Groups, Rings and Fields (Springer Undergraduate Mathematics Series)

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Springer Undergraduate Mathematics Series

Springer

London Berlin Heidelberg New York Barcelona Budapest Hong Kong Milan

Paris Santa Clara Singapore Tokyo

Advisory Board Professor P.J . Cameron Queen Mary and Westfield College Dr M.A.J. Chaplain University of Dundee Dr K. Erdmann Oxford University

Professor L.C .G. Rogers University of Bath Dr E. Stili Oxford University Professor J.F. Toland University of Bath

Other books in this series Basic Linear Algebra

T.S. Blyth and E.F. Robertson

3-540-76122-5

Multivariate Calculus and Geometry

Sean Dineen

�

3-540-76176

Elements of Logic via Numbers and Sets

D.L. Johnson 3-540-76123-3

Elementary Number Theory

Gareth A. Jones and J. Mary Jones

3-540-76197-7

Introductory Mathematics: Applications and Methods

G.S. Marshall 3-540-76179-9

Vector Calculus

P.C. Matthews 3-540-76180-2

Introductory Mathematics: Algebra and Analysis

Geoff Smith 3-540-76178-0

D.A.R. Wallace

Groups, Rings and Fields With 15 Figures

'Springer

D.A.R. Wallace, BSc, PhD, FRSE Dept. of Mathematics, Strathclyde University, Livingstone Tower, 26 Richmond Street,

Glasgow G 1 lXH, UK

Cover illrutration element> rtproduted by kind permiuion of Aptech Systems, Inc., Publishers of the GAUSS Mathematical aod Statistical System, 23804 S.E. ICent-KansJey Road, Maple Valley, WA 98038, USA. Tel: (206) 432- 7855 Pu (206) 432 - 7832 emm: [email protected]: www.aptech.com Americm Statistical Association: Chance Vol8 No I, 1995 article by KS aod KW Heiner 'Tree RiJl8S of the Northern Shawangunlu' page 32 fig 2 Springer-Verlag: MathematiC& in Education and Research Vol4 Issue 3 1995 article by Roman E Maeder, Beatrice Amrhein and Oliver Gloor 'Diustrated Mathematia: Visualization of Mathematical Objects' page 9 fig 11, originally published u a CD ROM 'Diustrated Mathematia' by TELOS: ISBN 0-387-14222-3, german edition by Birkhauser: ISBN 3-7643-510Q-4, Mathematics in Education and Research Vol 4 Issue 3 1995 article by Richard J Gaylord aod Kazume Nishidate "Iraffic Engineering with Cellular Automata' page 35 fig 2. Mathematica in Education and Research Vol 5 Issue 2 1996 article by Michael 'Irott 'The Implicitization o{ a 'Irefoil!Cnot' page 14. Mathematica in Education and Research Vol 5 Issue 2 1996 article by Lee de Cola 'Coins, '!r..., Bars aod Ilells: Simulation o{ the Binomial Process page 19 fig 3. Mathematica in Education and Research Vol 5 Issue 2 1996 article by Richard Gaylord and Kazume Nuhidste 'Contapous Spreading' page 33 fig I. Mathematica in Education aod Research Vol 5 Issue 2 1996 article by joe Buhler and Stan Wagon 'Secrets o{ the Madelung Constant' page 50 fig I.

ISBN 3-540-76177-2 Springer-Verlag Berlin Heidelberg New York British library Cataloguing in Publication Data Wallace, David A.R Groups, rings and fields. - (Springer undergraduate mathematics series; 3423) I. Group theory 2. Rings (Algebra) 3. Algebraic fields I. Title 512 ISBN 3540761772 Library of Congress Cataloging-in-Publication Data Wallace, D.A.R. (David Alexander Ross) Groups, rings, and fields / D.A.R. Wallace. p. em.- (Springer undergraduate mathematics series) Includes index. ISBN 3-540-76177- 2 (pbk.: alk. paper) I. Algebra, Abstract. I. Title. II. Series. QA162.W36 1998 512'.02-dc21

98-4961 CIP

Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in

the case of reprographic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. ©Springer-Verlag London Limited 1998 Printed in Great Britain

The use of registered names, trademarks etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use . The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Typesetting by BC Typesetting, Bristol Printed and bound at the Athenaeum Press Ltd., Gateshead, Tyne & Wear 12/3830-543210 Printed on acid-free paper

Con ten ts

Preface

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vii

Chapter 1 1.1 1 .2 1 .3 1.4 1 .5 1 .6

Sets and Mappings . .... ............................................................... Union and Intersection........................... ................................... Venn Diagrams.......................................................................... Mappings. . ..... ... ............. ....... ........... ....................... . ............ ...... Equivalence R.elations . .... .. ... .. .. ......... .................... .. ... .. ....... .. ... Well-ordering and Induction ..................................................... . Countable Sets . .. ... .. . . .

Chapter 2 2.1 2.2 2.3 2.4 2.5

The Integers . ...... ...... ........ ......................... .................... ...... ...... 4 7 Divisibility................................................................................. 47 Divisors...................................... ................................................ 50 Division Algorithm . . ...... ......... .............. .. ............ ........... ..... ...... . 52 Euclidean Algorithm ... .... .. ....... .......................... ............ .......... 56 Primes........................................................................................ 62

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1 1 8 17 27 34 41

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Chapter 3 Introduction to Rings.... .............................. ..... ......................... 3.1 Concept of a Polynomial ......... ................................................. 3.2 Division and Euclidean Algorithms........................................... 3.3 Axioms and Rings . . ... ...... ......... .............. .... ......... . ..... .. ... ........... .

v

71 72 76 82

vi

Contents

Chapter 4 Introduction to Groups . . . . . . . . .... . . . . . . . . . . . . ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 4.1 Seiiligroups . . .. . . . . . . .. . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . .. . . . . . . ... . . . . .. . . . . . . . . . . . . . . . . . . . . 94 4.2 Finite and Infinite Groups......................................................... 99 4.3 Subgroups................. ..... .. .. ............ ................ . .... ............. 1 1 5 4.4 Lagrange's Theorem, Cosets and Conjugacy.... .... ........ .. .... .. 127 4.5 Homomorphisms ....... .. ......... . ........... ...... .... .................. .... 134 ...

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Chapter 5 Rings .. ......... .. ..... ... . ... . ... . ........ . ... . ... ... ....... . 145 5. 1 Arithmetic Modulo n . . .. . . .. . ... .. . ....... .... 145 ... .. . . . ... .. 150 5.2 Integral Domains and Fields . 5. 3 Euclidean Domains .. .. . . .. .... . ..... .. .... .. .. 163 5.4 Ideals and Homomorphisms .... . .... . ... . ...... . .. .... ...... 168 5.5 Principal Ideal and Unique Factorization Domains ......... ..... . 177 5.6 Factorization in Q[x] . . . . . . 186 ..

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Chapter 6 6.1 6.2 6.3 6.4 6.5

Topics in Group Theory

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Suggestions for Further Study Index

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Permutation Groups . Generators and Relations Direct Products and Sums Abelian Groups .. . .... p-Groups and Sylow Subgroups .

Hints to Solutions

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Preface

This text is w ritten for students w ho are m eeting abstract algebra for the first tim e. Many of these students w ill take only one course of abstract al gebra, w hereas others m ay proceed to m ore advanced courses. The aim here is to pro vide an appropriate, interesting and entertaining text for those w ho require a rounded course as w ell as for those who w ish to continue w ith further studies in algebra. A fundam ental difficulty for beginning students is often the axiom atic nature of abstract algebra and the exacting need to follow the axioms precisely. The pre sent text is w ritten so that an axiom atic treatm ent should seem to evolve as a natural developm ent of intuitive ideas. To this end particular and extensive attention is paid to the integers, w hich are familiar objects of know ledge, and w hich, together with som e simple properties of polynomials, are used to give m otivation for the introduction of more abstract algebraic concepts. Historical allusions are m ade throughout the text in order to emphasise that abstract modern algebra has evolved from sometim es quite rudimentary ideas. Other rem arks are intended to broaden, in a light-hearted manner, the student's general education. Each section is provided w ith copious examples and ends w ith a set of exercises for w hich solutions are provided at the end of the book. In order to render the text suitable for self-study, any argum ents in the text have been carefully crafted to facilitate the understanding, and to prom ote the enjoyment, of the reader. The text is self-contained and no prerequisites are absolutely necessary although a 'nodding acquaintance' w ith complex numbers and w ith matrices w ould, on occasion, be of advantage. vii

viii

Preface

Finally, I mus t expres s my profound thanks to the members of the editorial team of Springer-Verlag and to my w ife, Rachel Henderson, for her many encour agements and for her indefatigable efforts in turning a scribbled manuscript into an elegant w ord-processed document.

D.A.R. Wallace Department of Mathematics University of Strathclyde January 1998

1 Sets and Mappings

The notions of a 'set' and of a 'mapping' are fundamental in modern mathe matics. In many mathematical contexts a perceptive choice of appropriate sets and mappings may lead to a better understanding of the underlying mathe matical processes. We shall outline those aspects of sets and mappings w hich are relevant to present purposes and, for the delectation of the reader, conclude w ith a few logical paradoxes in regard to sets.

1.1 Union and Intersection We are accustomed to speak of a 'collection' of books , or of an 'assembly' of people or of a 'list' of guests. In mathematics the w ord 'set' is us ed to denote the basic concept w hich is expressed by each of the collective nouns; in mathe matics we say simply a 'set' of books , or a 'set' of people or a 'set' of guests.

Defi nition 1 A collection or assembly of objects is called a set. Each object is said to be an

element of the set. Thus in the phrase 'a set of books ', each book involved in the set is an element of the set. Frequently the symbols cons tituting the elements of a (mathematical)

2

Groups, Rings and Fields

set w ill be letters or numbers. If a set A consists of, say, the elem ents a, b, c then we w rite A = {a, b, c} , the use of curly brackets being custom ary. Notice that as we are concerned only with m em bership of the set w e disregard any repetition of the sym bols and have no preferred order for writing them dow n.

Exam ple 1 Let A have 1 , 2, 3, 4

as

elem ents. Then A = { 1 , 2, 3, 4} = { 1 , 1 , 2, 2, 3, 4}=

{2, 1 , 3, 4} = { 1 , 3, 4, 2} etc.

Defi nition 2 Let A be a set. If A consists of a finite num ber of elements, say a1, £Z:!, ... , aN (w here the notation presumes the elements are distinct) then w e write

A = { al> £Z:!, . . . , aN} and A is said to be a finite set of cardinality N, written N = IAI. If A does not have a finite number of elem ents A is called an infinite set and is said to have infinite cardinality. The set consisting of no elements at all is called the empty set or the null set and is denoted by 0 (0 is a letter of the Norw egian alphabet). Evidently 101 = 0.

Exa m ples 2 1. 2. 3. 4.

0 = set of all unicorns. 0 =set of all persons living on the m oon in 1996.

1 { 1 , 2, 3, 4} 1 = 4. l {a, b, c, d, e } l = 5.

A set is often given by som e property w hich characterizes the elem ents of the set. Before giving a definition w e offer a colourful example.

Exa m ple 3 Let A be the set of the colours of the rainbow. Whether or not a given colour is in A is determined by a property, namely that of being one of the colours of the rainbow. Of course in this particular instance w e may write dow n the elements explicitly. Indeed

A = {red, orange, yellow , green, blue, indigo, violet}.

3

Sets and Mappings

Defi nition 3 Let P be some property. The set of elements, each of w hich has the property P, is written as

A = {x l x

has the property P},

w hich w e read as ' A is the set of all x such that x has the property P'.

Exa m ples 4

1.

Let A cons ist of the ,squares of positive integers. Then

A = {x l x = n2 , n = 0, 1, 2, . . . } .

The set A may als o be written as

A = { 1, 4, 9, . . . }, 2. 3.

but notice that this notation is ambiguous since it w ould only be an inference that A consists of squares of integers.

The set {xlx < 0 and x > 1} is the empty set since there is no number w hich is simultaneous ly less than 0 and greater than 1.

The set { x l x2 = 3, square equal to 3.

x

is an integer} is empty since there is no integer w ith

Defi nition 4 Let A be a set. Let x be an element. If xis an element of A we w rite

x E A, and if x is not an element of A w e w rite

x ¢ A. Exa m ple 5

a E {a, b, c}, b ¢ { 1, 2, 3} , 36 E {x l x = n 2 , n = 0, 1 , 2, . . . } , 37 ¢ {x l x = n2, n = 0, 1 , 2, . . . }. Defi nition 5 A set A is said to be a subset of a set B if every element of A is also an element of B. We write

A�B


4

to indicate that A is a subset of B and read the notation as 'A is contained in or equal to B'. The em pty set is deemed to be a subset of every set. If A is a subset of B but A ::j; B then A is said to be a proper subset of B. If A is not a subset of B then there exists at least one element of A w hich is not an element of B and we w rite A�B. If A�B and B � A then the sets A and B are equal and w e w rite A =B.

If A and B are not equal w e write A ::j;B. If A�B and A ::j;B w e w rite A c B. Given tw o sets A and B, to prove that A�B we have to show that x E A im plies x E B. To prove that A�B, we have to exhibit an x E A such that x ¢ B. Tw o sets A and B are equal if and only if they are elem entw ise indistinguish able.

Exa m ples 6

1. 2.

{a, c}, B = {a, b, c, d}, C = {a, b, c}. Then A�B, C �B. Let A= {xlx 2 = 1}, B= { -1, 1} . Although A and B are described in differ ent w ays they have the same elements and so A = B.

Let A=

Defi nition 6 Applications of the theory of sets usually take place w ithin some fixed set (w hich naturally varies according to the circumstances). This fixed set for the particular application is called the universal set and is often denoted by U. The subset of U consisting of the elements of U having the property P is denoted by

{x E Ulx

has the property P},

or sim ply as

{ xlx

has the property P}

if the universal set is obvious. Frequently explicit reference to a universal is om itted. We now introduce the operations of union and intersection on sets.

5

Sets and Mappings

Defi nition 7 Let A, B, C, . be sets. The set consisting of those elements each of which is in at least one of the sets A, B, C, ... is called the union of A, B, C, . . . and is w ritten .

.

AUBUCU. .. The set cons isting of those elements each of w hich is in all of the sets A, B, C, . .. is called the intersection of A, B, C, . . . and is w ritten as

AnBncn. . . The sets A, B, C, are said to be disjoint if the intersection of any tw o distinct setsis em pty, that is, AnB = AnC = BnC = ... = 0. The union A UBUC U is said to be a disjoint union if the sets A, B, C, .. are disjoint. .

. .

.

.

.

.

Exa m ple 7 Let U = { a, b, c, d, e, f,g,h,i} be the universal set. Let A = {a, b,c }, B=

{ d, f,i}, C = { a, b, c, d,j, g} , D = { a,e, g,h,i}, E

=

{e,g,h } .

Then

AUB = {a, b,c,d, f,i},

AnB = 0,

CUD = {a, b, c, d, e, J, g,h,i } = U, AUBUC= {a, b, c, d, f,g,i},

CnD = {a,g},

AnBnC = 0,

BUCUD = { a, b,c,d, e, f, g,h,i} = U,

BnCnD = 0,

U = AUBUE, w here this union is disjoint as AnB = AnE = BnE = 0. It is im m ediate that for sets A and B we have A nA = A, A UA = A and

AnB = BnA � A � AUB = BUA. We prove a useful lemma.

Lem ma 1 Let A, B, X be sets such that A � B. Then the following hold.

1 . Anx �Bnx. 2. AUX�BUX.

Grou ps, Rings and Fields

6

Proof 1 . We have to prove that every element of AnX is also in BnX. Let c E AnX. Then c E A and so, as A � B, c E B. But c E X and so c E Bnx. Hence AnX � Anx. 2. Let c E A UX. Then either c E A or c E X. If c E A then c E B and so, in either case, c E Bu X. Thus Au X � BUX. 0

Defi n ition

8

Let A be a subset of the universal set U. The subset of U consisting of all elements of U not in A is called the complement (note spelling! ) of A. The complement of A is denoted by U \ A or by A' if the universal set is clear. U \ A = A' = {x E U lx ¢A}

( Another notation for the complement, but not used in this text, is CA.) U = (U \ A) UA, (U \ A) nA = 0, and so

A = U if and only if A' = 0.

Exa m ple

8

Let U = { 1, 2, 3, 4, 5}, A = {2, 3} . Then U \ A = { 1, 4, 5} . The notion of complement extends to that of relative complement.

Defi nition 9 Let A and B be sets. The relative complement of A in B, denoted by B \ A, is the set of elements of B which are not in A.

