the readability of the book. W e should like to express our special thanks to Professor Martin Isaacs for his highly useful comments. W e are also grateful to Fred Flowers for his usual superb j o b of typing the manuscript, and to Mr. Gary W. Ostedt of the Macmillan C o m p a n y for his enthusiasm for the project and for bringing it to publication. With this we wish all the readers a happy voyage on the mathematical journey they are about to u n d e r t a k e into this delightful and beautiful realm of abstract algebra. I.N.H.

aES a£S SCT,TDS S=T

0 AUB AC)B {s E 5 | s satisfies P] A - B A' (a, b) AX B

U f:S-> T

m i: S —» S, i

s

l

r (A)

f°g, fg A(S)

s„ n\ Z 0(s)

a is an element of the set S, 3 a is not an element of the set S, 3 S is a subset of the set T, 3 The sets 5 and T a r e equal (have the same elements), 4 T h e empty set, 4 T h e union of the sets A and B, 4 T h e intersection of the sets A and B, 4 T h e subset of elements of S satisfying P, 4 The difference of the sets A and fi, 4 T h e c o m p l e m e n t of A, 5 O r d e r e d pair consisting of a, b (see also below), 5 T h e Cartesian product of A and B, 5 T h e set of real n u m b e r s , 8 Function from the set S to the set T, 8 Image of the element s under the function/, 8 T h e identity function on S, 9 Inverse image of t u n d e r f, 10 Inverse image of a subset A of T under / : S —» T, 10 Composition or product of functions / a n d g , 11,18 Set of 1-1 mappings from a set S to S, 16 Symmetric group of degree n, 16,109 n factorial, 17 Set of integers, 21 Orbit of s relative to mapping /, 21

xiv

S y m b o l List

N m |n m |n (a, b) C i,—i z = a + bi z = a - bi IIz |z| ?• (cos 9 + / sin 6) 9„ Q E |G | C(a) (a) Z(G) a ~ b a = b mod n a = b(n) [a] cl(a) o(a) i {H) Z„ U„ U. Given an element s E S, then g sends it into the element g(s) in T; so g(s) is ripe for being acted on by / Thus we get an element f(g(s)) E U. W e claim that this procedure p r o vides us with a mapping from 5 to U. (Verify!) W e define this m o r e formally in the Definition, if g : S --> 7' and f:T^>U, then the composition (or product), d e n o t e d by f°g, is the mapping f°g:S^> U defined by (f°g)(s) = f(g(s)) for every s G S . N o t e that to compose the two mappings / and g—that is, for f°g to have any sense—the terminal set, T, for the mapping g must be the initial set for the mapping /. One special time when we can always compose any two mappings is when S = T = U, that is, when we map S into itself. Although special, this case is of the utmost importance. W e verify a few properties of this composition of mappings. L e m m a 1.3.1. a

(f g)°

If /;: .S'

T, g : T

U. and / : (/ ^ V, then

=

ItProof. H o w shall we go about proving this lemma? To verify that two

mappings are equal, we merely must check that they do the same thing to every element. Note first of all that b o t h f° (g ° h) and ( / ° g)° h define m a p pings from S to V, so it m a k e s sense to speak about their possible equality. O u r task, then, is to show that for every s E S, (f°(g°h))(s) = ((/° g) ° )( )- W e apply the definition of composition to see that n

s

( / o (go h))( )= S

/((go /,)(,))=

f(g(h(s))).

T h i n g s F a m i l i a r a n d Less F a m i l i a r

12

Ch. 1

Unraveling 0

{(f g)°h)(s)

=

(f°g)(Ks))=f(g(h(s))),

we do indeed see that

(/»(r#) =

((f°g)°hm

for every s E S. Consequently, by definition, f° (g ° h) = (f° g)° /?. • (The symbol • will always indicate that the proof has been completed.) This equality is described by saying that mappings, under composition, satisfy the associative law. Because of the equality involved t h e r e is really n o n e e d for parentheses, so we write f° (g h) as f° g ° h. 0

Lemma 1.3.2. is also 1-1.

T a n d f : T-> U are both 1-1, t h e n / ° g : 5 - » U

Proof. Let us suppose that (f°g)(si) = ( / ° g ) ( s ) ; thus, by definition, f(g(*i)) = f(g( 2))S i n c e / i s 1-1, we get from this that g( ) = g(s ); however, g is also 1-1, thus s = s follows. Since (f°g)(si) = (f°g)(s ) forces S\ = s , the m a p p i n g / g is 1-1. • 2

s

2

Sl

x

2

2

0

2

W e leave the proof of t h e next R e m a r k to the reader. Remark. If g : S also onto.

T and / : T-> U are both onto, then f°g

: S -> U is

A n immediate consequence of combining the R e m a r k and L e m m a 1.3.2 is to obtain L e m m a 1.3.3. If g : S Uis also a bijection.