Exa m ple 9 Let A = {p, q,r }, B = {q,r,s,t} . Then B \ A = {s,t }, A \ B = {p} . We conclude this section with a fairly obvious but useful result.

7

Sets and Mappings

Theorem 1 Let A and B be sets. Then the following statements are equivalent .

1 . A� B. 2. AnB=A. 3. AUB=B. Proof

We prove the equivalence of 1 and 2, leaving the equivalence of exercise. Suppose A�B. Then, by Lemma 1, A=AnA� AnB� A, and thus

1

and 3

as

an

A=AnB. Conversely if A=AnB then certainly A� B.

0

We conclude this section by introducing the 'Cartesian product' of sets. The adjective 'Cartesian' derives from the name ofR. du P. Descartes ( 1596-1650) , mathematician and philosopher whose philosophical outlook is enshrined in his famous aphorism "je pense, done je suis", in English "I think, therefore I am". (Quotation from Discours de la Methode [Leyden, 1637, 4th part]).

Defi nition 10 Let A and B be two sets. The Cartesian product Ax B of A and Bis defined to be the set of all ordered pairs of the form (a, b) , a E A, b E B; Ax B={ (a, b) l a E A, b E B}. The Cartesian product A1x A2 x ... x An of the sets A1, A2 , ... , An (in this order) is defined to be the set of all ordered n-tuples of the form

(a1, �, ... , an ) , a 1 E A1, a2 E A2 , ... , an E An ,

Note that if the sets A and B are distinct, then Ax B =I Bx A.

Exa m ple 10 Let A and B be finite sets ofm and n elements respectively. Then Ax Bis a set ofmn elements since in the ordered pair (a, b) there arem possibilities for a, and n possibilities for b.

G roups, Rings and Fields

8 Exercises 1.1

1 . Let the universal set U be the set {u , v , w, x, y, z}. Let A= { u , v , w}, B={w, x, y}, C={x, y, z}. Write down the subsets: A U B, AnB, AnC, A U G, A\ B, B\ A, A\ C, C\A, A', B', C', A\ (BU C) , B\ (An C) , (AnB) U ( B\A). 2. Let the universal set U be the set {a, b, c, 1, 2, 3}. Let X= { a, b, 1 }, Y={b, 2, 3}, Z={ c, 2, 3}. Write down the subsets: X U Y, X n Y, X U Z, XnZ, X\ Y, Y\X, X\Z, Z\X, X', Y', Z', (XnY)', X' U Y', (X U Z )', X' nZ'. 3. Let the universal set U be the set { 1 , 2, 3, 4, 5, 6}. Let A= { 1 , 2, 4, 5 } . Let X be a subset of U such that A U X= { 1 , 2, 4 , 5, 6} and AnX= { 2 , 4 }. Prove that there is only one possibility for X, namely { 2, 4, 6} . 4. Let A and B be sets. Prove that A � B if and only if B= A U B (Theorem 1 ) . 5. Let A be the set {3, 5, 8, . . . }. Is A the set {3, 5, 8, 13, . . . } or the set {3, 5, 8, 12, . . . } or neither? 6. Let � be a finite set of ki elements, i = 1, 2, ... , n. Prove that A1

X

A2

X

...

x

An is a flnite set of k 1 k 2 ... kn elements.

1.2 Venn Diagrams In this section we shall describe a technique which, in certain circumstances, enables us to visualize sets and their unions and intersections etc. by drawing pictures on paper. The resulting pictures are known as Venn diagrams, after J. Venn ( 1834-1 923). We proceed as follows. For the universal set U we draw a rectangle and for arbitrary subsets A, B of U we draw shapes as below: u

Figure 1.1.

Sets and Mappings

9

The shaded area represents A U B and the dark-shaded area A n B. Conveni ently we may assign numbers to the regions: u

� �

4

Figure 1.2.

Thus region 1 represents A n B, region 2 represents A\ (An B) , region 3 represents B\ (An B) , region 4 represents U \ (A U B) . The Venn diagram makes clear that A is the disjoint union of the two subsets A\ B and AnB, and A U B is the disjoint union of the three subsets A\ B, B\ A and AnB. Wise use of Venn diagrams may lead to general results for which a rigorous proof may later be obtained. We give two examples of this procedure in the following examples.

Exa m ple 1 1 Using the above Venn diagram we may observe the following: A

is represented by regions 1 , 2

A' B

3, 4 1, 3

B'

2, 4

{ An B)'

2, 3, 4

A' U B'

2, 3, 4

Thus, from our Venn diagram representation, we conclude that (AnB)'=A' U B'.

10


Similarly we conclude that (A U B)'=A' n B'. A rigorous proof will be given in Theorem 3. If now C is a third subset of U then the Venn diagram becomes: u

Figure 1.3.

The shaded region represents A n Bn C. For four or more subsets of U the method has fundamental limitations ( as those attempting to draw satisfactory diagrams of four subsets will discover) and should be avoided.

Exa m ple 12 Let A, B, C be subsets of the universal set U and suppose A�C, B�C. The appropriate Venn diagram is: u

Figure 1.4.

Exam ple 13 Let us draw a Venn diagram to illustrate the result that if X� Y then AnX� AnY.

11

Sets and Mappings

u

Figure 1.5.

Consider a Venn diagram with three subsets A, B,C of the universal set U and with regions assi gned numbers as follows: u

8

Figure 1.6.

region 1 represents AnBnC region 2 represents (AnB) \ (AnBnC) region 3 represents (AnC) \ (AnBnC) region 4 represents (BnC)\ (AnBnC)

Exa m ple 14 In the above diagram, suppose we are asked to determine the regions correspond ing to the following subsets:

BUG,An (BUC),AnB,AnC, (AnB) U (AnC). Then is represented by regions BUG An(BUC) AnB Anc (AnB) U (AnC)

1 , 2, 3, 4, 6, 7 1, 2, 3 1, 2 1, 3 1 , 2, 3


12

We notice that, from our Venn diagram representation, An (B U C) =(An B) U (An C) . Similarly we may conclude that A U (B n C) =(A U B) n (A U C) . A rigorous proof is given in Theorem 2.

Theorem 2 Let A, B, C be sets. Then 1 . A n (B U C) 2. A U (BnC)

= =

(An B) U (Bn C) (A U B) n (B U C)

(These results are called the Distributive Laws for sets. )

Proof We prove 1 , the proof of 2 is similar. Since B�B U C and C�B U C, we have An B�A n (B U C) and An C�A n(B U C) from which (AnB) U (AnC) �A n (B U G) . We now have to prove that An(B U C) �(An B) U (AnC) . For this part of the proof we consider elements and not simply subsets. Therefore let x EAn(B U C) . Then x EA and x EB U C and so x is in either B or C. If we suppose x EB then x EB and x EA from which we have x EA n B and so X E (An B) u (An C). The alternative supposition that X Ec leads to the same conclusion. Consequently we have A n (B U C) � (An B) U (An C) This completes the proof.

D

The next result ( see Example 1 1 ) gives the so-called De Morgan Laws, after A. De Morgan (1806-71 ).

Theorem 3 De Morgan Laws Let A, B be sets. Then the following statements hold. 1. (AnB)' =A' U B'. 2. (A U B)' =A' n B'.

13

Sets and Mappings

Proof We prove 1 , the proof of 2 is similar. Let X E (AnB) ' . Then X ¢ (AnB) . This means that X ¢A or X ¢B. Ifx ¢A then x E A' and so x E A'UB ' . Ifx ¢B then x E B ' and so x E A' UB ' . Thus (AnB) ' �A' UB ' . Let x E A' UB ' . Thenx E A' orx E B ' . Ifx E A' thenx ¢A and so certainly x ¢AnB and similarly if x E B' then x ¢AnB. Thus x ¢AnB and so x E (AnB) ' . Thus A' U B ' �(AnB) ' . Hence we obtain the desired result. D Many of the sets with which we shall be concerned will be infinite, but for finite sets we now introduce some counting techniques. If A and B are both finite sets then certainly AUB is finite since there cannot be more elements than in A and B considered separately. We shall derive a formula for the number of distinct elements in AUB. We recall that l A I is the number of elements in A (Definition 2).

Theorem 4 Let A and B be finite sets. Then AUB is a finite set and

l AUBI = IA I + IBI-IAnBI.

Proof Let us examine the Venn diagram for AUB. As we remarked

A = (A \ B) U(AnB) , B = (B \ A) U(AnB) , AUB = (A \ B) U(B \ A) U(AnB) are disjoint unions of subsets of A or B. These subsets are necessarily finite and so, by obvious (and legitimate) counting, we have

l A I = l A \ B I + IAnBI, IB I = IB \ A I + IAnBI and also

l A u BI= l A \ BI + IB \ A I + IAnBI = [ l AI - l AnB l] + [IBI - l AnB l] + l AnB l = IAI + IBI - l AnB l .

D


14

The following two examples illustrate the use of this formula.

Exa mple 15 Let A = {a, b, c, d}, B= {a, c, e, f, g}. Then AUB = {a, b, c, d , e, f, g}, AnB = {a, c}. As we expect

l AUBI= 7 = 4 + 5 - 2 = IAI + IBI-IA nBI.

Exa m ple 16 We are told that in a party of 95 English-speaking schoolchildren there are 20 who speak only English, 60 who can speak French and 24 who can speak German. We require to determine how many speak both French and German. The 'universal set' consists of 95 schoolchildren. We let F and G be the subsets consisting of those schoolchildren who can speak French and German respec tively. Then IF I= 60, IG I = 24. FUG is the subset of those schoolchildren who speak either French or German. Since 20 speak only English we have

IF u G I = 95 -20= 75. Then IF u G I = IF I + I G I -IFnG I and so 75 = 60 + 24 - IFnG l . Thus IFnGI= 9. This says that precisely 9 schoolchildren can speak both French and German. We may extend Theorem 4 to consider three or more sets, but we confine our extension to the case of three sets.

Theorem 5 Let A, B and C be finite sets. Then AUBUC is a finite set and IAUBUCI= IAI + IBI + ICI - IAnBI-IAnCI - IBnCI + IAnBnC I.

Proof We apply Theorem 4. We note that

(AnB) n ( AnC ) = AnBnC.

15

Sets and Mappings

Then, on letting P

= BUC, we have

IAUBUCI = IAUPI

= l AI + IPI- l AnPI = IAI + IBUCI-IAnPI = IAI + IBI + ICI-IBnCI - IAn(B u C )I = l AI + IBI + ICI-IBnCI-I( AnB) u (An C )I (by Theorem 2) = IAI + IBI + ICI - IBnCI - [IAnBI + IAnCI -I( AnB ) n(AnC ) I] = IAI + IBI + ICI - IAnBI - IAnCI - IBnCI + IAnBnCI. D

Applications of this result are often made to problems of which the following is typical.

Exa m ple 17 In a gathering of 136 fairly athletic students, 92 engage in gymnastics, 68 are swimmers and 78 are tennis players. Of the 136 students, 41 do gymnastics and swim, 43 do gymnastics and play tennis and 24 swim and play tennis. If 4 students participate neither in gymnastics, swimming nor tennis, how many indulge in all three of these sporting activities? How many play tennis but do not have other sporting activities? We have to reduce the situation of the problem to manageable mathematics. Let U be the set of 1 36 students, let G be the set of gymnasts, let S be the set of swimmers, and letT be the set of tennis players. We are given the information that

IGI = 92, l SI = 68, I T I = 78, IG nS l

= 41 , IGnT l = 43, ISnT l = 24. We are also given that IU \ (GUSUT ) l = 4 from which we deduce that IGUSUT I

=

IUI - IU \ ( GUSUT )I

= 1 36 - 4 = 132

Grou ps, Rings and Fields

16

Now applying Theorem 5 we have 132 = IGUSUT I = IGI + l S I + IT I - IGnSI- IGnT I + IGnSnTI

= 92 + 68 + 78 - 41 - 43 - 24 + 1 GnsnT 1. Hence IG nSnT l = 2, and so 2 students engage in all three activities. The number of students who play tennis but have no other sporting activities is given by IT \ ( G US ) l. The easiest way to determine this number is by means of a Venn diagram in which IG nSnT l = 2, a is the number of students who are gymnasts and play tennis but do not swim, and b is the number of students who swim and play tennis but are not gymnasts.

Figure 1.7.

We have a =

I( GnT ) \ S l = I( GnT ) \ ( GnSnT ) l = IGnT I - IGnSnT I = 43 - 2 =

41.

Similarly

b = I( SnT) \ GI = 24 - 2 = 22. Then, from the diagram, the number of students who play tennis but have no other sports i"l

ITI - a - b - 2 = 78 - 41 - 22 - 2 = 13.

17

Sets and Mappings Exercises 1. 2

1 . Let A and

B be sets.

Establish the De Morgan Law (AUB) ' = A'nB'

(i) by a Venn diagram, (ii) by a set-theoretic proof.

2. Let A, B and C be sets. Establish the Distributive Law AU( BnC )

=

(AUB) n( AUC )

(i) by a Venn diagram, (ii) by a set-theoretic proof.

3. Let A, B and C be sets. Prove, by constructing an appropriate example (often called a counter-example) that the result AU(BnC )

=

( AUB) nC

does not necessarily hold. H A �C does the result hold? Is the con dition A� C necessary and sufficient for the result to hold?

4. In a certain village of 1000 well-educated people, 250 read the Economist, 41 1 read the Bangkok Post and 315 read the Straits Times. Of these people, 72 read the Economist and the Bangkok Post, 51 read the Economist and the Straits Times and 31 read the Bangkok Post and the Straits Times. If only one person reads all three of the publications, how many do not read any of the publications? 5 . A party of 200 schoolchildren is investigated with regard to likes and dislikes of three items, namely, ice cream, sweets and fizzy lemonade. It is found that of those children who like ice cream only 7 dislike sweets. 1 10 children like sweets and 149 like fizzy lemonade. 80 children like both sweets and fizzy lemonade and 36 like both ice cream and lemon ade. If 31 children consume all three items avidly, how many children like none of these items?

1.3 Mappings

2

In our mathematics we have probably evaluated expressions such as x , sin x, and Jx 2 - 4 for given 'values of x'. We may have described these expressions loosely as being functions of x. While this description is somewhat vague it does encapsulate a concept which we would wish to make rather more precise. For a given value of x we expect unique evaluations of the expressions; thus


18

2 ( - 2) = 4, sin 11"/4 = 1/./2, but it is not permissible to put x =lin v'x 2 - 4 if we wish to obtain a real square root. This suggests that we have to specify how the 'values of x' may be chosen. We begin to appreciate that, in each case, we need to specify a set of elements and to have the function defined exactly on this set. We make the following definition.

Definition 1 1 Let A and B b e sets. Let there be a rule or prescription, denoted by f, by which to each element of A there is assigned a unique element, denoted by f(b) , of B. Then the rule is said to be a mapping or map or function from A to B. We write

a� f(a)

(a

E

A)

and, to indicate that f is a mapping on sets, we use both of the following notations:

f: A---+B and A � B. A is called the domain of f and B is called the codomain of f. [The designations 'map' and 'function' are perhaps more common in topology and analysis respectively. We shall use the term 'mapping'.] For a E A, f(a) is called the image of a and, likewise, the subset of B given as { f(a)ia

E

A}

is called the image of A or, sometimes, the range of f. We also denote this subset by f(A) . Finally we note that / ( 0 ) = 0 -

Exa mple 18 Let A = { a, b, c} and B = { 0, 1}. We define a mapping f where f: A� B by letting

!(a)

= 0, !(b) = 1 , /(c) = 1 .

Then f has domain A , codomain B and range B. We define a second mapping g where g : A� B by letting

1 , g(c) = 1 . Then g has domain A , codomain B and range {1 } . We note that f and g are g(a) = 1 , g(b) =

unequal since, for example,

f(a) = 0 # 1 = g(a) . The condition under which two mappings are equal should be fairly obvious but, for the sake of completeness, we state the condition formally.

19

Sets and Mappings

Remark Two mappingsf and g are equal if and only iff and g have the same domain A and f(a) = g (a) for all a E A.

Exa m ple 19 (Continued from Example 3) Let A be the set of the colours of the rainbow. Let B={1, 2, 3, 4, 5 , 6, 7}. We define a mapping f where f : A --+ B by the rule that the image of a given colour of the rainbow is the number of letters in that colour. We need not go further to specify the mapping although, in fact, we have ! (red)= 3,f(orange) = 6, ! (yellow)= 6,f(green) = 5, ! (blue)= 4, ! (indigo)=6, !( violet)= 6. We now introduce certain important sets of which we shall have more to say in the next chapter.

Notation The set of natural numbers, which is denoted by N (N for 'natural') ,

also the set of strictly positive integers, is

N={1, 2,3, . ..}.

The set of integers is denoted by Z (Z for 'Zahl', the German word for 'number') ,

Z={ . . . ,-2,-1,0,1,2, . . . }.

The set of rational numbers, which is also the set of quotients of integers, is denoted by Q (Q for 'quotient'),

E Z, n:rf O}. Thus Q consists of all fractions such as!,�, -is· Q= {m/nlm, n

The set of real numbers is denoted by lR and the set of complex numbers is denoted by C, Evidently

C={ a + ib : a, b N�

E IR} .

Z � Q � lR � C.

A real number which is not rational is said to be irrational, lR \ Q is the set of irrational numbers. We give three examples of mappings involving these sets before giving some us eful terminology for describing mappings.


20

Exam ples 20

1.

2.

3.

f: N --+ { -1, 1} given by f(n) = (-1 ) n (n E N) has domain N, codomain { -1 , 1} and range { - 1 , 1}. The mapping f: R --+ R given by f(x) = x2 + 1 (x E IR) has domain IR, codomain lR and range {y I y � 1}. The mapping f: N --+ Q given by f(x) = !n (n E N) has domain N, codomain Q and range {!, 1 , �, 2, . . . } = NU{�In E N}.

The mapping

Definition 12 Let

A and B be sets. Let f: A --+ B be a mapping.

The mapping / is said to be injective, or one-one, if whenever f(a1) = !( �) for a 1 , � E A then necessarily a 1 = �(ii) The mapping f is said to be surjective or onto if for each b E B there exists a E A such that f(a) = b. (iii) The mapping f is said to be bijective or one-one and onto if f is both injective and surjective.

(i)

Exam ples 2 1

1.

2.

3.

A = {a, b, c} and B = {p, q}. We define f: A --+ B by f(a) = p, f (b) = p, f (c ) = p. Then f is not injective since f(a) = f(b) but a-::/; b. We also have that f is not surjective since f(A) = {p} -::/; B. Let A = {a, b} and B = {p, q, r}. We define f: A --+ B by f(a) = p, f(b) = q. Then f is injective since f (a) -::/; f( b) but is not surjective since f(A) = {p, q} =/=B. Let A = { a, b, c} and B = {p, q}. We define f :A --+ B by f(a) = p, f(b) = p, f(c) = q. Then f is not injective since f(a) = f(b) but is surjective since f(A) = {p, q} = B.

Let

21

Sets and Mappings

[The reader may begin to suspect (correctly) that a bijective mapping exists between two finite sets if and only if they have the same number of elements. In particular the reader may rightly conclude that a bijective mapping cannot exist between a finite set and a proper subset; for infinite sets the situationis less restrictive, as the next example shows.]

4. Let 2N denote the set of strictly positive even integers, 2N

= {2nln EN}.

Now 2N is a proper subset of N but, nevertheless, the mapping f: N given by

f (n)

= 2n

�

2N

(n EN)

is bijective.

5. A mapping f is defined on N by (n odd) (n even) . Suppose we are asked to describe f. We note first that f is defined so that f: N � N. Then f is not injective since, for example, /(3) = 3 = /(6). But f is surjective since if k EN then f ( 2 k) = k. Frequently we have to consider the effect of applying mappings successively. More precisely we have the following definition.

Definition 13 Let A , B and C be sets and let f: A� B and g: B � C be given mappings. We have

where the range offis a subset of the domain of g. Then for each a E A we have a mapping, denoted by go f, called the circle-composition of the mappings g and f and given by

(go f) ( a)

=

g(f(a) )

in which w e first apply the mapping f t o a EA and then apply the mapping g to !(a) EB. If no ambiguity will arise we often write gf forgo f.


22

Theorem 6 1. 2. 3.

:

:

C be sets and let f A-+ B and g B-+ C be mappings. If f and g are surjective then g o f is surjective. Iff and g are injective then g o f is injective. Iff and g are bijective then g o f is bijective.

Let A, B and

Proof We prove 1 and leave 2 to be proved in subsequent Exercises. ! and 2 imply 3. Let c E C. Then since g is surjective we have b E B such that g(b) =c. Since f is surjective we have a E A such that f(a) = b. Then

(g o f) (a) =g(f(a) ) = g(b) = c and so

g o f is surjective.

D

Exa m ples 22 1. Let A={a, b, c, d}, B={1 , 2, 3}, C = {p, q, r, s}. Let mappings f, g be defined by f(a) = 1, f(b) =1 , /(c) = 2, f(d) =3, g(l) =p, g (2) = r, g(3) =r. Then we have with, for example, 0

(g !) (b) = g(f( b)) = g(l) =p, (g o f)(d) =g(f(d)) = g(3) =r. f and g be defined by g(x) = x 2 (x E IR) . f(x) = sinx (x E IR), Then we may form g o f and also f o g :

2. Let mappings

g

lR

f ---+ IR---+IR ,

lR

f ---+IR---+ IR ,

g

Now

(g o f) (x) =g(f(x)) =g(sin x) = (sin x) 2 =sin2 x but

(f o g) (x) =f(g(x)) = f(x2 ) = sin x2 .

23

Sets and Mappings

since itis not g o f and f o g have the same domain but are different mappings 2 2 true, contrary to an occasional misguided belief, that sin x = sinx for all X EIR. Consider the case of four sets A, B, C, D and three mappings f, g, h where f :A --+ B, g: B--+ C and h C --+ D. We may construct the mappings g o f and h o g and then the further composi tions h o (g o f) and (h o g) o f, giving the pictures below. :

gof

� -4D �

(h o g)

h 0 (g 0 f )

o

f

Figure 1.8.

Then h o (g o f) and (h o g) o f are both mappings from been constructed differently. Could they be equal?

A to B but they have

We consider a specific example before proving the general result.

Exa m ple 23 Let

f, g, h be the mappings given by

f(x) = x2 + 1 (x EZ) , 2

g(n) = 3 n ( n EN), h(t) = #+1 (t EQ), and so,


24

But

+ 1) (x2 + 1) (x [h o (g o J)] (x) = h G (x2 + 1)) G <x2 + 1)r +1, (h o g) (n) = h(g(n)) = h G n) � (n G (x2 + 1)r +1 (x (h o g)( J (x)) = (h o g) (x 2 + 1)

(g o J) (x) = g(f(x)) = g(x2

=

2 3

EZ) ,

=

=

[(h o g) o Jl (x) = Thus

EN ) ,

=

[h o (g o f)] (x)

=

Ez).

[(h o g) o f] (x) (x EZ)

from which 0

0

0

0

h (g /) = (h g) f. Theorem 7 Associativity of C ircle-Com position

J, g, and h be mappings such that f A---+ B, g B---+ C, h C---+ D.

Let A, B, C, D be sets and let

:

:

:

Then 0

0

0

0

h (g /) = (h g) f. P roof The proof simply entails repeated and careful application of Definition a EA. Then

[h o (g o f)] (a) = h((g o f) (a)) h(g(f(a))) =

and

[(h o g) o f] (a)

= =

Thus

[h o (g o /)] (a)

=

(h o g) ( J (a)) h(g(f(a)) ) .

[(h o g) o f] (a)

13.

Let

Sets and Mappings

and since

a

25

is arbitrary 0

0

h (g f)

=

We may now write, unambiguously, tions h o (g o f) or ( h o g) o f.

0

0

(h g) f.

D

h o g o f to mean either of the composi

Rema rk Mappings may appear to be necessary but somewhat humdrum objects of study. However particular mappings may give rise to curious problems. Consider, for example, the mapping f : N --+ N given by

f(n)

=

{

!n ! (3n + 1 )

(n even) (n odd) .

Thus /(26) = 13, /(25) = H 75 + 1 ) = 38. Suppose we iterate this mapping several times and, by way of illustration, we follow the effect of the iterations on the number 10. Then /(1 0)

=

5,

( ! 0 /) ( 10)

=

/(5)

(f o f o /) (10) = /(8) (! f f /) ( 1 0) = /(4) f f f /) (10) = 1 .

(! 0

0

0

0

0

0

0

=

8,

=

4,

=

2,

After five iterations on 10 the number 1 appears. The reader may care to verify that commencing with 65 there are 1 9 iterations before 1 first appears. Other integers may be tried at random as test cases, but the reader is advised to culti� vate patience as the number of iterations before 1 first appears may be quite large. The conjecture, that for any n E N and for sufficiently many iterations the number 1 always appears, remains to be proved or disproved.


26 Exercises 1. 3

1 . Let A, B, C be the sets {a, b, c}, {p, q, r}, { 0, 1} respectively. Let the mappings f: A __. B and g : B ->C be defined by

f(a) = p, f(b) = /(c) = q, g(p) Evaluate

2.

Let

0, g(q)

= 1 , g(r)

=

0.

(g o f) (a), (g o f) (b) , (g o /) (c) .

A, B, C, D be sets and let p, q, r, s be mappings such that p : A--> B, q: B--> C, r: A--> B, s: D--> C.

Which of the following are defined:

3.

=

q o p, p o q, p o r, p o s, r o s, q o s?

f, g, h be mappings of Z into Z given by f(n) = 2n, g(n) = 3n + 5, h(n) = -6n (n E Z) . Which of the following pairs of mappings are equal: f o g, g o f; f o h, h f; g h, h g? Let

0

0

0

4. Let A, B and C be sets and let f: (i) (ii) (iii)

A--> B and g : B--> C be mappings. Iff and g are injective prove that g o f is injective. If g o f is injective, are g and f injective? If g o f is surjective, are g and f surjective?

5. Let X, Y be sets and let f: of X. (i) (ii) (iii)

6.

X--> Y be a mapping. Let A, B be subsets

If A�

B prove that f(A) � f(B). Prove that f(A UB) = f(A) U/(B) . Prove that f(AnB) � f(A) nf(B) .

Let A be the subset of Z consisting of the even integers and let B be the subset of Z consisting of the odd integers. Let a mapping f: Z--> Z be defined by

f(n) = n2 - 3n + 5 (n E Z). Prove that f(A) n/ (B) .

3 E f(A)nf(B)

and so deduce that

f(AnB) ::f;

27

Sets and Mappings

1 .4 Equivalence Relations In the context of humans, animals and plants, the notion of relationship is all pervasive; two individuals are regarded as 'related' if they share a common parent, grandparent or even a more distant ancestor. ThUB coUBins are said to be related but we would not normally regard a cat and a dog as being related. In these illUBtrations we encounter the idea of a universal set, whether composed of humans or of domestic animals, and of the relationship which members or ele ments of the set may bear to one another under some particular criterion. Since the underlying concept of relationship is very useful in mathematics we discuss a more mathematical example before proceeding to a precise definition.

Exa m ple 24 We consider a set A and a relationship which exists between some pairs of elements of A. Let X be a set and let f : A--+ X be a mapping. We shall say that elements a and b of A are 'related' iff(a) =f(b). By this definition any two elements of A may or may not be related. For convenience we write for a, b EA, a,...., b if and only if a, b are related; the presence of the symbol ,...., (pronounced 'twiddles') denoting that a and b are related. We note three obvious facts: 1 . For all a E A, a is related to a sincef(a)

=

f(a) implies a,...., a (for all a EA) .

2. If a, b EA are such that a is related to b, then b is related to a, since a,...., b impliesf(a) =f(b) and sof(b) = f(a) and b,...., a. 3. If a, b, c EA are such that a is related to b and b is related to c, then a is related to c since a,...., b impliesf(a) = f(b) and b,...., c impliesf(b) = f(c) from which f(a) = f(b) f(c) and so a,...., c. =

The definition we seek basically repeats the conclUBions of this example.

Definition 14 Let A be a set. A relation, denoted by,...., , is defined between some pairs of ele ments of A subject to the following (named) conditions. 1 . For all a EA, a,...., a (reflexivity) . 2. If a, b EA are such that a,...., b then also b,...., a (symmetry). 3. If a, b, c EA are such that a,...., b and b,...., c then also a,...., c (transitivity).


28

A relation"' on A which is reflexive, symmetric and transitive as above, is called an equivalence relation. Symbols R, p, u are frequently used to denote equivalence relations. We note in passing that the relation of the previous example is an equivalence relation.

Exam ples 25 1. On any set A the relation of equality is an equivalence relation since we have a= a for all a EA, we certainly have a= b implies b =a( a, b EA) and if a=b and b= c(a, b, c E A) then a=c. An equivalence relation between the elements of a set generalizes the notion of equality for elements of the set. 2. On Z a relation"' is defined by a"' b if and only if a, b are both even or a, b are both odd. Thus 4"' 4 and 5"' 5 but we do not have 4"' 5. Now, in general, a"' a and certainly a"' b implies b"' a (a, b EZ ) . Suppose now a"' b and b"' c (a, b, c EZ) . Then, if b is even, a and c must also be even and a"' c, whereas if b is odd, a and c must also be odd and a"' c. Thus"' is an equivalence relation on Z. Notice that we could have defined this equivalence relation by means of a mapping f: Z -+ {0, 1} for which

f(n) =

{

0 1

(n even) (n odd )

(n EZ).

3. LetS be the set of all words in a given English dictionary. Define two words to be related by p if they begin with the same letter and end with the same letter. Thus we have 'atom p alum' , 'pleasure p perseverance' and 'a p aroma' . Then p is easily seen to be an equivalence relation on S. 4. LetS be the set of points composing the circumference of a given circle. For P, Q ES define a relation u by P u Q if and only if there is a diameter on which P, Q are points. Thus if P u Q then either P, Q coincide or P, Q are at oppo site ends of a diameter. Simple geometrical considerations show that u is an equivalence relation. 5. Let X be the set Q x Q of ordered pairs of rational numbers. For a= (a1, a2 ) E Q x Q and b (b1, b2) EQ x Q we define a relation"' by =

a"' b if and only if a1 + a2

=

b1 + b2 •

We claim that"' is an equivalence relation on X. Certainly"' is reflexive.

29

Sets and Mappings

If for some a, b E X we have a "' b then a = (a 1 , a2 ) , b = (b 1 , � ) , where ai , bi E Q (i = 1 , 2) and a 1 + a2 = b 1 + b2 . But then b1 + b2 = a 1 + a2 and so b "' a and "' is symmetric. Suppose now for some a, b, c E X we have a "' b and b "' c. Then writing a = (a 1 , a2 ) , b = (b1 , b2 ) , c = (c1 , c2 ) we have a 1 + a2 = b1 + b2 = c1 + � and so a "' c and "' is transitive. Hence "' is an equivalence relation. 6. A relation, p, is defined on Q by letting, for a, b E Q, apb if and only if we have a - b E Z. By this relation we have 43 5 43 5 38 - p - as - - - = - = 2 E Z 19 19 19 19 19

but 1 1 -p2 3

is

false

as

1 1 5 - - = - ¢ Z. 2 3 6

-

We may prove that p is an equivalence relation

as

follows:

(i) For all a E Q, a - a = 0 E Z and so a "' a (reflexivity) . (ii) If for some a, b E Q, a "' b then a - b E Z which immediately implies that b - a = - (a - b) E Z and so b "' a (symmetry) . (iii) If for some a, b, c E Q, a "' b and b "' c then evidently a - b E Z and b - c E Z from which a - c = (a - b) + (b - c) E Z and so a "' c (transi tivity) . It is sometimes important, in regard to a set of conditions specifying a mathema tical entity such as an equivalence relation, to know whether the stipulated con ditions are themselves independent or not . We show that the three conditions of reflexivity, symmetry and transitivity for an equivalence relation are indepen dent by exhibiting t hree examples in each of which two of the conditions are satisfied but not the remaining condition.

Exa m ples 26 1 . Let a relation "' be defined on Z by a "' b if and only if ab =I 0 (a, b E Z) . Then the relation is symmetric since a "' b implies ab =I 0 and so ba =I 0. Thus b "' a (a, b E Z) . The relation is also transitive since a "' b, b "' c implies ab =I 0, be =I 0. Hence a =I 0, b =I 0, c =I 0 and thus ac =I 0 giving a "' c. On the other hand the relation is not reflexive since we do not have a "' a for all a E Z, in particular we do not have 0"' 0.


30

2. The relation of inequality, �. on lR is reflexive and transitive since we have, for all a E IR, a � a and if a � b, b � c then a � c (a, b, c EIR) . We cannot infer from a � b (a, b EIR) that b � a and so this relation is not symmetric. 3. Let a relation "' be defined on Z by a"' b if and only if 2 divides a - b or 3 divides a - b. Such a relation is not transitive since, for example, 7"' 5 since 7 - 5 = 2 and 5"' 2 since 5 - 2

=

3

but 7"' 2 is false since 7 - 2

=

5 which is divisible by neither 2 nor 3.