T and f:T^

U are both bijections, then

x

If / i s a 1-1 correspondence of S o n t o T, then the "object" f~ :T-^S defined earlier can easily b e shown to be a 1-1 mapping of T o n t o S. In this case it is called the inverse of /. In this situation we have Lemma 1.3.4. I f / : S —>• T is a bijection, t h e n / ° / ~ ' = / and / ^ ° / = i , where i and i are the identity mappings of S and T, respectively. r

s

s

r

l

Proof. W e verify o n e of these. If / G T, then {f°r ){t) = f{f~\t)). But what is / ( 0 ? By definition, / ( f ) is that element s G S such that _ 1

_ 1

Q

Sec.

Mappings

3

13

- 1

t = / ( s ) . So / ( / - ' ( f ) ) = / O o ) = f. I n other words, ( / ° / ) ( f ) = r for every t £ 7 ; h e n c e / ° / = / , the identity mapping on T. • 0

_ 1

r

W e leave t h e last result of this section for the reader to prove. L e m m a 1.3.5. I f / : S —> T and i is the identity mapping of T o n t o itself and i is that of S onto itself, then i °f = / a n d / i = f. T

0

s

T

s

PROBLEMS

Easier Problems 1. For the given sets S, T determine if a mapping / : S —» T is clearly and unambiguously defined; if not, say why not. (a) S = set of all women, T = set of all men, f(s) = husband of s. (b) S = set of positive integers, T = S, f(s) = s - 1. (c) S = set of positive integers, T = set of nonnegative integers, f(s) = s - 1. (d) S = set of nonnegative integers, T = S, f(s) = s - 1. (e) 5 = set of all integers, T = 5, / ( s ) = 5 — 1. (f) 5 = set of all real n u m b e r s , T = S, f(s) = V s . (g) 5 = set of all positive real n u m b e r s , T = S, f(s) = 'Vs. 2. In those parts of Problem 1 where / does define a function, determine if it is 1-1, onto, or both. *3. If / is a 1-1 mapping of S o n t o T, prove t h a t / o n t o S. *4. I f / i s a 1-1 mapping of S onto T, p r o v e t h a t /

- 1

_ 1

is a 1-1 mapping of T

° / = /y.

5. Give a proof of the R e m a r k after L e m m a 1.3.2. *6. If / : S -> T is onto and g:T^U

and h:T—>U are such that

g°f=h°f,

prove that g = h. *7. If g : S -> r, then g =

:>S

T, and i f / : T

U

is 1-1, show that if / ° g =

f°h,

8. L e t 5 be t h e set of all integers a n d T = {1, —1); / : S -> T is defined by / ( s ) = 1 if 5 is even, / ( j ) = - 1 if 5 is odd. (a) D o e s this define a function from S to 7 7 (b) Show that / ( $ ! + j ) = f(si)f(s ). W h a t does this say about the integers? (c) Is f(s s ) = f{si)f(s ) also true? 2

1

2

2

2

14

T h i n g s F a m i l i a r a n d Less F a m i l i a r

Ch. 1

9. Let S be the set of all real numbers. Define f:S-^> g:S->Sbyg(s)=s + l. (a) F i n d / ° g . (b) Find (c) I s / ° g = g o / ?

2

S by f(s)

= s , and

10. Let S be the set of all real n u m b e r s and for a,b G S, w h e r e a ¥= 0; define (a) Show that f„ °f _ = /„ „ for some real u, v. Give explicit values for u, v in terms of a, b, c, and d. (b) I s / °/,, = f , °/„, always? (c) Find all f such that / ° / = / ° f ,, • (d) Show that /7J, exists and find its form. b

f l ; 6

rf

c

c d

a>b

d

6

a > i

u

u

c

b

11. Let S be the set of all positive integers. Define / : S -> 5 by / ( l ) = 2, / ( 2 ) = 3, / ( 3 ) = 1, and f(s) = s for any other s G 5. Show that f°f°f = i . W h a t i s / " in this case? 1

s

Middle-Level Problems 12. Let S be the set of nonnegative rational numbers, that is, S = {m/n | m, n nonnegative integers, n + 0}, and let Tbe the set of all integers. (a) D o e s f:S—> T defined by f(m/n) = 2"'3" define a legitimate function from S to r? (b) If not, how could you modify the definition of / so as to get a legitim a t e function? 13. Let S be the set of all positive integers of the form 2"'3", where m > 0, n > 0, and let T be the set of all rational n u m b e r s . Define / : 5 T by /(2'"3") = m/n. Prove that / defines a function from S to T. (On what properties of the integers does this d e p e n d ? ) 14. Let f:S-> S, where S is the set of all integers, b e defined by f(s) = as + b, where a, b are integers. Find the necessary and sufficient conditions on a, b in order t h a t / / = i . 0

s

15. Find a l l / o f the form given in P r o b l e m 14 such t h a t / ° / ° / = 16. I f / i s a 1-1 mapping of S onto itself, show that ( /

_ 1

)

_ 1

i. s

= /.

17. If S is a finite set having m > 0 elements, how m a n y mappings are there of S into itself? 18. In Problem 17, how many 1-1 mappings are there of S into itself? 19. Let S b e the set of all real n u m b e r s , and define / : S —» S by f(s) = s + as + b, where a, b are fixed real numbers. Prove that for n o values at a, b can / be onto or 1-1. 2

Sec.

3

Mappings

15

20. Let S be the set of all positive real n u m b e r s . F o r positive reals a, c and nonnegative reals b, d, is it ever possible that the mapping / : S S defined by f(s) = (as + b)/(cs + d) satisfies f°f= z ? Find all such a, b, c, d that do the trick. s

21. L e t S be the set of all rational n u m b e r s and let f : S —> S be defined by f (s) = as + b, where a # 0, b are rational numbers. Find all f of this form satisfying f f„ = /„, ° / , for every / . a

b

ab

C(i

a

c> d

ib

b

c

d

fl

b

22. L e t S be the set of all integers and a, b, c rational numbers. Define f:S—>S by f(s) = as + bs + c. Find necessary and sufficient conditions on a, b, c, so t h a t / d e f i n e s a mapping on S [Note: a, b, c n e e d not b e integers; for example, f(s) = %s(s + 1) = ^s + §s does always give us an integer for integral s.} 2

2

Harder Problems 23. L e t S be the set of all integers of t h e form 2"'3", m > 0, n > 0, and let T be the set of all positive integers. Show that there is a 1-1 correspondence of S onto T. 24. Prove that there is a 1-1 correspondence of the set of all positive integers onto the set of all positive rational n u m b e r s . 25. L e t S be the set of all real n u m b e r s and T the set of all positive reals. Find a 1-1 mapping / of S o n t o T such t h a t / I s , + j ' ) = / ( s ) / ( s ) for all s s G S. 2

u

]

2

2

26. F o r the / i n Problem 25,

-1

find/

explicitly.

27. If / g are mappings of S into S and f° g is a constant function, then (a) W h a t can you say about / if g is onto? (b) W h a t can you say about g if / i s 1-1? 28. If S is a finite set and / is a mapping of S onto itself, show that / must be 1-1. 29. If S is a finite set and / is a 1-1 mapping of S into itself, show that / must be surjective. 30. If S is a finite set a n d / i s a 1-1 mapping of 5, show that for some integer n > 0, . / ° / ° / ° • • • ° / = ^. n times 31. If 5 has 7 « elements in Problem 30, find an n > 0 (in terms of m) that works simultaneously for all 1-1 mappings of S into itself.

16

Ch. 1

T h i n g s F a m i l i a r a n d Less F a m i l i a r

4 . A(S)

(THE SET O F 1-1 M A P P I N G S OF S O N T O ITSELF)

W e focus our attention in this section on particularly nice mappings of a nonempty set, S, into itself. Namely, we shall consider the set, A(S), of all 1-1 mappings of S onto itself. Although most of the concern in the b o o k will b e in the case in which S is a finite set, we do not restrict ourselves to that situation here. W h e n S has a finite n u m b e r of elements, say n, then A (S) has a special n a m e . It is called the symmetric group of degree n and is often d e n o t e d by S„. Its elements are called permutations of S. If we are interested in the structure of S„, it really does not m a t t e r much what our underlying set S is. So, you can think of S as being the set {1, . . . , n). Chapter 3 will be devoted t o a study, in some depth, of S„. I n the investigation of finite groups, S plays a central role. T h e r e are many properties of the set A (S) on which we could concentrate. W e have chosen to develop those aspects here which will motivate the notion of a group and which will give the reader some experience, and feeling for, working in a group-theoretic framework. G r o u p s will be discussed in Chapter 2. W e begin with a result that is really a compendium of some of the results obtained in Section 3. n

L e m m a 1.4.1. (a) fgGA

A (5) satisfies the following:

(S) implies that f°gEA

(b) /, g, h G A(S)

implies that (f°g)°

(S). h = f° (g° h).

(c) T h e r e exists an e l e m e n t — t h e identity mapping i—such z o / = / for every / G A(S). (d) Given / G A(S), 8°f=

there exists a g G A(S)

(g =

that f° i =

such that f°g

=

i-

Proof All these things were d o n e in Section 3, either in the text m a t e rial or in the problems. W e leave it to the reader to find the relevant part of Section 3 that will verify each of the statements (a) through (d). • W e should now like to k n o w how many elements t h e r e are in A(S) when S is a finite set having n elements. To do so, we first m a k e a slight digression. Suppose that you can d o a certain thing in r different ways and a second independent thing in s different ways. In how many distinct ways can you do both things together? T h e best way of finding out is to picture this in

Sec. 4

A{S)

( T h e S e t of 1-1 M a p p i n g s o f S O n t o Itself)