But the relation is reflexive since for all a EZ, a - a = 0 and so a"' a and the relation is symmetric since a"' b (a, b EZ) implies that a - b, and thus b - a, is divisible by 2 or by 3 and so b"' a . We may consider the totality of the inhabitants of a given town as being dis tributed into the families which dwell in the town. Any two members of the same family are somehow related but no two members of different families are related. An inhabitant of the town belongs to only one family and so identifies uniquely the family to which he or she belongs. In a similar way a set with an equivalence relation will be seen to be the union of disjoint subsets called equivalence classes, each equivalence class consisting (like a family) of all elements which are related to one another and each equivalence class will be uniquely identifiable by any element belonging to it. We make these ideas more precise.

Defi n ition 15 LetS be a non-empty set with an equivalence relation"'· Let a ES. The subset Sa. ofS given by S4 = {x ES :x"' a} is said to be the equivalence class determined by, or containing, a (note that as a"' a, a ES4) . We now prove a result which we shall have frequent occasion to

use.

31

Sets and Mappings

Theorem 8 In the notation of Definition 15 the following hold:

U Sa. aES 2 . Sa = Sb if and only if a "' b (a, b E S ) . 3. Either Sa = Sb or Sa n Sb = 0 (a, b E S ) . 1. S =

Proof 1 . Since a E Sa it follows that S �

U Sa and so S = U Sa. aeS

aeS

2. If Sa = Sb then a E Sb and so a "' b. Conversely suppose a "' b. We prove that Sa � sb . Let X E Sa. Then X "' a and a "' b implies that X "' b and so X E Sb. Thus Sa � Sb. Since, by symmetry, b "' a we also have Sb � Sa and so Sa = Sb.

3. Suppose Sa n Sb '::/: 0· We show that Sa = Sb . Let c E Sa n Sb and so c"' a and c"' b. But then a "' c and c"' b which implies, by transitivity, that a "' b. By (ii) we conclude that Sa = Sb. D It is a consequence of Theorem 8 that from the union equivalence classes that formally coincide, we may write

S=

U Sa, on eliminating

aeS

U Sa

aeT

where T is a subset of S such that Sa (a E T) are distinct, and Sa n Sb = 0 (a, b E T, a '::/: b) . The non-empty set S is thereby expressed as a disjoint union of distinct equivalence classes.

Exa m ples 27 (Continued from Examples 25 ) We examine the examples immediately following Definition 14 to determine the appropriate equivalence classes. 1. With equality as the equivalence relation on a set each equivalence class con si<Jts of a single element . 2 . There are two equivalence classes, namely the subset of Z consisting of the even integers and the subset of Z consisting of the odd integers. 3. The Engli'lh alphabet has 26 letters: a, b, c, . . . , x, y , z. Each word begins with, 2 and end'l with, one of these letters. We now define 26 = 676 subsets each defined for any pair of letters of the alphabet as follows:


32

Let Saa be the subset of all word� beginning with a and ending with a. Let Sab be the subset of all word� beginning with a and ending with b. Let Szz be the subset of all words beginning with z and ending with z. The reader will immediately observe that some of the subsets 800 , Sa.b• . . . , Szz are empty (Szz = 0 but, perhaps surprisingly, Sa.z # 0) . Any one of these subsets, if non-empty, is an equivalence class. S is the union of these subsets and, on omitting empty subsets, becomes a disjoint union of equivalence classes. (We leave it to the aspiring lexicographer to determine which subsets are non-empty.) 4. Every diameter determines an equivalence class consisting of the two end points of the diameter. 5. For every q E Q there is an equivalence class given by Xq = { (a�> a2 ) E Q Each equivalence class is of this form.

x

Qlal + � = q}.

We have seen that if a set admits an equivalence relation then that set is a dis joint union of particular subsets called equivalence classes. Conversely if a set is a di�j oint union of subsets then we shall show that, correspondingly, an equiva lence relation may be defined on the set. We are led to make the following definition.

Defi n ition 16 Let S be a non-empty set. Let { S>. : >. E A} be a collection, indexed by an index set A, of non-empty subsets of S such that 1 . S>. n SP. = 0 (>., J.L E A, >. # J.L ) and 2. S>. = S. >. e A The collection of subsets is said t o form a partition o f S.

U

Suppose we have a non-empty set S with a partition as above. For a, b E S define a relation "' by letting a "' b if and only if a, b belong to the same subset S>. (say) of the partition. Thus any two elements in an S>. are related but no element in S>. is related to an element in SP. (>. # J.L ) . Then it may readily be verified that "' i� an equivalence relation on S and that the equivalence classes are the subsets S>. (>. E A) . A partition therefore induces an equivalence relation and an equivalence relation induces a partition, subsets of the partition forming the equivalence classes of the equivalence relation.

33

Sets and Mappings

Exa m ple 28 Let 8 = {a, b, c, d, e , J, g }. Let 81 = {a , b, d} , 82 = {c, g} , 83 = { e, !} . Then the subsets 81 , 82 , 83 form a partition {81 , 82 , 83 } of 8. We define the corresponding equivalence relation "' on 8 as follows: a "' a , a "' b, a "' d, b "' a, b "' b, b "' d,

c "' c, e "' e , c "' g, e "' f, d "' a, d "' b, d "' d, g "' c, f "' e , g "' g, f "' f.

Exercises 1 . 4

1 . A relation p is defined on lR by a p b i f and only i f a2 = b2 (a, b E IR) . Prove that p is an equivalence relation on JR. Identify the equivalence classes . 2. A relation R is defined on Z by aRb if and only if a - b is even (a, b E Z) . Prove that R is an equivalence relation on Z. Identify the equivalence classes. 3. A relation r is defined on Z by a r b if and only if a - b is divisible by 6 (a, b E Z) . Prove that r is an equivalence relation and determine the equivalence classes.

4. What are the equivalence classes in Examples 25, no. 6 immediately following Definition 14? 5. Let Oxy be the two-dimensional coordinate plane. Let P, Q be points of Oxy.

(i) P, Q are said to be related by a if the points 0, P, Q are collinear. Is a an equivalence relation? (ii) P, Q are said to be related by r if there exists a rotation about 0 which sends P into Q. Is r an equivalence relation? Sketch, if possible, the equivalence classes. 6. How many distinct equivalence relations may be defined on a set of one, two, three or four elements? 7. A relation p is defined on Z by a p b if and only if a - b is divisible either by 5 or by 7 (a, b E Z) . Is p an equivalence relation?


34

8. Let S = {a, b, c, x, y, z} . Two relations p, u are defined on S as follows: a p a, b p b, cp c, x p x, y p y, z p z, a p b, b p a, a p e, cp a, b p c, cp b ; a u a, bu b, c u e, x u x, yu y, zu z, a u x, x u a, b u y, yu b, c u z, z u c , a u y, y u a .

Do either p or u define an equivalence relation on S? 9. Let S be a set on which a relation "' is defined. The relation "' is symmetric and transitive. Loose thinking would suggest that a "' b (a, b E S ) implies b "' a (by symmetry) and so a "' a ( by transitivity) thus giving reflexivity, and consequently"' would be an equivalence relation. What is wrong with thls loose thinking? 1.5 Well-ordering and Induction The elements of the set N, that is the set of strictly positive integers, may be written down in ascending order with a repeated inequality sign, thus: 1 0) and a = ( - b) q + r = b( -q) + r (b < 0) ,

of which only the second statement is wholly new. If a < 0 then -a = I b l q + r and so a = - I b l q - r which is almost in the desired form except that we require to have a positive remainder. We make a small but significant adjustment by writing a = - l b l q + r = l b l ( - 1 - q) + ( l b l - r ) . Now 0 < r < l b l implies that 0 < l b l - r < l b l and so a = bq ' + r '

where r ' = I b l - r and q ' = - 1 - q (b > 0) , q ' = 1 + q (b < 0) . This completes the proof. D

Exa m ples 9 1 . a = - 574, b = 34. By the Division Algorithm (First Version) we have 574 = 34. 1 6 + 30,

from which we have - 574 = 34. (-16) - 30 = 34(-16 - 1 ) + (34 - 30) = 34(- 17) + 4

giving q = - 1 7, r = 4. 2. a = -64 12, b = -97. By the Division Algorithm (First Version) we have 6412 = 97. 66 + 10,

from which we have - 6412 = ( -97) .66 - 10 = ( -97) . (66 + 1 ) + (97 - 10)

giving q = 67, r = 87.

=

( -97) .67 + 87

56


Exercises 2. 3

1 . Given integers a and b, find integers q and r such that a = bq + r (0 ::=; r < l b l) where: (i ) a = 8, b = 6; ( ii) a = 27, b = 8; (iii ) a = 241, b = 35; ( iv ) a = -2513, b = 46; (v ) a = 54321 , b = -761 ; ( vi ) a = - 52148 , b = -732.

x and y be real numbers. Prove ( i ) l xy l = l x i i Y I and ( ii ) I x + y I ::=; I x I + I y I · Construct an example in which, for appropriate choice of x and y, l x - Y l � l x i - I Y I ·

2. Let

2.4 Euclidean Algorithm

We now describe an effective method, the so-called Euclidean Algorithm, for finding the G.C.D. of two integers. The Euclidean Algorithm depends on repeat ing the Division Algorithm for a finite number of times until a particular con clusion, yielding the G.C.D., i� reached. The method appeared originally in a geometrical form as Proposition 2 of Book VII of 'The Elements' - a book which was composed by Euclid of Alexandria around 300 BC and which survived as a textbook for over 2000 years. First we prove a brief but useful result.

Theorem 3 Let a, b and d be non-zero integers and let d be a positive common divisor of a, Suppose there exist integers x and y such that

b.

ax + by = d. Then d is the greatest common divisor of a and b. P roof Let c be a common divisor of a, b. We require to show that c divides d. Now we certainly have a = ca ' , b = cb' for suitable a', b' E Z. Hence

ax + by = (ca' )x + (cb' )y = c(a'x) + c(b'y) = c(a'x + b ' y) . Hence c divides d and so d = (a, b) . d=

0

57

The I ntegers

Exa m ple 10 It is easy to verify that 97 is a common divisor of 291 and 388 but it is less easy, without the use of the Euclidean Algorithm, to find an equation of the form given in Theorem 3. One such equation is 291 . 1 5 + 388( - 1 1 )

=

97,

and from this equation we may conclude that 97 is the G.C.D. of 191 and 388. We shall give the explicit procedure of the Euclidean Algorithm but, by way of preliminary illustration, we begin with two particular examples of its use.

Exa m ples 1 1 1 . Suppose we wish to find the G.C.D. of 4947 and 1552. Our procedure is to apply the Division Algorithm systematically as follows. Applying the Division Algorithm three times we have 4947 = 1552.3 + 291

(1 )

1552

=

291 .5 + 97

(2)

291

=

97.3

(3)

We claim that 97 is the required G.C.D. First we have to prove that 97 is indeed a divisor. We consider the equations in reverse order. Certainly from equation (3) , 97 divides 291 . But then from equation (2) 97 divides 1 552. In turn, from equation ( 1 ) , 97 divides 4947. Thus 97 is a common divisor of 1 552 and 4947. Now to prove that 97 is the G.C.D. we let d be a divisor of 1 552 and 4947. By equation (1) d divides 291 and then from (2) d divides 97. Hence 97 is the G.C.D. We also write 97 in the form of Theorem 3 as follows. From equation (2) 97 = 1552 - 291 .5 and hence, from equation ( 1 ) , 97

=

1552 - (4947 - 1552.3) .5

= =

A soon as we have established that

1552 - 4947.5 + 1 552.3.5 4947. (-5) + 1552.16.

97 is a common divisor then the represen tation of 97 in the form above gives an alternative proof that 97 is indeed the G.C.D.

58


2.

Suppose we wish to find the G .C.D. of 163059 and 80004. Then applying the Divi<Jion Algorithm four times we have

163059 = 80004.2 + 3051

(1)

80004 = 3051 .26 + 678

( 2)

3051 = 678.4 + 339

(3)

678 = 339.2

(4)

We consider these equations in the reverse order to show that 339 is a common divisor of 163059 and 80004. From (4) 339 divides 3051. And so from ( 3) 339 divides 3051. But then (2) implies that 339 divides 80004 and then finally from (1) , 339 divides 163059. Hence 339 i"l a common divisor of 163059 and 80004. By considering the equations in the normal order we may conclude, as in the previous Example, that 339 is, in fact, the G.C.D. We may also establish thi"l conclusion by writing 339 in the form 163059x + 80004y as follows. From (3) we have

339 = 3051 - 678.4 (2) we have 339 = 3051 - (80004 - 3051 .26).4 = 3051 . 105 - 80004.4 Applying (1) now gives 339 = (163059 - 80004.2)105 - 80004.4 163059.105 + 80004.( -214) and so from

=

as

we desired.

Keeping these two Examples in our minds we may now proceed to give the Euclidean Algorithm in its proper generality.

The Euclidean Algorithm

Given two non-zero integers a and b, we shall describe how the Algorithm will yield, in a finite number of steps, the G.C.D. of a and b. We follow the procedure of the two examples above. For convenience we may suppose a > 0, b > 0 and introduce the notation ao = a, al = b. By the Division Algorithm we obtain integers q1 and a2 such that ao =

al ql + a2,

0 � a2 < al .

59

The I ntegers

If a2 = 0 then ao = a1q1 and we have immediately that a1 is the G.C.D. of ao and a1 • If � =I 0 then we continue 88 follows. We know, again by Division Algorithm, that we obtain integers q2 , a3 such that

0 � a3 < a2. If a3 = 0 then we halt the procedure but if a3 ::/; 0 we continue and obtain inte a1

=

a2q2 + a3 ,

gers q3 , a4, such that Either a4

=

0 or if a4 ::/; 0 then we continue 88 in the examples above. ao

=

a1 q1 + a2

{ 1)

a2

=

a2q2 + aa

{ 2)

a2

=

a3qa + a4

{ 3)

aa

= a4q4 + as

{ 4)

Now continuing we obtain a finite or infinite sequence ab a2, a3 , such that a1 > a2 > a a > . . . �

•

•

•

of integers

0.

This is only possible if the sequence is finite and for some n, an + l = 0. Thus suppose we halt at the nth equation. As in the examples above we have the following equations:

{1 ) ( 2) { 3) {4) (n - 2 ) (n -

1)

(n) We claim that an is indeed the G.C.D. of ao = a and a1 = b We prove first that an is a common divisor of a0 and a1 by considering the equations in the reverse order. From (n) an divides an - l · Then from (n - 1) an divides an _ 2. Continuing we have finally, on considering { 1 ) , that an divides a2 and al and so an divides ao . .


60

Thus an is a common divisor of ao and a 1 • Let now c be a common divisor of ao and a 1 . By (1) c divides ao - a 1 q1 = a2 . Then from (2 ) c divides a 1 - a2 q2 = a3 . Continuing we find that c divides an _ 2 - an _ 1 qn - l = an · Thus, as we claimed, a is the G.C.D. of ao and a1 . We are also now able to write an in the form ao x + a 1 y for some x, y E Z. From ( n -

1)

From ( n -

2)

and so, substituting from (n - 2 ) into (n -

From ( n - 3)

1) , an = an - 2 - (an -3 - an - 2 qn - 2 ) qn - l = (1 + qn - 2 qn - d an - 2 - an- 3 ·

and so, substituting from (n - 3 ) into the equation for an we have

an = (1 + qn- 2 qn- d an- 2 - an-3 = (1 + qn 2 qn_ t ) [an 4 - an qn ] - an-3 -3 3 = [-qn-3 (1 + qn - 2 qn- l ) - 1]an-3 + (1 + qn- 2 qn - d an- 4 = -(1 + qn -3 + qn -3 qn- 2 qn- l )an-3 + (1 + qn - 2 qn - l )an - 4 ·

Continuing this procedure we shall eventually obtain an equation of the form for some

x, y E Z.

We may summarize thi'l last result in the form of a converse to Theorem 3.

Theorem 4 Let a and b be non-zero integers and let d be the greatest common divisor of a and b. Then there exist integers x and y such that d = ax + fry.

We remark in passing that x and y are not unique since d = that d = a(x - bt) + b(y + at) for any integer t.

ax + fry implies

61

The I ntegers

We conclude this section with one further example which, in view of our extended discussion above, we shall give in as brief a form as is necessary for calculation.

Exa m ple 12 We wish to find the G.C.D. of 108810 and 108810x + 93346y for some x, y E Z.

93346 and to write the G.C.D.

108810 = 93346. 1 + 15464

(1)

93346 = 15464.4 + 562

{2)

15464 = 562.27 + 290

(3)

562 = 290.1 + 272

(4)

290 = 272.1 + 18

(5)

272 = 18. 15 + 2

(6)

18 = 9.2 Thus 2 is the G.C.D. of 108810 and 93346.

Reversing the order of the equations we have

(93346, 108810)

as

2 = 272 - 18.15 (from (6) ) = 272 - (290 - 272) .15 (from (5)) = -290.15 + 272.16 = -290.15 + (562 - 290).16 (from (4) ) = 562.16 - 290.31 = 562.16 - (15464 - 562.27) .31 (from (3) ) = -15464.31 + 562.853 = -15464.31 + (93346 - 15464.6).853 (from (2)) = 93346.853 - 15464.5149 = 93346.853 - (108810 - 93346) .5149 (from (1)) = 93346.6002 - 108810.5149 =

which gives the desired result.

(7)


62 Exercises 2. 4

1. 2.

Let a, b, x, y and n be integers such that n divides both a and b. Prove formally that n divides ax + lry.

Use the Euclidean Algorithm to find the greatest common divi 2. We shall argue by induction and suppose that the result is true for n - 1 and for all m such that 2 � m � n - 1 . We prove first that n is a

product of primes. If n is a prime then, as we remarked, the as.�ertion is true. Suppose that n is composite and that n = n1n2 where n1 , n2 E N, 1 < n1 < n, 1 < n2 < n. By our assumption each of n1 and n2 is a product of primes and so therefore is n. Suppose now that n = P 1 P2 · · · Pr

=

qlq2 · · · qs .

Then p1 divides q1 q2 . . . qn. By Theorem 5 , p1 must divide one of q1 , q2, . . . , q8 • By renumbering the qi 's if necessary we may suppose p1 divides q1 • But p1 and q1 are primes so p1 = ql . Thus we have from which P2P3 · · · qr = q2q3 · · · qs

By our induction assumption we have

r = s

and with suitable renumbering

Pi = qi ( i = 2, 3, . . . , n ) . This completes the proof.

D

67

The I ntegers

Arising out of this result we may collect together equal prime divisors and so we conclude that n may be written as n = P101 P202 · · · PtOt

where p1 , p2, . . . , Pt are distinct primes and ai > 0 (i = 1 , 2, . . . , t) . Many fascinating, but sometimes still unresolved, questions may be raised in regard to primes . Euclid answered one obvious initial question, namely 'how many?'

Theorem 9 ( 'The Elements' , Book IX, Proposition 20) There are infinitely many primes.

P roof Let p 1 , p 2 , . . . be the primes in ascending order (of course p 1 = 2, p2 = 3, p3 = 5 , etc. ) . We argue by contradiction and so suppose there is only a finite number N of primes; pN is then the largest prime. Let M

cannot be a prime since M > PN But, since M is a product of some of the primes p 1 , p 2 , . . . , pN , at least one of these primes must divide M, say p 1 divides M. But we have ·

M 1 - - P2 Pa · · · PN = - . P1 P1

where the number on the left-hand side of this equation is an integer whereas the number on the right-hand side is not. We have reached a contradiction and so there is not a finite number of primes and we obtain the result. D Various simple extensions of this result are known. We shall give one such extension after some preliminary remarks. Any odd integer is either of the form 4n + 1 or 4n + 3 for suitable n E Z, for example 21 = 4 . 5 + 1 and 23 = 4.5 + 3. In particular every prime other than 2 is of the form 4n + 1 or 4n + 3. We observe that the product of odd integers, all of which are of the form 4n + 1 , is again an integer of this form since ( 4n1

+ 1 ) (4� + 1 )

= 4 (4n1 �

+ n1 + n2 ) + 1 .


68

Exa mple 17 The first ten primes, of the form 4n + 3, are 3, 7, 1 1 , 19, 23, 31, 43 , 47, 59, 67.

Theorem 10 There are infinitely many primes of the form 4n + 3 .

P roof Let p1 , p2, be the primes of the form 4n + 3 in ascending order (of course p1 = 3, p2 = 7, p3 = 1 1 , etc.) . We argue by contradiction and so suppose there is only a finite number N of such primes; PN is then the largest such prime. Let . • .

M

=

4 ( P2P3 . . · PN ) + 3

Then, by an argument similar to that used in the proof of Theorem 9, M is not divi�ible by any of p2, p3 , , PN . Further M is not divisible by 3 since 3 does not divide 4 ( P2p3 PN ) . Thus M is of the form 4n + 3 but no prime factor of M is of this form. Thus all the prime factors of M must be of the form 4n + 1 . But from the remark above the product of such prime factors is again of the form 4n + 1 and not of the form 4n + 3. Thus we obtain a contradiction and so the result is proved. D • • •

•

•

•

We have developed some elementary properties of integers in order to whet the appetite for further study in number theory and, more importantly for our pur poses, to prepare the ground for the axiomatic treatment of groups, rings and fields which is to follow. However, before leaving the present context we offer some random remarks on these objects of wonder the primes.

Sieve of Eratosthenes

Eratosthenes (c. 275-195 BC) gave a method for detecting primes which, for fairly obvious reasons, is known as a 'sieve'. The implementation of the method requires Lemma 3 above. Lemma 3 implies that if we wi�h to determine whether a positive integer n is prime then we may confine our attention to possible prime divisors less than or equal to yn. We illustrate the method by detecting those primes less than 100. We write down the integers from 1 to 100 and successively eliminate those divisible by 2, 3, 5 , 7, these being the primes � v'fOO = 10 as follows. We circle 2 as being prime and stroke out every 2nd number thereafter. We circle 3

69

The I ntegers

as the next prime and stroke out every 3rd number thereafter. Similarly we circle

5 , and then 7, and stroke out every 5th, and then every 7th, number. We obtain the following picture:

1 11

1-2"

.2t

®

@

.4'

@

(J)

.6'

.8'

M

l5'

.16"

17

18"

31 41

22

13 23

.24

26

.26'

2!r

M

.3-3'

M

.35'

.36'

M

.20'

.28' .J8'

.39'

.46'

.48'

Mr

.as-

59

..56' .66'

.68'

$'

::ro

M

.45'

.46"

M

43 53

M

..55'

M

M

M

M

M

.65

.6t)

67

X

�

$

:rr

�

M

M

.86

.87'

M

.95

.96"

61 71

:n

.81

.82'

73 83

M

M

..93'

97

..w

19 29

37 47

.M

.9'

..36'

.so-

.88'

79 89

.98'

.99'

100

.96"

The numbers without strokes, other than 1 , are 11 , 13, 17, 19, 23, 29, 31 , 37, 41 , 43, 47, 53, 59, 61 , 67, 71, 73, 79, 83, 89, 97 which are therefore together with 2, 3, 5, 7 the primes less than 100. The method, while systematic, has obvious limitations.

The Distribution of the Primes

The primes are distributed irregularly amongst the integers but, nevertheless, we should like some measure of the irregularity. Two results which are of especial interest will be quoted without proofs (which are analytic rather than algebraic) . be the ascending sequence of primes. J. Bertrand surmised Let p1 , p 2 , (1845) and P.L. Chebyshev (1821-94 ) proved that •

•

•

Pn + l < 2pn ( n E N) .

For the second result we introduce a function commonly denoted by 7r (thus 7r i'i here not 3. 14159 . . . ) and defined for x E N by letting 1r ( x ) be the number of primes .:S x. Quick calculation shows that 7r(1 ) = 0, 1r (2) = 1 , 7r(3 ) = 7r(4 ) = 2, 7r(5) = 1r (6) = 3 , 7r(7) = 1r (8) = 7r(9) = 7r(10) = 4, etc. K.F. Gauss (1777-1855) formulated the remarkable conjecture that 1r (x ) log x ___;_--'-� --X

--+

1

as

x

--+

oo .


70

This was established independently by C.J. de la Valle- Poussin (1866-1962) and J. Hadamard (1865-1963) , the longevity of whom cannot solely be attributed to an interest in number theory.

Goldbach's Conjecture

This conjecture originated with C. Goldbach (1690-1764 ) and is to the effect that every even integer, other than 2, is the sum of two primes; for example, 4 = 2 + 2, 6 = 3 + 3, 8 = 5 + 3, 10 = 3 + 7 = 5 + 5 . The difficulty of tackling the conjecture seems to lie in the fact that primes are clearly concerned with factorization but not obviously with addition. Various allied results have been obtained but the conjecture itself remains to tantalize. After this brief excursion into aspects of number theory we shall develop the previously mentioned axiomatic treatment through a discussion of polynomials. Exercises 2. 5

1. 2. 3. 4. 5.

Write down formal proofs that

v'3 and v'6 are irrational.

Prove that V2 + v'3 is irrational.

The sum of two rational numbers i� rational. L� the sum of two irrational numbers always irrational? Write down a formal proof of Theorem

7.

Two primes p and q are said to be a 'prime pair' if q = p + 2, for example 17 and 19 is a prime pair. How many prime pairs are there between 1 and 100? (It is unknown whether or not there are infinitely many prime pairs. )

6.

Apply the Sieve of Eratosthenes to find the primes between

7. 8.

Calculate 7r (10n ) , n =

200.

100 and

1 , 2, . . . , 10. In how many ways may 144 be written as the sum of two primes?

3 In troduction to Rings

In the previous chapter we have seen that the integers possess a division algorithm and that from this division algorithm there may be derived a Euclidean Algorithm for finding the greatest common divisor of two given integers. 'Polynomials' share many properties in common with the integers, having a division algorithm and a corresponding Euclidean Algorithm. As our treatment of polynomials proceeds, initially somewhat informally, it will become apparent that we need to consider much more precisely the extent to which integers and polynomials share common features. In this way we shall be led to enunciate axioms for an algebraic system called a 'ring' and for a ring of a particular type called an 'integral domain' which incorporates some of the features common to integers and polynomials. Axioms in mathematics are never constructed arbitrarily but are designed to focus attention on significant aspects of the system or systems under considera tion. While axioms are famously present in Euclid's 'Elements' their modern extensive use stems from the mathematical work of the nineteenth century, a decisive influence in their use being that of the great D. Hilbert (1862-1943) . The concept of a ring was introduced by R. Dedekind (1831-1916) but the first set of axioms for a ring, although not quite equivalent to those in use today, was published in 1914 by A.H. Fraenkel ( 1891-1965).

71


72

3.1 Concept of

a

Polynomial

What does the term 'polynomial' mean to us? We think of algebraic expressions such as 2x + 1 or x 2 + 5x + 6 or 2x3 + x 2 - 2x - 1 etc. We speak of these as being 'polynomials in x ' and say that 2x + 1 has 'degree' 1, x 2 + 5x + 6 has 'degree' 2 and 2x3 + x 2 - 2x - 1 has 'degree' 3. We know how to add, subtract and multiply such polynomials:

(3x + 5) + ( 4x2 - 2x - 1) = 4x2 + x + 4, (2x2 - 6) - (x2 - 5x - 1) = x 2 + 5x - 5, (x2 + 2x + 4)(3x + 1) = x 2 (3x + 1) + 2x(3x + 1 ) + 4(3x + 1) = (3x3 + x 2 ) + (6x 2 + 2x) + (12x + 4) = 3x3 + 7x 2 + 14x + 4. As in the case of the integers we may perform long division, for example 3x + 1 does not divide 9x3 - 3x 2 + 6x + 4 but leaves a remainder when we employ long divi"lion as follows:

3x2 - 2x + �3 3x + 1)9x3 - 3x2 + 6x + 4 9x3 + 3x 2 - 6x2 + 6x - 6x2 - 2x 8x + 4 8x + �3 � 3

--

Thus we write

4 9x3 - 3x 2 + 6x + 4 8 3 2 = 3x - 2x + - + 3x + 1 3 3x + 1 '

-------

--

or, more usefully,

(

9x3 - 3x 2 + 6x + 4 = (3x + 1) 3x2 - 2x +

�) + �,

which is in a form to be expected from the existence of a division algorithm. For convenience we gather together in a more formal manner some of the aspects of polynomials mentioned above.

I ntrod uction to Rings

73

Defi nition 1 By a po]ynomial in z over Ql we understand an expression a( x) of the form m a(x) = a0 + a 1 x + . . . + amx where the so-called coefficients a0 , a 1 , . . . , am are rational numbers, and x is sometimes called an 'indeterminate'. If all of the coefficients are 0 the polynomial is said to be the zero polynomial, denoted by 0. If a( x) is not the zero polynomial and if am =/; 0 then a(x) is said to have degree m, briefly deg a(x) = m. If a(x) has degree m and am = 1 then a(x) is said to be monic. A polynomial which is either the zero polynomial or has degree 0 is said to be a constant polynomial and is therefore of the form a(x) = a0 (a0 E Ql). Polynomials of degrees 1 , 2 or 3 are sometimes called linear, quadratic or cubic respectively. Two polynomials a(x) and b(x) where m a (x) = a0 + a 1 x + . . . + amx , n b(x) = b o + b 1 x + . . . + b n x , are deemed to be equal if and only if m = n and a o = bo , a 1 = b b . . . , a n = b n . In particular a(x) = 0 if and only if all coefficients of a(x) are 0. The addition and multiplication of a(x) and b(x) are defined as a(x) + b(x)

=

a(x)b (x)

=

=

( a o + bo ) + (a 1 + b l )x + . . . ,

(ta ) (ta ) a r xr

b 8 a8

2 a o b o + (a o b l + a 1 bo )x + (a o b 2 + a 1 b 1 + a 2 b o )x m n + . . . + amb nx + .

The set of all polynomials in x over Q is denoted by Ql [x] . Similarly Z [x] , IR [x] , C [x] are defined by choosing the relevant coefficients for the polynomials.

Exa m ple 1 1 + ../2x 2 + v'3x5 is a polynomial in x over 1R of degree 5 . 5 + 2y + 3y2 is a quad ratic polynomial in y over Z ( and also over Q, lR and C) . 2 + � :? + z1 is a monic cubic polynomial in z over Q. A key property of polynomials in x over Q, say, is the possibility of substitut ing for the x. We may replace x in any equation involving polynomials in x by


74

any number, not necessarily in Q, and thereby obtain a valid equation involving that number. Thus if a(x) = ao + a 1 x + . . . + am xm EQ[x] and if c ElR we have a{c) = a0 + a 1 c + . . . + am cm ElR and if f(x), g(x ), h(x) EQ[x] are such that f(x)g(x) = h(x) then f(c)g(c) = h{c) .

Exa m ple 2 Let f(x) = x 2 + x + 1 , g(x) = 2x - 3, h(x) = 2x3 - x 2 - x - 3. Then we may verify that J(x)g(x) = h(x) . If we substitute .;2 for x then J{ v'2) = ( ../2) 2 + v'2 + 1 = 3 + v'2, g( ../2) = 2 ../2 - 3,

h( ../2) = 2 { ../2) 3 - ( ../2) 2 - v'2 - 3

= 3 ../2 - 5, and, as expected, J( ../2)g( v'2) = h{ v'2) . We shall investigate polynomials in Q[x] . With obvious and appropriate changes our result.� apply also to IR[x] and C[x] but not always to Z[x] .

Defi n ition 2 Let f(x) and g(x) be polynomials in Q[x] , g(x) =I 0. Then we say that g(x) divides f(x), or that g(x) is a divisor of f(x), if there exists a polynomial h(x) EQ[x] such that f(x) = g(x)h(x) . Notice that the zero polynomial 0 is divisible by any polynomial g(x) =I 0 since 0 = g(x) O and that any polynomial f(x) =I 0 has the trivial divisors c and cf(x) for any c EQ, c =I 0.

Exa m ples 3 1. 2x 2 + 2 EQ[x] has the divisors c and c{x 2 + 1 ) (c =1 0) since 2 2 2x + 2 = - c{x2 + 1 ) . c 2 2 3 2. 2x + 3x + 1 divides 2x + x - 2x - 1 since 2x3 + x 2 - 2x - 1 = {2x 2 + 3x + 1 ) (x - 1 ) .

I ntroduction to Rings

3.

75

In Q[x], x 2 - 2 has only the trivial divisors c, c (x 2 however there is the non-trivial factorization

- 2) , c E Q, c ::/; 0. In IR[x]

x2 - 2 = (x - v'2) (x + v'2). This example shows that it is important to know in which system the factor ization is to take place.

Lem ma 1 Let f(x), g(x) and h(x) be polynomials in Q[x] such that g(x ) and h(x) are non zero. Let g(x) divide f(x) and h(x ) divide g(x). Then h(x) divides f(x).