17

a concrete context. Suppose that there a r e ;• highways running from Chicago to Detroit and s highways running from Detroit to A n n Arbor. In how m a n y ways can y o u go first to Detroit, then to A n n A r b o r ? Clearly, for every road you t a k e from Chicago to D e t r o i t y o u have s ways of continuing on to A n n A r b o r . Y o u can start your trip from Chicago in r distinct ways, hence you can complete it in s + s + s + • • • + s = rs r times different ways. It is fairly clear that we can extend this from doing two independent things t o doing in independent ones, for an integer m > 2. If w e can d o the first things in \ \ distinct ways, t h e second in r ways, . . . , the m t h in r,„ distinct ways, then we can do all these together in i\r .. . r,„ different ways. Let's recall something m a n y of us have already seen: 2

2

Definition. If n is a positive integer, then nl (read "n factorial") fined by nl = 1 • 2 • 3 • • • n. L e m m a 1.4.2.

is de-

If S h a s n elements, then A (S) has nl elements.

Proof. Let / G A(S), w h e r e S = \x , x , . . . , x„}. H o w many choices d o e s / h a v e as a place to send x l Clearly n, for we can send x u n d e r / t o any element of S. B u t now / is not free to send x anywhere, for since / is 1-1, we must have f(xA ¥= f(x ). So we can send x anywhere except onto f(xi). H e n c e / can send x into n — 1 different images. Continuing this way, w e see that / can send x, into n — (i — 1) different images. H e n c e t h e n u m b e r of s u c h / ' s is n(n ~ l)(n - 2) • • • 1 = nl • 1

2

x

l

2

2

2

2

Example T h e n u m b e r n\ gets very large quickly. T o be able to see the picture in its entirety, we look at t h e special case n = 3, where nl is still quite small. Consider A(S) = S , where S consists of the three elements x x ,x . We list all the elements of 5 , writing out each mapping explicitly by what it does to each of x x , x . 3

h

3

u

1.

i: Xi

2

3

x —> x , x —> x .

—» X\,

2

2. 3.

2

3

3

f:x\—>x ,x -^>x- ,x ^X\. 2

g:x±^

x ,x ~+ 2

2

u

4. g ° / : * i ->x x -*x ,x ->x . u

2

3

2

x x ^> 3

2

3

i

2l

x. 3

(Verify!)

2

3

18

T h i n g s F a m i l i a r a n d Less F a m i l i a r

5. f° g : A " ! - > a - , a : 3

6.

/ " /

: a c

1

^ a c

3

, a c

x , A ' - > a c j . (Verify!)

2

2

Ch. 1

3

2

^ a c

, a -

1

- > a :

3

2

(Verify!)

.

Since we have listed h e r e six different elements of S , and S has six elements, we have a complete list of all t h e elements of S . W h a t this list tell us? T o begin with, we n o t e that f°g^g°fso one familiar of the kind of arithmetic we have b e e n used to is violated. Since g G S g G S , we must have g°g also in S . W h a t is it? If we calculate g°g, easily get g ° g = i. Similarly, we get 3

3

3

3

3

3

(f°g)°(f°g)

= ' =

(g°f)°(g°.f)-

1

N o t e also that f°(f°f) r e a d e r to show that g°f=

only does rule and we

= i, hence f f^ °g.

= f°f.

Finally, we leave it to t h e

1

It is a little cumbersome to write this p r o d u c t in A (S) using t h e °. From now on we shall drop it and write f° g merely as fg. Also, we shall start using t h e shorthand of exponents, to avoid expressions like f°f°f°---°f. We define, for / G A(S), f° = i, f = f°f = ff, and so on. F o r negative exponents —n we define by f~" = ( / ) " , w h e r e n is a positive integer. T h e usual rules of exponents prevail, namely / ' / • ' = f and (f') = W e leave these as exercises—somewhat tedious ones at t h a t — f o r the reader. 2

_ 1

r+s

s

Example D o not j u m p to conclusions that all familiar properties of exponents go over. F o r instance, in t h e example of the / , g G S defined above, we claim that (fg) ^ f g . T o see this, we n o t e that 3

2

2

2

fg '•

X

X

X

l ~^ 3'

X

2

X

2> 3

2

X

~* L) 2

so that (fg) : x —> x x ~> x , x —> x , that is, (fg) hand, f / and g = i, hence f g = f /, whence (fg) t

u

2

2

2

2

2

3

3

2

2

2

= i. On t h e other ^ fg in this case. 2

2

However, some other familiar properties do go over. F o r instance, if / , g, h are in A (S) and fg = fh, then g = h. W h y ? Because, from fg = fh we have f~\fg) = f-\fh); therefore, g = ig = (f^ftg = / ^ ( / g ) = r\fh) = (f f)h ih = h. Similarly, gf = hf implies that g = h. So we can cancel an element in such an equation provided that we do not change sides. In S3 our /, g satisfy gf = f~ g, but since / f~ we cannot cancel the g h e r e . 1

=

1

l

PROBLEMS

Recall t h a t / g stands for f°g will be a n o n e m p t y set.

and, also, w h a t / ' " means. S, without subscripts,

Sec. 4

19

( T h e S e t o f 1-1 M a p p i n g s of S O n t o Itself)

A(S)

Easier P r o b l e m s 1. If s ¥= s are in S, show that there is a n / G A(S) {

such that f(s )

2

2. Ifsi G S, let / / = { / G A (5) (a) / G //. (b) I f / . g G W . t h e n / i , ' G //. (c) I f / G # , t h e n / G # .

{

= s. 2

= s,}. Show that:

_ 1

3. Suppose that s\ + s are in S and f(s{) = s , w h e r e / G A (5). Then if H is as in P r o b l e m 2 and K = {g G A ( 5 ) | g ( s ) = s } , show that: 2

2

2

2

(a) If g&K, then f-tgfGH. (b) If h G //, then there is some g G K such that /z = 4. If/, g, h G. A(S), show that ( / about ( r ' g / y ?

_ 1

g / ) ( / ~ V ) = f~ (gh)f. x

1

f~ gf. W h a t can you say

5. If /, g G A (5) a n d / g = g / show that: (a) (b>

2

{fgf-fg . ( / g r w - v

1

-

6. Push the result of P r o b l e m 5, for t h e same / and g, to show that (fg)'" f"'g"' for all integers m. s

+

*7. Verify the rules of exponents, namely ff / G A (S) and positive integers r, s. 2

8. If / g G A (5) and (fg)

2

= / ' ' ' and (f'Y

= / " for

2

= / g , p r o v e t h a t / g = gf.

9. If 5 = {x^ x , x , x } , l e t / , g G S be defined by 2

3

4

4

and g :

- » x , x -> x

Xj

2

2

1 ;

x

3

x , x -> x . 3

4

4

Calculate: 2

3

4

(a)/ ,/ ,/ . 2

3

(b) g , g . (c)/g. (d) g / 3

(e) (fg)\ (f)

(gf) .

1

r .*

-

1

10. I f / G S , show t h a t / 3

6

=

= z.

11. Can you find a positive integer m such that /"' = i for all f G S ? 4

20

T h i n g s F a m i l i a r a n d Less

Familiar

Ch. 1

Middle-Level Problems * 12. If / G S„, show that there is some positive integer k, depending o n / , such that/''" = i. (Hint: Consider the positive powers o f / . ) * 1 3 . Show that there is a positive integer / such that / ' = i for all f G S„. 14. If m < n, show that there is a 1-1 mapping F: S„, —> S„ such that F(fg) F(f)F(g) for a l l / g G 5 .

=

m

15. If 5 has three or more elements, show that we can find / g G / I (5) such that/g # g / . 16. Let S be an infinite set and let M C A(S) be the set of all elements / G A (S) such that f(s) # s for at most a finite n u m b e r of s G S. Prove that: ( a ) / g G M implies t h a t / g G M. (b) / G Mimplies t h a t / ' G M . -

17. For the situation in P r o b l e m 16, show, if / G A (5), that / £

G

^}

m

u

s

t

_

1

M/

=

equal Af.

18. Let S T and consider the subset [7(7) = { / G A (5) | / ( f ) G T f o r every t G T}. Show that: (a) / G U(T). (b) / g G U(T) implies t h a t / g G U(T). 19. If the 5 in P r o b l e m 18 has n elements and T has m elements, h o w many elements are there in U(T)7 Show that there is a mapping F: U(T) S,„ such that F(fg) = F(f)F(g) f o r / g G U(T) and F i s o n t o c 20. If /7? < «, can i in P r o b l e m 19 ever be 1-1? If so, when? 7

21. In S„ show that the mapping / defined by

[i.e., /(*;) = A . - if i < n, f(x„) = A^] can b e written as / = g j g • • • g„_! w h e r e each g G 5„ interchanges exactly two elements of S = {x , . . . , A'„), leaving the other elements fixed in S. 2

/+]

f

x

Harder Problems 22. I f / G S„, show t h a t / = h\h '' ' *23. Call an element in S a transposition ing the others fixed. Show that any sitions. (This sharpens the result of 2

n

2

for some E S„ such that /r = /. if it interchanges two elements, leavelement in S is a product of transpoProblem 22.) n

24. If n is at least 3, show that for some / in S„, / cannot be expressed in the f o r m / = g for any g in S„. 3

T h e Integers

Sec. 5

21

3

25. If / G S„ is such that / it i b u t / = j , show that we can n u m b e r the elements of S in such a way that / ( . t j ) = .v , f(x ) = -v , f(x ) = X ) , / ( x ) = x ,f(x ) = x , f(x ) = x , ... ,f(x ) = x ,f(x ) = x , f(x ) =x for some k, and, for all the other x, G 5, f(x) = x . 2