P roof This follows closely the proof of the corresponding lemma in Chapter 2.

Defi n ition 3 Let f(x) and g(x) be non-zero polynomials in Q[x] . A polynomial h(x) is a common divisor of f(x) and g(x) if h(x) divides f(x) and h(x) divides g(x) . A common divisor d(x) which is monic and which is such that d(x) is divisible by any common divisor of f(x) and g(x) is called the greatest common divisor (G.C.D.) of f(x) and g(x) .

Exa m ple 4 The quadratic polynomials 2 x 2 + 7x + 3 and . x + 21 smce 2 x 2 + 7x + 3

=

( 2 x + 1)(x + 3)

=

6x2 + x - 1 2

in

Q[x]

have G.C.D.

(x + D (x + 3),

( D (3x - 1 ) .

6x2 + x - 1 = (2 x + 1 ) (3x - 1) = 2 x + Exercises 3. 1

In each of the following find the G.C.D. of the pair of polynomials

1. x 2 + x - 2, x 2 + 3x + 2 . 2 . 6x 2 + x - 2, 15x 2 + 13x + 2.

76


3. x 3 + x 2 - x - 1, 4. x 3 + x 2 + x + 1 ,

2 x3 + 2x + x. 2x3 - x 2 + 2x - 1 .

3.2 Division and Euclidean Algorithms

As we remarked above, we shall prove results only for Q[x] but they apply with trivial modifications for IR[x] and C [x] .

Theorem 1 The Division Algorith m Let a(x) and b(x) be polynomials over Q[x] , b(x) being non-zero. Then there exist polynomials q(x) and r(x) with coefficients in Q satisfying the condition a(x)

where either r( x) degree of b(x) .

=

=

b(x)q(x) + r(x)

0 or if r(x) =/= 0 then the degree of r(x) is strictly less than the

Proof Let a(x) and b(x) be the polynomials of degree m and n respectively given by m a(x) = a o + a 1 x + . . . + amx , am =/= 0, n b n =/= 0. b(x) = bo + b 1 x + . . . + bn x , We first dispose of a trivial case , namely deg a( x) < deg b(x). We simply write a(x)

=

Ob(x) + r(x)

where q (x) = 0 and r(x) = a(x), which gives the result. We shall use the Principle of Induction by arguing in regard to deg a(x ) . If deg a(x) = 0 then a(x) = a0• If also deg a(x) = deg b(x) then b(x) = bo and so a(x)

where q =

=

b(x)q

:: On the other hand if deg a(x) < deg b(x) then we have the result .

by the trivial case above. Let now deg a(x) > 0. Suppose the result is true for all polynomials of degrees strictly less than the degree of a( x). We wish to prove the result for a( x). If deg a( x) < deg b( x) then again we have the trivial case above. Suppose therefore that deg b(x) � deg a(x). Then we aim to construct a polynomial of degree strictly less than deg a(x) and to apply the induction assumption to this poly nomial. Let

Introduction to Rings

77

f(x) = a(x) -

�: xm - n b(x)

(by convention

x 0 = 1) .

Then f(x) has degree strictly less than deg a(x ) since the terms in xm cancel as we see below:

�

(a o + a 1 x + . . . am x m ) - m (bo + b 1 x + . . . + bn x n )xm - n . n By the induction assumption there exist polynomials p(x) and r(x ) such that f(x) = b(x)p(x) + r(x) where either r(x ) = 0 or if r(x ) -::j; 0 then deg r(x) < deg b (x) . Hence b(x)p(x) + r(x) = f(x) = a(x ) - xm - n b(x) f(x)

=

�:

and so

a(x) where q(x) deg

=

p(x) +

=

(

b(x) p(x) +

�: xm - n) + r(x)

b(x) q(x) + r(x) m x m - n and either r(x) = O or if r(x) -::j; 0 then deg r(x) n

�

=

b(x) . This completes the proof.

< D

The following useful result comes as a consequence of the Division Algorithm.

Theo rem 2 The Remainder Theorem

f(x) E Q[x] and let c E Q. 1 . There exists q(x) E Q[x] such that f(x) = (x - c) q (x) + f(c) . 2. x - c divides f(x) if and only if f(c) = 0.

Let

P roof

1.

2.

By the Division Algorithm there exist q(x) and r(x) such that

f(x) = (x - c)q(x) + r(x) where either r(x) = 0 or ifr(x) -::j; 0 then deg r(x) < deg (x - c) = 1. In either case r(x) = r0 where r0 is a, possibly zero, constant. Then f(c) = (c - c)q(c) + ro = Oq(c) + r0 = r0 . 2 is an immediate consequence of 1. D


78

Exa m ple 5

f(x) = 2x 5 + x4 + 7x3 + 2x + 10 is divisible by x + 1 since

/ (-1 ) = 2(-1 ) 5 + (-1) 4 + 7(-1) 3 + 2(-1) + 10 = -2 + 1 - 7 - 2 + 10 = 0.

Exa m ples 6 For the polynomials

a(x)

and

b(x ) below we find q(x)

and

r(x)

such that

a(x) = b(x)q(x) + r(x) where r(x ) = 0 or if r(x) :F 0 then deg r(x) < deg b(x). We employ long division which is, in fact, the basis of the Division Algorithm.

1 . a(x) = x3 + 4x 2 + 5x + 7, b(x) = x + 1.

x + 1) x3 + 4x2 + 5x + 7 x 3 + x2 3x2 + 5x 3x2 + 3x 2x + 7 2x + 2 5 Thus

a(x) = b(x) q (x) + r(x) where q(x) = x 2 + 3x + 2, r(x) = 5.


79

2. a(x) = 2x3 + 7x2 + 1, b(x) = 3x + 2 1 x 2 + ll9 x - M7 2 3 3x + 2)2x3 + 7x 2 +1 2 2x 3 + !x 3 !l x 2 3 ll x 2 + M9 x 3 �x + 1 �x - � l!2 27 Thus

a(x ) = b(x) q(x) + r(x) where q(x)

=

ix2 + !jx - M, r(x) = Pr ·

We turn now to the matter of finding the G.C.D. of two polynomials. The Euclidean Algorithm used here to find the G.C.D. is exactly analogous to the Euclidean Algorithm used in the case of the integers, we merely have to modify some details appropriately. In these circumstances we shall content our selves therefore by illustrating the method by means of an example.

Exa m ples 7

1.

Suppose we wish to find the G.C.D. of

a(x) = 2x3 - 5x 2 - 2x - 3 and b(x) = x3 - x 2 - x - 15. Then as in the Euclidean Algorithm for finding the G.C.D. of two integers we use the Division Algorithm which in this example must be applied three times as follows:

2x3 - 5x - 2x - 3 = 2(x3 - x 2 - x - 15) + ( -3x 2 + 27) ,

(

)

x3 - x2 - x - 15 = (-3x2 + 27) - 31 x + 31 + (8x - 24 ) ,

(

)

1 9 -3x 2 + 27 = (8x - 24) - -8 x - -8

°


80

Thus we would expect that 8x - 24 would be 'almost' the G.C.D. However, it will be recalled that, for definiteness, we have chosen that the G .C.D . should be monic. But 8x - 24 = 8(x - 3) and so x - 3 is the required G .C.D. Furthermore by considering the above equations in reverse order we have

(

X

1 8(x - 3 ) = (x3 - x 2 - x - 15) - ( -3x2 + 27) - 3 + 3

)

( -i + �) = a(x) [- ( - i + D] + b(x) [1 + 2 ( - i + D] 2 = a( [ i - � ] + b( [- ; + �] ,

= b(x) - [a(x) - 2b(x)]

X

X

)

)

from which we have

1 1 x - 3 = a(x) 24 (x - 1 ) + b(x) 24 (-2x + 5).

2.

Suppose we are given the polynomials a(x) = 2x4 + 3x3 + 5x 2 + 6x + 2 and b(x) = x4 + 5x 2 + 6, and we wish to find the G.C.D. d(x) and to write d(x) in the form

d(x) = a(x)f(x) + b(x)g(x) for suitable polynomials f(x) and g(x ) . We have

a(x) = 2b(x) + (3x3 - 5x2 + 6x - 10) ,

( ) ( )( ) X

)

104 52 2 + b (x ) = ( 3x3 - 5x 2 + 6x - 1 0) - + -5 + -x 3 9 9 9 ' 104 27 45 52 x2 + 3x3 - 5x 2 + 6x - 10 = - . 9 9 -x 52 - 52

(

Then, since

52 2 104 52 2 -g x + g = -g (x + 2), the G .C.D. is

x 2 + 2.

Also

81


(

x 2 + 2 = � 52 x 2 + 104 52 9 9 =

)

:2 [ b(x) - (3x3 - 5x2 + 6x - 10) (� + �)]

+

= 521 [9b(x) - (3x 3 - 5x 2 + 6x - 10) (3x 5)]

= 1 [9b(x) - (a(x ) - 2b(x) ) (3x + 5)] 52 + 19) -(3x + 5) = 52 a ( ) + (6x 52 b( ) X

Thus

+

X .

+

- (3x 5) g = 6x 19 52 ' ( ) 52 is a possible solution for f(x) and g(x). (Recall that we know that f(x) g(x) are not uniquely determined. ) ! (X ) =

X

and

Exercises 3. 2

1.

For the given polynomials a(x) and b(x) in Q [x] , find q(x) and r(x) in

Q[x] such that

a(x ) = b(x)q(x) + r(x ) where either r(x) = 0 or if r(x) # 0 then deg r(x) < deg b(x). ( i ) a(x) = x 3 + 3x2 + 1, b(x) = x4 + 1 . ( ii ) a(x) = 3x3 + 3x 2 + 2x + 1 , b(x) = 3x 2 + 2. ( iii) a(x) = 2x 2 + 5x + 1 , b(x) = 5x - 1. ( iv ) a(x) = x3 + 4x 2 4x 1 , b(x) = x + 3. (v ) a(x) = x 4 + 3x 2 1 , b(x) = 2x 2 + 1 . 2. For the given polynomials a(x) and b(x) in Q[x], find the G.C.D. d(x) and write d( x) in the form d(x) = a(x)f(x) + b(x)g(x) for suitable polynomials f(x) and g(x) in Q[x]. (i ) a(x) = x4 + x 3 x 1 , b(x) = x 2 + x 1. ( ii ) a(x) = 2x3 + 10x 2 2x 10, b(x) = x3 - 2x 2 x - 2. ( iii ) a(x) = x4 - 4x 3 2x - 4, b(x) = x 3 + 2. ( iv) a(x) = x 3 + 5x2 7x 2, b(x) = x 3 + 2x 2 - 2x - 1 .

+ + +

++ + + + + +

+

+


82

3.3 Axioms and Rings

From our deliberations we have seen that the system of polynomials behaves in a similar manner to the system of the integers; it would therefore seem advisable to study further whatever aspects these two systems have in common. In order to understand better what apparently similar systems have in common it is often convenient to lay down a common set of axioms which is satisfied by the systems. A set of axioms is not invented arbitrarily but is constructed to reproduce aspects of common and significant interest across the systems. Once the axioms are laid down the axiomatic system then delineated takes on, as it were, a life of its own and may then be explored both for its own sake and to achieve a better overall understanding of previously considered systems. In regard to the integers and polynomials we observe that there are in each system two crucial operations, one is the addition of two numbers or polynomials and the other is the multiplication of two numbers or polynomials. These opera tions are said to be binary since they involve two numbers or polynomials. The operations satisfy certain conditions which are tacitly assumed in elementary treatments but which must be made explicit when we pass to an axiomatic treat ment. We shall give appropriate axioms for the system, later defined to be a ring, in which both of these operations occur. For convenience of reference and in com mitting the axioms to memory, we give the definition of a ring in a succinct form but then follow the definition with a commentary upon the axioms of the defini ition. Finally, we conclude with the definition of an integral domain which is a ring of a particular kind and which brings together many of the more obvious features of integers and polynomials.

Defi nition 4 Let R be a non-empty set in which there are defined two binary operations called addition and multiplication. For a, b E R the outcome of the addition of a and b, called the sum of a and b, is denoted by a + b ('a plus b') and the outcome of the multiplication of a by b, called the product of a by b, is denoted by the simple juxtaposition ab ('a times b') . Then R is called a ring if the following axioms hold. 1.

Axioms of Addition

1 .1 For all a, b E R, a + b E R (closure) . 1 . 2 For all a, b, c E R , (a + b) + c = a + (b + c) (associativity) . 1 .3 There exists a distinguished element denoted by 0 such that for all a E R, a + 0 = 0 + a = a (0 is 'zero') .

83

I ntroduction to Rings

a E R there exists an element denoted by -a E R such that a + (-a) = (-a) + a = 0 (the inverse of addition) . For all a, b E R, a + b = b + a (commutativity) .

1 .4 For any 1 .5 2.

AxioiDB of Multiplication

2.1 For all a, b E R, ab E R (closure) . 2.2 For all a, b, c E R, (ab)c = a(bc) (associativity) . 3 . AxioiDB of Distributivity

3.1 3.2

For all a, b, c E R, a(b + c) = ab + ac (distributivity) . For all a, b, c E R, (a + b)c = ac + be (distributivity) .

Com menta ry 1.1 1 .2

1.3

The closure o f addition is t o ensure that the sum of a and b also belongs to R. It would be a bizarre system if this fact were not to be so . The associativity of addition enables us to dispense with brackets in addition. Thus we may write simply a + b + c where the meaning is (a + b) + c or, equally, a + (b + c) . This axiom only states that a zero exists. On the face of it there could be more than one such zero. However, if 0' were to denote a second zero we would have for all a E R,

a+0 = 0+a = � a+� = �+ a = � In particular we would have 0' + 0 = 0 + 0' = 0', 0 + � = 0' + 0 = 0 from which 0' = 0. Henceforth we may speak of the zero of R. 1 .4

Now that we know that the zero element is unique, by carefully applying the axioms we may establish that for each a E R the inverse of addition -a is uniquely determined by a; if we suppose that we have b E R such that a + b = b + a = 0 then we may show that b = -a as follows. By 1 . 2 (associativity)

b + (a + (-a)) = (b + a) + (-a). But

b + (a + (-a)) = b + O = b


84

and

1.5

2.1

2.2 3.1/3.2

(b + a) + (-a) = 0 + (-a) = (-a) , from which we have b = -a. Henceforth we may speak of the inverse of addition -a for given a E R. Notice that as a + (-a) = (-a) + a = 0 we have a = - (-a) . This axiom states that the sum o f a and b i s independent o f the order in which a and b are added to one another; a and b are said to 'commute' .

(The words 'commutativity' and 'commute' stem from electrical engineering in which a commutator is an apparatus for reversing the current. ) The closure of multiplication is t o ensure that the product of a and b belongs to R (see comment on 1.1). Note, however, that we must main tain the order of a followed by b since, in general, ab and ba are not neces sarily equal. The associativity of multiplication enables us to dispense with brackets in multiplication. Thus abc is defined unambiguously, being either (ab)c or a(bc) . These axioms give conditions linking addition and multiplication. (The reader may recall the remarks in Chapter 2 on Goldbach's Conjecture.)

In these axioms there are certain tacit assumptions in regard to the order in which the bracketing and the two binary operations are to be considered. Thus the axiom a(b + c) = ab + ac means that, on the left-hand side, we add b and c within the brackets to give b + c and then we perform a multiplication by a whereas, on the right-hand side, we multiply a and b and also multiply a and c before performing the addition of ab and ac. These remarks may see m superfluous but only because we are so accustomed to the particular order of pre cedence of brackets and operations (an order which in years gone by was encap sulated in the mnemonic BODMAS = bracketsjofjdivisionjmultiplication/ addition/subtraction) . For example, by a + be, we understand that b and c are first multiplied to give be and then a and be are added. A convention also arises in regard to the minus sign ' - ' . In writing a - b we are actually simply writing a followed by the additive inverse of b; however by a b we understand that in fact we intend a + (-b) , the + sign being implied but normally omitted. If we now relate these axioms either to the set of integers Z or to the set of poly nomials Q[x] with the usual addition and multiplication operations, we should observe that the axioms are clearly satisfied; for example, the integers satisfy 1 .5 since we have assumed from early childhood that the sum of two integers was the same irrespective of the order in which they were added. Nevertheless -

85


the reader should convince himself or herself, certainly for the integers and pos sibly for polynomials, that the other axioms are equally obviously satisfied. We shall henceforth speak of the 'ring of integers' and of the 'ring of polynomials' (although this latter terminology is still a trifle imprecise) .