4

5

5

3k+3

6

6

4

2

3k+1

3

3k+2

3

3k+2

3 k + 3

3 k + 1

t

26. View a fixed shuffle of a deck of 52 cards as a 1-1 mapping of the deck o n t o itself. Show that repeating this fixed shuffle a finite (positive) number of times will bring the deck back t o its original order. *27. If / G A(S), call, for s G S, the orbit of s (relative to / ) the set 0(s) = { f (s) | all integers / } . Show that if s, t G S, then either 0(s) n 0(t) = 0 or 0(s) = 0(t). j

28. If 5 = [xi, x ,..., x ] a n d / G 5 is defined by/(x,) = x , if i = 1, 2, . . . , 11 a n d / ( x ) = x , find the orbits of all t h e elements of 5 (relative to / ) . 2

n

12

1 2

+ 1

1

29. If / G A (S) satisfies /

3

= ;', show that the orbit of any element of S has

one or three elements. *30. Recall that a prime number

is an integer p>l

such that p cannot b e fac-

tored as a product of smaller positive integers. If / G A(S) satisfies f

p

~ i,

what can you say about the size of the orbits of the elements of S relative to / ? W h a t property of the prime numbers are you using to get your answer? 3 1 . P r o v e that if S has m o r e than two elements, then the only elements / „ in A(S)

such t h a t / / = ff 0

0

for a l l / G A(S) m u s t satisfy f

0

= i.

*32. W e say that g G A(S) commutes with / G A(S) if fg = gf. Find all the elements in A(S) that c o m m u t e with f:S -> S defined by /(*,) = x ,f(x ) = x , a n d / ( s ) = s if s # x x . 2

2

x

u

2

33. In 5„ show that t h e only elements commuting with / defined by f(x) = x i if / < n, f{x,) = X i , are the powers off, namely i = f°, f f ,.. . ,/""'. 2

j+

34. F o r / G A(S), let C ( / ) = [g G A(S) \fg = gf). Prove that: (a) g, h G C ( / ) implies that gh G C(.f). (b) g G C(f) implies that g G C ( / j . (c) C ( / ) is not empty. - 1

5. THE INTEGERS

T h e mathematical set most familiar to everybody is that of the positive integers 1 , 2 , . . . , which we shall often call Equally familiar is t h e set, Z, of all integers—positive, negative, and zero. Because of this acquaintance with Z, we shall give here a rather sketchy survey of the properties of Z that we shall use often in t h e ensuing material. Most of these properties are well k n o w n to all of us; a few are less well known. T h e basic assumption we m a k e about the set of integers is the

22

Ch. 1

T h i n g s F a m i l i a r a n d Less F a m i l i a r

Well-Ordering Principle. has a smallest member.

A n y n o n e m p t y set of nonnegative integers

M o r e formally, what this principle states is that given a n o n e m p t y set V of nonnegative integers, there is an element v G V such that v < v for every v G V. This principle will serve as the foundation for our ensuing discussion of t h e integers. T h e first application we m a k e of it is to show something we all k n o w and have taken for granted, namely that we can divide o n e integer by another to get a r e main der that is smaller. This is known as Euclid's Algorithm. W e give it a m o r e formal statement a n d a proof based on well-ordering. Q

0

Theorem 1.5.1 (Euclid's A l g o r i t h m ) . If m and n are integers with n > 0, then there exist integers q and r, with 0 < r < such that m = qn + r. Proof. Let W be the set of m - tn, where t runs t h r o u g h all the integers, i.e., W = [m — tn 11 G Z}. N o t e that W contains some nonnegative integers, for if t is large enough and negative, then m — tn > 0. L e t V = {v G W | v ^ 0}; by t h e well-ordering principle V has a smallest element, r. Since r G V, r £ 0, a n d r = m - qn for some q (for that is t h e form of all elements in W D V). W e claim that r < n. If not, r = m - qn > n, hence m - (q + > 0. B u t this puts m - (q + l)n in V, yet m - (q + l ) n < r, contradicting the minimal n a t u r e of r in V. With this, Euclid's A l g o r i t h m is proved. • Euclid's Algorithm will have a host of consequences for us, especially about the notion of divisibility. Since we are speaking about t h e integers, be it understood that all letters used in this section will be integers. This will save a lot of repetition of certain phrases. Definition. Given integers m 0 a n d n we say that m divides n, written as m | n, if n = cm for some integer c. Thus, for instance, 2 [ 14, ( - 7 ) | 14, 4 | ( - 1 6 ) . If m | n, we call m a divisor or factor of n, and n a multiple of m. T o indicate that m is not a divisor of n, w e write 7771n; so, for instance, 3 f 5. T h e basic elementary properties of divisibility are laid out in Lemma 1.5.2. (a) 1

I

T h e following are true:

n for all n.

(b) If 777 * 0, then m \ 0.

Sec.

5

T h e Integers

23

(c) If m | n and n | q, then m \ q. (d) If m | « and m | q, then m | (un + vq) for all it, v. (e) If m 11, then m = 1 or m = — 1 . (f) If m | n and « | m, then 777 = ±n. Proof. T h e proofs of all these parts are easy, following immediately from the definition of m \ n. W e leave all but Part (d) as exercises but prove Part (d) h e r e to give the flavor of h o w such proofs go. So suppose that m | n and m \ q. T h e n n = cm and q = dm for some c and d. Therefore, un + vq — u(cm) + v(dm) = (uc + vd)m. Thus, from the definition, m \ (un + vq). • Having the concept of a divisor of an integer, we now want to introduce that of the greatest common divisor of two (or m o r e ) integers. Simply enough, this should be the largest possible integer that is a divisor of b o t h integers in question. However, we want to avoid using the size of an integer— for reasons that may become clear m u c h later when we talk about rings. So we m a k e the definition in what may s e e m as a strange way. Definition. Given a, b (not b o t h 0), then their greatest common sor c is defined by: (a)

divi-

c>0.

(b) c

I

a and c \ b.

(c) lid\a

and d \ b, then d \ c.

W e write this c as c — (a, b). In other words, the greatest common divisor of a and b is the positive n u m b e r c which divides a and b and is divisible by every d which divides a and b. Defining something does not g u a r a n t e e its existence. So it is incumbent on us to prove that (a, b) exists, and is, in fact, unique. The proof actually shows more, namely that (a, b) is a nice combination of a and b. This combination is not unique; for instance, (24,9) = 3 = 3 - 9 + ( - 1 ) 2 4 = ( - 5 ) 9 + 2 - 2 4 . T h e o r e m 1.5.3. If a, b are not b o t h 0, then their greatest common divisor c = (a, b) exists, is unique, and, moreover, c = rn^a + n b for some suitable mo and n . 0

n

T h i n g s F a m i l i a r a n d Less F a m i l i a r

24

Ch. 1

Proof. Since not b o t h a and fa are 0, the set A = {ma + nb \ m, n G Z) has nonzero elements. If .v G A and x < 0, then —x is also in A and —x > 0, for if x = m a + n b, then —x = (—m )a + (—n^fa, so is in A T h u s has positive elements; hence, by the well-ordering principle t h e r e is a smallest positive element, c, in A. Since c G / i , by the form of the elements of A we k n o w that c = m a + n b for some / 7 J f ) , We claim that c is our required greatest c o m m o n divisor. First note that if rf | « and d \ fa, then d \ (m a + n b) by Part (d) of L e m m a 1.5.2, that is, d | c. So, to verify that c is our desired element, we need only show that c \ a and c | b. By Euclid's Algorithm, a = qc + ;•, where 0 < /• < c, that is, « = q(m a + /? /3) + r. Therefore, ;• = —gn b + (1 — qm )a. So r is in A. B u t r < c and is in A, so by the choice of c, r cannot be positive. H e n c e r = 0; in other words, a = qc and so c | a. Similarly, c | b. For the uniqueness of c, if t > 0 also satisfied r ] a, 11 /j> and d | f for all d such that d | a and d | fa, we would have t \ c and c 11. By Part (f) of L e m m a 1.5.2 we get that t = c (since both are positive). • x

{

{

0

Q

Q

Q

0

0

a

0

Let's look at an explicit example, namely a = 24, fa = 9. By direct examination we know that (24, 9) = 3; note that 3 = 3 - 9 + ( - 1 ) 2 4 . W h a t is ( - 2 4 , 9)? H o w is this done for positive numbers a and fa which may be quite large? If fa > A, interchange a and fa so that a > fa > 0. Then we can find (a, fa) by 1. observing that (a, fa) = (fa, r) w h e r e a ~ qb + r with 0 < r < fa (Why?); 2. finding (fa, r), which now is easier since one of the n u m b e r s is smaller than before. So, for example, we have (100, 28)

(28,16)

since 100

3 (28) + 16

( 28, 16)

(16, 12)

since

28

1 (16) + 12

( 16,12)

(12, 4)

since

16

1 (12) +

4

This gives us (100, 28)

(12, 4) = 4.

It is possible to find the actual values of m

a

4 = m

0

100 + n

0

28

and n such that 0

Sec. 5

The Integers

25

by going backwards through the calculations m a d e to find 4: Since

16 = 1 (12) +

4,

4 =

16 + (

1) 12

Since 28 = 1 (16) + 12,

12=

28 + (

1) 16

Since 100 = 3 (28) + 16,

16 = 100 + ( 3) 28

B u t then 4 = 16 + ( - 1 ) 12 = 16 + ( - 1 ) ( 2 8 + ( - 1 ) 16) = ( - 1 ) 28 + (2) 16 = (™1) 28 + (2)(100 + ( - 3 ) 28) = (2) 100 + ( - 7 ) 28 so that m = 2 and « = —7. This shows how Euclid's Algorithm can be used to compute (a, b) for any positive integers a and b. W e shall include some exercises at the end of this section on other properties of (a, b). W e come to the very important 0

Definition.

n

W e say that a and b are relatively prime if (a, b) = 1.