Exa m ple 8 Rings arise in many ways. Below we consider a ring the elements of which are mappings. Let X be a non-empty set. Let R be the set of mappings from X to JR. We introduce an addition and multiplication, denoted by ' . ', into R as follows. Let f : X -+ JR and g : X -+ JR. Define the sum f + g and the product f. g as mappings from X to JR by, (! + g) (x) = f(x) + g(x)

(x

E X) ,

(f. g) (x) = f(x)g(x)

(x

E X) .

We note that on the right-hand side of the first equation the plus sign + refers to addition in JR and on the right-hand side of the second equation the product is in JR. Consequently the sum and product of the mappings f and g are well-defined. We claim that R is a ring with the given definitions of sum and product . We require to verify all the axioms for a ring. We use the same numbering system as above. 1 . 1 Certainly, by definition, R is closed under addition. 1 . 2 Let J, g, h E R. We want to show that (! + g) + h = f + (g + h) . Now for

xEX

[(! + g) + h] (x) = (! + g) (x) + h(x)

( defmition)

= [f(x) + g(x)] + h(x)

(defmition)

= f(x) + [g(x) + h(x)]

( associativity in JR)

= f(x) + (g + h) (x)

(definition)

= [/ + (g + h)] (x)

(definition) .

Hence, by the condition for the equality of mappings (! + g) + h

=

f + (g + h) .


86

1 .3

The zero of R is given by the mapping 0 given by O(x ) all f E R and x E X

=

0 (x E X) since for

( ! + O) (x) = f(x) + O(x) (definition) (definition of 0 E R) = f(x) + 0 = f(x) (0 E IR) = 0 + f(x) (0 E IR) = O(x) + f(x) (definition of 0 E R) = (0 + f)(x) (definition). Hence, again by the condition for the equality of mappings,

f + O = f = O + f. 1.4

The additive inverse of f E R is given by - f where (- f) (x) -f(x)(x E X). This is well-defined since on the right-hand side we have the negative of a real number. We must verify that -/ + f = 0 f + (-f) . Now for x E X =

=

(-f + f)(x) = ( - f)(x) + f(x) = -f(x) + f(x) =0 = O(x)

(definition) (definition of

- f)

(definition of 0

E R) .

Thus

-! + ! = 0. Similarly f + (-f)

1.5

Let

=

o.

f, g E R. Then for x E X ( ! + g) (x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x)

(definition) (commutativity of addition in IR) (definition) .

Thus

f+g=g+f 2.1

and so addition in R is commutative. By definition R is closed under multiplication.

87


2.2 Let J, g, h E R. We want to show that

(f. g) .h = f.(g.h) . Now for x E X

[(f.g).h] (x) = ( f. g) (x)h(x) = [f(x )g(x )]h(x) = f(x) [g(x)h(x)] = f(x ) (g.h) (x) = [f.(g.h)] (x)

(definition) (definition) (associativity in IR) (definition) (definition) .

Thus 3.1

(f.g) .h = f. (g. h). Let f, g, h E R . We prove that f. (g + h ) = f.g + f.h. Let x E X. Then (definition) [f.(g + h)] (x) = f(x) (g + h) (x) (definition) = f(x) [g(x ) + h(x)] = f(x)g(x ) + f(x)h(x) (distributivity in IR) = (f.g) (x) + (f.h) (x) (definition) (definition) . = [( !.g) + (f.h)] (x) Thus

f. (g + h) = f.g + f. h. The other distributivity axiom is proved similarly. We have now shown that R is a ring. We may remark, partly as an aside, that as the ring of integers Z is an example of a system that satisfies the axioms of a ring then these axioms are necessarily self-consistent. We shall not concern ourselves unduly in determining to what extent the axioms of this, or of any other, system are independent of one another. We have not, however, adequately considered all of the distinctive features of the ring of integers and of the ring of polynomials. In Z there are three properties which are so familiar that ordinarily we do not make specific reference to them: there is a number 1 such that 1a = a1 = a for all a E Z, the order in which we multiply integers is irrelevant (ab = ba for all a, b E Z) and the product of two non-zero integers is non-zero. We formalize these concepts as follows.

Defi nition 5 Let

R be a ring.

1 . An element 1 E R such that 1a = a1 = a for all a E R is called an identity element or identity or unity of R.


88

2. Let a and b be elements of R. Then a and b are said to commute if the products ab and ba are equal. If any two elements of R commute then R is said to be commutative. 3. Let 0 be the zero element of R. Let a and b be elements of R such that ab = 0. Then a and b are called divisors of zero and if both a =I 0 and b =I 0 then a and b are called proper divisors of zero. We have previously shown above that the zero element of a ring is unique. By an entirely similar argument we may show that if a ring has an identity element (which need not be the case ) then it has a unique identity element .

Defi nition 6 A commutative ring with an identity called an integral domain.

1 (1 =I 0)

and no proper divisors of zero is

( Note that the use here of the word 'domain' in the phrase 'integral domain' no connection with the 'domain' we encountered previously in the mapping of one set into another. Note also that some texts, especially of an advanced nature, do not insist upon 'commutativity' as a necessary condition for an integral domain. ) has

Exa mple 9 Under the usual operations of addition and multiplication the following are all integral domains, Z, Q, .IR, C. As part of this introduction to rings we require to prove some elementary results, results which in themselves are obvious for the integers but which require proof in an axiomatic context.

Theorem 3 Let

R be a ring. Then the following hold.

If a + b = a + c (a, b, c E R) then b = If a + b = c + b (a, b, c E R ) then a = 2. For all a E R, aO = Oa = 0.

1.

c.

c.

89

I ntroduction to Rin gs

3. For all a, b E R, a( -b) = ( -a)b = - (ab) , (-a) ( -b) = ab. If R has an identity 1 then ( - 1 ) ( - 1 ) = 1 . 4. For all a, b E R, a(b - c) = ab - ac , (a - b)c = ac - be.

P roof 1 . From a + b = ativity

a+c

we have

(-a) + (a + b) = (-a) + (a + c) .

By associ

((-a) + a) + b = ((-a) + a) + c and so

0 + b = 0 + c and b = c. The second assertion is similarly proved. 2. aO + aO = a(O + 0) = aO = aO + 0 and so aO = 0. Similarly Oa = 0. 3. We have 0 = a0 = a(b + (-b)) = ab + a(-b) and so ab + -(ab) = O = ab + a(-b) from which - (ab) = a(-b). Similarly we prove -(ab) = (-a)b. Let a' = -a. Then, as we observed in the Commentary above, -a' = - (-a) = a. Hence

(-a) (-b) = a'(-b) = (-a')b = ab. If 1 E R then ( - 1 ) ( - 1 ) = 1 1 = 1 . 4 . a(b - c) = a(b + (-c) ) = ab + a( -c) = ab + - ( ac ) = ab - ac . Similarly (a - b)c ac - be. =

D

Exa m ple 10 Let D be an integral domain. Let a, b, c E C, c =I 0, be such that ac = be. We want to prove that a = b . We have (a - b)c = ac - be = 0 and since D is an inte gral domain either a - b = 0 or c = 0. But c =I 0 and so a - b = 0 and a = b. We have considered polynomial rings in x in which the polynomials have coefficients from Z, Q or IR, yielding Z [x] , Q[x] or IR[x] respectively. But similarly we may have a polynomial ring in x in which the coefficients belong to an integral domain D yielding D[x] . We have the following result.

Theore m 4 Let D be an integral domain. Then the polynomial ring domain.

D [x]

is an integral


90

P roof Certainly D[x] is a commutative ring with the identity of D as the identity of D[x] . Thus we have to show that D[x] has no proper divisors of zero. Let

f(x) = ao + a 1 x + . . . + am xm (am =I 0), g(x) = bo + b 1 x + . . . + b n xn (b n =/; 0) , be two non-zero polynomials in D[x]. Then f(x)g(x) is a polynomial of highest term am bn xm + n . But since D is an integral domain a m bn =I 0. Thus f(x)g(x) =I 0 and D[x] is an integral domain. D Corolla ry Let

f(x)

and g(x) be non-zero polynomials in D[x]. Then deg

f(x)g(x) = deg f(x) + deg g(x) .

In many situations we have to consider a ring which is itself contained within a larger ring, for example Z is a ring within the ring Q and Q is a ring within the ring JR. We sometimes have to consider subsets of a ring which may be rings, or 'subrings' as they will be called, with respect to the operations of the ring.

Defi nition 7 Let S be a non-empty subset of the ring R which is also a ring under the addition and multiplication in R. Then S is called a subring of R. This definition is too descriptive to be really useful. We therefore derive a criterion which is more readily applicable.

Theorem 5 S u bring Criterion A non-empty subset S of a ring R is a subring of R if and only if the following axioms of a ring are satisfied.

1.1 S is closed under addition. 1.4 For all a E S, -a E S. 2.1 S is closed under multiplication. Proof Certainly if S is a subring the axioms quoted above must be satisfied.


91

Suppose now that the axioms above are satisfied by S. Then certainly we have 0 = a + ( -a) E S. We claim that the remaining axioms for a ring are auto matically satisfied in S because they are satisfied in R. Consider, for the sake of argument, the associativity of addition. We have (a + b) + c = a + ( b + c) for all a, b, c E R and therefore for all a, b, c E S as S � R. The other remaining axioms for a ring are similarly satisfied since S�R. Hence we infer that S is a subring. D

Exa m ple 1 1 It may be shown that R = M2 (Q) , the set of 2 x 2 matrices over Q , is a ring under the usual matrix operations. Suppose we want to show that

( � :) (: :) � ( : ) ( ) ( ) ( )( ) ( )

is a subring of R. We apply the subring criterion. We have, for x, y, z, u, v, w E Q, x

+

-

y+v

z+w

X y 0

z

V

X

Y

U

O

z

O w

(

u

=

=

-X

-y

0

-z

E S,

XU XV + YW 0

zw

E S,

E S.

)

where we are using the closure etc. of addition and multiplication in Q to ensure x+y y+v . . . that, for example, 15 mdeed m S. 0 z+w We conclude that S is a subring of R. Exercises 3. 3

1 . Prove that if a ring has an identity then it has only one identity.

2. Verify that Z, Q, lR and C with the usual operations of addition and multiplication are integral domains. 3. Show that the set R of one element 0, R = {0} , is a ring if addition and multiplication are given by 0 + 0 = 0 and 00 = 0 respectively.

4. Let R be a set of two elements, 0 and 1 , R = {0, 1 } , in which addition and multiplication are given by the following tables:

92


Product 0 0 0 1 0

1 0 1

Prove that R is an integral domain. 5. Let 2Z denote the set of even integers with the usual operations of addition and multiplication. Prove that 2Z is a ring. Is 2Z an integral domain?

6.

Let R be a ring. (i) Let a, b, c E R be such that a + b = c + b. Prove that a = c. (ii) Let a, b E R. Prove that - (ab) = ( -a)b.

7.

Let M2 (Q) be the set of 2 x 2 matrices over Q. Under the usual opera tions prove that M2 (Q) is a ring.

8.

Let R1 and � be subrings of a ring R. Prove that R1 n� is a subring of R. Can you generalize this result?

4

In troduction to Groups

In Chapter 3, we began to consider an axiomatic treatment of certain algebraic concepts and operations; in particular in the case of a ring we needed to consider two binary operations. Since simplicity usually has some advantages we shall in this chapter consider only one binary operation. We begin therefore with the notion of a 'semigroup' from which we shall be led to consider a 'group', one of the most fundamental structures in modern mathematics. The evolution of the concept of an abstract group owes much to the labours of many mathematicians of whom only a few will be mentioned here. The origins of the concept may be traced from the work of P. Ruffini ( 1765-1 822 ) and E. Galois ( 1 81 1-32 ) through to that of L. Kronecker who developed ideas for what we now call an Abelian group ( 'Abelian' after N.H. Abel, 1 802-29 ) . The abstract concept of a finite group was first formulated in 1854 by A. Cayley ( 1821-95 ) but its sig nificance was not properly appreciated until 1 878. W. von Dyck ( 1 856-1934 ) and H. Weber ( 1 842-1913 ) were influential in the development of group theory, the latter giving the first definition of an infinite group in 1 893. These remarks can give at most a brief indication of the history of the evolution of the concept of an abstract group which "emerged not from a single act or the creation of a single scholar but was rather the outcome of a process of abstraction with certain dis cernible steps. These were not essentially logical in nature; rather they repre sented the variable extent to which the process of abstraction was carried through. " ( quotation from The Genesis of the Abstract Group Concept by H. Wussing, translated MIT Press, 1984, page 234 ) . It is perhaps psychologically salutary to reflect that the concepts of a group and, indeed, of other parts of mathematics, have not always evolved directly but frequently by fits and starts. 93

94


4.1 Semigroups

Defin ition 1 Let S be a non-empty set on which there is defined a binary operation denoted by * · For a, b E S the outcome of the operation between a and b is denoted by a * b. Then S is called a semigroup if the following axioms hold. (i) For all (ii) For all

a, b E S, a * b E S (closure) . a, b, c E S, (a * b) * c = a * ( b * c)

(associativity) .

Exa mple 1 Z under the operation of + is a semigroup since for all a, b E Z, a + b E Z and for all a, b, c E Z, (a + b) + c = a + (b + c). In future, in general considerations of semigroups, we shall omit the symbol for the binary operation unless an especial need arises for it to be shown. We shall use, as far as possible, a multiplicative notation in which the outcome of the binary operation between two elements is shown by the simple juxtaposition of the elements; we shall speak of the multiplication (sometimes addition) of these elements. Semigroups are common in mathematics and easy to create as the following examples show.

Exa mples 2 1 . Let S be a set of one element, S =

{a}, say. Define aa = a.

Then S becomes a semigroup. 2. Let S be a non-empty set. Define a binary operation on S by

ab = b for all a, b E S. We claim that S is a semigroup under this operation. Certainly we have for all a, b E S, ab = b E S. We also have for all a, b, c E S, (ab)c = be = c and a(bc) = ac = c and so (ab)c = a(bc) . Thus S is a semigroup. 3. Let S be a non-empty set and let operation on S by

s

be a fixed element of S. Define a binary

ab = s for all a, b E S.

95

Introduction to Groups

Then certainly for all a, b E S, ab E S and then we have for all a, b, c E S, ( ab)c = sc = s and a(bc) = as = s from which ( ab)c = a(bc) . Thus S is a semigroup. None of the examples above may be of great intrinsic interest but nevertheless in each example the axioms of a semigroup are satisfied. After the next lemma we consider a particular type of semigroup.

Lem ma 1 Let S be a semigroup. Let e, f E S be such that ex = Then e = f and so ex = xe = x for all x E S.

x

and x f = x for all x E S.

Proof By the condition on e we have, in particular, e f = f . Similarly e f = e. Thus = f and the rest is clear. 0

e

Defi nition 2 A semigroup S is called a monoid if there exists e E S such that ex all x E S. e is said to be the identity element or identity of S.

= xe = x

for

By Lemma 1 above a semigroup has at most one identity element.

Associativity and Index Law

Let a, b, c, d , . . . be elements of a semigroup S. Expressions such as ( ab) ( cd) , a((bc)d) , a(b(cd) ) are formally distinct but are in fact equal since, by associativity, (ab) ( cd ) = a(b(cd ) )

=

a((bc) d ) ) .

Thus we could write, without uncertainty of meaning, simply abed. Further we see that in more complicated expressions of elements and brackets we may simi larly dispense with the bracketing provided the elements appear in the same order in each of the expressions.

Gro ups,

96

Rings and Fields

Consequently we may now define the powers of a

E S unambiguously as a1 = a, a2 = aa, a3 = aaa, . . . , an = aa . . . a, where a appears precisely n times. It follows that for m, n = 1 , 2 , . . . am an = (aa . . . a) (aa . . . a) = (aa . . . a) = am+ n where within the brackets we have a appearing m times, n times and m + n times respectively. If further S is a monoid with identity e we may define a0 = e for all a E S and

then

am an = am+ n ( m, n = O , 1 , . . . ) . Exa m ples 3 1 . Let X be a non-empty set and let S(X) be the set of mappings of X into X. We have already met the circle composition of mappings of X into X, namely for J, g E S(X) we defined f o g by

( ! o g) (x) = f(g(x)) (x E X). Furthermore we showed (Theorem 7, Chapter 1 ) that for J, g, h 0

0

0

E S(X) then

0

( ! g) h = f (g h). The set S(X ) of mappings o f X into X is therefore a semigroup under the circle composition of mappings. S(X) has an identity element given by the mapping t where t (x) = x(x E X) since for all f E S(X ) we have (t o f) (x) = t(f(x)) = f(x) = f(t(x)) = ( ! o t) (x) (x E X) and so

t o f = f = f o t. 2.