So the integers a and b are relatively p r i m e if they have no nontrivial c o m m o n factor. A n immediate corollary to T h e o r e m 1.5.3 is T h e o r e m 1.5.4. The integers a and b are relatively prime if and only if 1 = ma + nb for suitable integers in and n. T h e o r e m 1.5.4 has an immediate consequence T h e o r e m 1.5.5.

If a and b are relatively prime and a | be, then a \ c.

Proof. By T h e o r e m 1.5.4, ma + nb = 1 for some in and n, hence (ma + nb)c = c, that is, mac + nbc = c. By assumption, a \ be and by observation a | mac, hence a | (mac + nbc) and so a \ c. • Corollary. If b and c are both relatively prime t o a, then be is also relatively prime to a. Proof. W e pick up the proof of T h e o r e m 1.5.5 at mac + nbc = c. If d = (a, be), then d | a and d\ be, hence d | (mac + nbc) = c. Since d \ a and d \ c

26

Ch. 1

T h i n g s F a m i l i a r a n d Less F a m i l i a r

and (a, c) = 1, we get that d = 1. Since 1 = d = (a, be), we have that be is relatively prime to a. • W e now single out an ultra-important class of positive integers, which we m e t before in P r o b l e m 30, Section 4. Definition. A prime number, or a prime, is an integer p > 1, such that for any integer a either p \ a or p is relatively prime to a. This definition coincides with the usual one, namely that we cannot factor p nontrivially. For if p is a prime as defined above and p = ab w h e r e 1 < a < p, then (a,p) = a (Why?) and p does not divide a since p > a. It follows that a = 1, so p = b. O n the other hand, if p is a prime in the sense that it cannot b e factored nontrivially, and if a is an integer not relatively prime to p, then (a, b) is not 1 and it divides a and p. But then (a, b) equals p, by our hypothesis, so p divides a. A n o t h e r result coming out of T h e o r e m 1.5.5 is T h e o r e m 1.5.6. If p is a prime and p\(a a some i with 1 < ; < ; ? . l

2

• • • a,), then p \ a for f

Proof. If p | a there is nothing to prove. Suppose that p\a ; then p and ffj are relatively prime. B u t p \ a (a • • • a„), hence by T h e o r e m 1.5.5, p | a • • • a . R e p e a t the argument just given on a , and continue. • u

{

l

2

n

2

2

T h e primes play a very special role in the set of integers larger than 1 in that every integer n > 1 is either a prime or is the product of primes. W e shall show this in the next t h e o r e m . In the t h e o r e m after the next we shall show that there is a uniqueness about the way n > 1 factors into prime factors. T h e proofs of both these results lean heavily on the well-ordering principle. T h e o r e m 1.5.7. primes.

If n > 1, t h e n either n is a prime or n is the product of

Proof. Suppose that the t h e o r e m is false. Then there must be an intger m > 1 for which the t h e o r e m fails. Therefore, the set M for which the t h e o r e m fails is nonempty, so, by the well-ordering principle, M has a least element m. Clearly, since m G M, m cannot be a prime, thus m = ab, where 1 < a < m and 1 < b < m. Because a < m and b < m and m is the least element in M, we cannot have a G M or b G M. Since a ^ M, b £ M, by the definition of M the t h e o r e m must b e true for both a and b. T h u s a and b are

Sec.

T h e Integers

5

27

primes or the product of primes; from m = ab we get that m is a product of primes. This puts m outside of M, contradicting that m G M. This proves the t h e o r e m . • W e asserted above that there is a certain uniqueness about the decomposition of an integer into primes. W e m a k e this precise now. T o avoid trivialities of the kind 6 = 2 • 3 = 3 • 2 (so, in a sense, 6 has two factorizations into the primes 2 and 3), we shall state the t h e o r e m in a particular way. T h e o r e m 1.5.8. Given n > 1, then there is one and only one way to write n in the form n = p ^p • • • p , where p < p < • • • < p are primes and the exponents a , a • • •, a are all positive. a

c

ak

2

x

k

2>

x

2

k

k

Proof. W e start as we did above by assuming that the t h e o r e m is false, so there is a least integer m > 1 for which it is false. This m must have two distinct factorizations as m = pVpT ' ' ' Pt q \ 2 ' ' ' e are primes and where t h e exponents a , ... , a and b , .. . , b are all positive. Since p \ pi • • • p = q\ • • • q ', by T h e o r e m 1.5.6 p \ qf> for some /; hence, again by T h e o r e m 1.5.6, p | q hence p = cp. By the same t o k e n q = p- for some /; thus p ^ /?• = q < q, = p . This gives us that p = q . N o w since m/p < m, ni/p has the unique factorization property. But m/p = p" ^ p - • • • p" = Pi