Let a binary operation * be introduced into Z by defining for all

a, b E Z

a * b = a + b + ab. Then we claim that Z with this operation is a monoid. We certainly have a * b E Z for all a, b E Z. Let a, b, c E Z, then

(a * b) * c = (a + b + ab) * c = (a + b + ab) + c + (a + b + ab)c = a + b + ab + c + ac + be + abc = a + b + c + ab + ac + be + abc,

97

I ntroduction to Groups

a * (b * c) = a * (b + c + be) = a + (b + c + be) + a(b + c + be) = a + b + c + be + ab + ac + abc = a + b + c + ab + ac + be + abc, from which we have Now for all

a E Z we have 0 * a = 0 + a + Oa = 0 + a + 0 = a, a * 0 = a + 0 + aO = a + 0 + 0 = a.

Thus 0 plays the role of an identity element when Hence Z is a monoid Wlder * ·

* is the binary operation.

3. The set M2 {Q ) of 2 x 2 matrices with entries from Q is a monoid Wlder the usual matrix addition, namely

( Xt Yt ) + ( X2 Y2 ) ( Xt + x2 �

�

�

�

=

�+�

Y1 + Y2 �+�

).

The sum of two 2 x 2 matrices with entries from Q is again a 2 x 2 matrix over Q and so the operation of addition is closed. The associativity of addition is a well known property of matrix addition. Further

M2 {Q ) is a monoid. The set M2 {Q ) of 2 x 2 matrices with entries from Q is a monoid Wlder the

is the identity element for addition and so 4.

usual matrix multiplication, namely

( x l Y t ) + ( x 2 Y2 ) = ( x 1 x 2 + Y 1 � •

Zt t l

Z2 t2

)

X 1 Y2 + Y1 t2 · ZtX 2 + t l z2 Zt Y2 + t l t2

Closure is obvious and matrix multiplication is known to be associative. The identity element for multiplication is

We conclude this section with two results which will be important in our study of groups.


98

Lem ma 2 Let S be a monoid with identity element e. Let a E S. Let a' , a" E S be such that a' a = e = aa " . Then a' = a " .

Proof The result depends crucially on the associativity axiom. We have ( a I a ) a II = ea II = a II , a ' ( aa" ) = aI e = a I and so a I = aII .

D

Theorem 1 Let S be a semigroup with the following two conditions. 1 . There exists e E S such that ea = a for all a E S. 2. For each a E S there exists a' E S such that a' a = e. Then S is a monoid with identity e and a ' a = aa ' = e.

Proof We show first that e is the identity of S. Let a E S. Then, by twice applying condition 2, there exists a ' E S such that a' a = e and there exists a " E S such that a" a ' = e. But by condition 1 and associativity we have ( ) = ( a" a' ) ( ) = a" ( a' ( ae )) = a" (( a' a ) e ) ae ae = e ae = a " ( ee ) = a "e = a " ( a' a )

=

( a" a' ) a = ea = a.

Consequently e is the identity of S. But then a ' a Lemma 1, that a = a" and so we complete the proof.

=

e = a " a ' implies, by D

We remark that in the two conditions of this theorem e 'acts' on the left of any a E S to give ea = a and for given a E S a' 'acts' on the left to give a ' a = e. If

we were to replace these left-handed conditions by corresponding right-handed conditions then the conclusion of the theorem would still hold. If, however, one condition is on the left and the other condition is on the right then the con clusion of the theorem need not hold. To see this last assertion we consider the following example.

99


Exa m ple 4 Let S be the two-element semigroup, S = { e, !} with multiplication, ee = e, e f = f, ff = f , fe = e. S is a semigroup of the type described in Examples 2 no. 2, where e acts on the left and satisfies condition 1 of the last theorem. We also have ee = e and fe = e so that , from the right we would have, e' = e and !' = e. Nevertheless e is not an identify element of S. Exercises 4. 1

1 . A binary operation * is defined on Z by a * b = ba that , under this operation, Z is a monoid.

(a, b E Z) .

Prove

2. A mapping !a.,b of lR into lR is defined for a, b E lR by !a.,b : x --+ a + bx (x E IR). Prove that the set of all such mappings is a monoid under the circle composition of mappings. Hint: Prove that 0

fc,d fa.,b = fc+a.d,bd· 3. Let t be a given integer. On Z an operation * is defined by

a * b = a + b + tab (a, b E Z) . Prove that Z under this operation is a monoid. 4. Let S be a semigroup with the following two conditions. (i) There exists e E S such that ae = a for all a E S. (ii) For all a E S there exists a' E S such that aa' = e.

a' a = e. (Hard) Let S be a finite semigroup and let a E S. By considering the subset {am I n E N} prove that there exist p, q E N, 2q < p such that aP = aq. If b = ap - q prove that b2 = b.

Prove that e is the identify of S and that 5.

4.2 Finite and Infinite Groups

Definition 3 Let G be a non-empty set on which there is defined a binary operation so that the outcome of the operation between a and b (a, b E G) is denoted by ab. Then G is called a group if the following axioms hold.


100

1. 2. 3. 4.

For all a, b E G, ab E G (closure) . For all a, b, c E G, (ab)c = a(bc) (associativity) . There exists e E G such that for all a E G , ea = a (existence o f identity) . For each a E G there exists an element denoted by a- 1 such that a- 1 a = e (existence of inverse) .

By Theorem 1 e is the identity of G and, for each a E G, we have a - l uniquely determined by a and such that, a- 1 a = aa- 1 = e, a- 1 is called the inverse of a. Notice that from a- 1 a =

aa- 1 = e we may deduce that (a- 1 ) - 1 = a (a E G).

Partly for later convenience we give two alternative criteria by which a semi group is a group.

Theorem 2 L

9 S be a semigroup. Suppose that for any a, b E S there exist x, y E S such that

ax = b and ya = b. Then S is a group. Proof

Let a E S. By assumption there exists e E S such that ea = a. We cannot imme diately assert that e is the identity for S since, on the face of it, this e depends on the particular element a. We consider an arbitrary c E S. Then again by assump tion there exists u E S such that au = c. Hence ec = e(au) = (ea)u = au = c. Furthermore by assumption there exists c' E S such that c' c = e. Hence, finally , S is a group. D

Theorem 3 Let S be a finite semigroup in which cancellation exists, that is if a, b, x E S and ax = bx then a = b and if a, b, y E S and ya = yb then a = b. Then S is a finite group.

P roof Let a, b E S. We show that there exists x E S such that ax = b. Let S have n dis tinct elements s 1 , s2 , . . . , sn . Consider as 1 , as2 , . . . , asn . These elements of S are also distinct since asi = asi implies si = si and so i j. Thus

=

{as l , as2 , . . . asn } = {s 1 , s2 , . . . , sn } ·


101

Hence, as b is one of the s 1 , s2 , . . . , sn , there exists sk such that ask = b. Similarly we may find y E S such that ya = b. Thus, by the previous result, S is a group. D Since the group concept is widely encountered in mathematics we shall give a varied selection of examples, not all of which need to be fully comprehended on a first reading. For convenience in this chapter, and elsewhere, we shall use G to denote a group and e t o denote the identity of G without further explanation.

Exam ples 5 1 . The set of non-zero elements of Q, that is Q \ {0} , is a group under the usual multiplication. To verify that Q \ {0} is a group we note the following immediate facts. If a, b E Q \ {0} then ab E Q \ {0} . Multiplication in Q, and so in Q \ {0} , is associative. 1 is the identity as 1a = a1 = a for all a E Q \ {0} . If finally a E Q \ {0} then a has a reciprocal � such that a a=a = 1 and so a- 1 = .

G) (�)

�

In a similar manner we may prove that IR \ {0} and C \ {0} are groups. 2. The set of integers Z under multiplication is not a group. We know that this set is a monoid under multiplication but, for example, the number 2 does not have an inverse in Z as ! E z. 3. The set of integers Z is a group under the usual addition. Here we have to insert the specific symbol + for the binary operation but the group axioms are easily verified. Thus if a, b E Z then a + b E Z and for all a, b, c E Z (a + b) + c = a + (b + c) . The element 0 plays the role of the identity element for addition since O+a= a+O = a for all a E

Z and the inverse of a E Z with respect to addition is -a, since (-a) + a = a + (-a) = 0.

We now gather together some elementary results on groups.

Theo rem 4 Let G be a group. 1 . Cancellation Let a, b, x E G be such that ax = bx. Then a = b. (Similarly ya = yb (y E G ) implies a = b. )


102

2. Unique Solution of Equation Let a, b E G. Then the equation ax = b has the unique solution x = a - 1 b. (Similarly ya = b has the unique solut ion y = ba - l . )

P roof 1 . There exists x - 1 E G such that xx - 1 = e. Then (ax)x - 1 = (bx)x - 1 and so a(xx - 1 ) = b(xx- 1 ) from which ae = be and a = b. 2. Certainly a- 1 b is a solution since a(a - 1 b) = (aa- 1 ) b = eb = b. On the other hand, ax = b implies that a - \ ax) = a- 1 b from which we conclude that (a- 1 a)x = a- 1 b and so x = ex = a- 1 b. D

Defi nition 4 A group G is said to be finite if the set G is a finite set, otherwise the group is said to be infinite. If G is finite the cardinality of G (which is the number of elements in G) is called the order of G, written I G I , G then has finite order. An infinite group is often said to have infinite order.

Exa m ple 6 Z under addition and Q \ { 0} under multiplication are examples of groups of infinite order. In the case of a finite group of very small order we may conveniently display the binary operation in the tabular form of a so-called 'Cayley table'. Thus if G has order n, G = { a 1 , a 2 , , a n } say, then we write the table as • • •

al

a2

ai

an

al a2

a l al a2 a l

a l a2 a2 a2

a l ai a 2 ai

a l an a2 a n

ai

aia l

aia 2

aiai

a ia n

an

ana l

a na 2

a n ai

a na n

where the entry in the (i, j)th position is aiai . By the cancellation result in Theorem 4 the elements aia 1 , ai a 2 , . . . , ai a n are distinct and so any row of the Cayley table must have distinct elements in the row. Similarly any column of the table must consist of distinct elements.

103


Exa m ples 7 1 . Let G have order 1 . Then G = { e } and the Cayley table is

As a concrete example of this group we have { 1 } as a multiplicative group. 2. Let G have order 2. Then G = { e, a} , say. Now a2 = a or e and a2 = a implies aa = a2 = ae and so a = e which is false. Thus a2 = e. The Cayley table is

As a concrete example we have { 1 , - 1 } under.. multiplication. 3. Let G have order 3, say, G = { e, a , b } . Consider ab. If ab = a or ab = b we have, by cancellation, that b = e or a = e respectively and either inference is false. Thus ab = e. Hence b = a- 1 and so also ba = e. We obtain the par tially completed Cayley table e

e e

a b

a b

a

b

a

b

e e

But each row and column must contain distinct elements and so the Cayley table is e a b

e e

a

b

a

b

a

b

e

b

e

a

The set of matrices

{0 D·G �). (� D} 0 1 0

1 0 0

0 0 1


104

( ) ( ) ( )

forms a group with the same table if we put e =

1 0 0 0 1 0 0 0 1

,a=

0 1 0 0 0 1 1 0 0

,b =

0 0 1 1 0 0 0 1 0

4. Let G have order 4. Then G = { e, a, b, c} , say. We mUBt distinguish two cases. (i ) Suppose there exists an element in G which is not its own inverse and let a be such an element. ThUB a- 1 =I a. For the sake of argument suppose a - 1 = b. ThUB ab = ba = e. But then ac =I a and ac =I c, as otherwise c = e or a = e respectively, hence we mUBt have ac = b. Similarly ca ::p e, a, c and so ca = b . We now obtain the incomplete Cayley table e

a

b

c

e

e

a

b

c

a b

a b

e

b

e

c

c

b

But the elements in the second row mUBt necessarily be distinct and so we have a2 = c. Also ac = b implies bac = b2 and so, as ba = e, we obtain c = b2 • ThUB maintaining the distinctiveness of the elements in any given row or column we obtain the Cayley table e

a

b

c

e

e

b

c

a

e

b

b

a b

a c e

c

a

c

c

b

a

e

A possible realization is the group { 1 , i, - 1 , -i } . (ii ) Suppose now that every element is its own inverse. Then a2 = b2 = c2 = e. Then ab =I e, a, b and so ab = c, also ba =I e, a, b and so ba = c. Similarly we deduce that ac = ca = b and be = cb = a. The Cayley table is e b c a e e b c a e b c a a b b e c a c b e c a

105


This abstract group is known as the Klein four-group (after F. Klein, 1 849-1 925) . A realization is afforded by a group of four matrices, namely

Theorem 5 Let a , b E G. Then

P roof Inserting brackets for clarity we have (b- 1 a - 1 ) (a b) = b - 1 (a - 1 (ab) ) = b - 1 ( (a- 1 a)b)

=

b- 1 (eb) = b- 1 ( b) = e.

Hence D

Corol lary Let a 1 1 a 2 , . . . , a n E G . Then ( a 1 a 2 . . . an ) - 1

=

an- 1 a n-_1 1 . . . a -1 1 .

Proof A simple induction suffices.

D

We now wish to extend the index law for powers of a E G, namely am an = am + n (m, n = O, 1 , . . . ) to the case in which ( n > 0) .

m

and

n

may be negative. First we have to define a-n

Defi n ition 5 Let a E G. Let

n

E N. Then a-n is defined by a-n = (a- 1 ) n

Using this definition, which is entirely natural if we think of reciprocals of rational numbers, we may extend the index law above. We give examples which show how the extension is used.


106

Exam ple 8 Let a E G. (a3 ) (a- 1 ) 2 = (aaa ) (a- 1 a- 1 ) = aa(aa- 1 )a- 1 = aaea- 1 = aaa - 1 = a(aa- 1 ) = ae = a. ( a- 3 ) ( a ) 2 = ( a - 1 a- 1 a- 1 ) ( aa) = a- 1 a- 1 ( a- 1 a ) a = a- 1 a- 1 ( e ) a = a -1 a- 1 a = a- 1 ( a - 1 a) = a - 1 e = a - 1 .

Theorem 6 I ndex Law for Grou ps Let a E G. Then

Proof ( may be omitted on a first reading) We may clearly confine our argument to the cases in which one, or both, of m and n is negative. We shall only consider here the case in which m > 0 and n < 0. Let n = -p where p > 0. Then am an = am a-p = am (a- 1 )P = (a . . . a) (a- 1 a- 1 ) • •

•

where in the first bracket we have m a's and in the second bracket we have p a - 1 's. But (a . . . a) (a-1 . . . a- 1 ) = (a . . . a) (aa- 1 ) (a - 1 = (a . . . a)e(a- 1

• • •

• • •

a- 1 )

a- 1 )

= (a . . . a) (a- 1 . . . a- 1 )

where on the right-hand side we have in the first bracket m - 1 a's and in the second bracket p - 1 a- 1 's. Continuing in this way to replace aa- 1 by e we eventually have m - p a's

(p < m) (m = p) (m < p) .

Thus for p < m we have

107


for m = p we have and for m < p we have

am an = ( a- 1 ) p - m = a - (p - m) D

Exam ple 9 Let x, y, z E G and suppose we are required to simplify g = (yx ) - 1 (yx - 1 ) 2 (xyz) (x - 3 y2 z) - 1 . we have = X - 1 X - 1 yyy- 2 X3 = X - 2 y2 y- 2 X3 = X - 2 X3 = X.

Defi nition 6 Two elements a, b E G are said to commute if ab = ba. If any two elements of G commute then G is said to be commutative or Abelian, otherwise G is said to be non-commutative or non-Abelian. If the operation in a group is written as multiplication then the group may or may not be Abelian. If however an additive notation, with +, is employed then it is usual to presume that the operation is commutative.

Exa m ple 10 We prove that if a2 = e for all a E G then G is necessarily Abelian. By assump tion for all a, b E G (ab ) 2 = e. Thus

We now give several further examples of groups.

Exam ples 1 1

(

)

1 . Let 0(2) consist of all matrices of the form T8 where

T8 =

cos (} sin (} - sin (} cos (}

(0 E IR) .

We claim that under matrix multiplication 0(2) is a group. We require some elementary trigonometry. We have

T8T.., = = =

( ( (

sin (}

cos (}

- sin (} cos (}

)(

cos cp

sin cp

- sin cp cos cp

)

cos (} sin cp + sin (} cos cp

cos (} cos cp - sin (} sin cp

- sin (} cos cp - cos (} sin cp - sin (} sin cp + cos (} cos cp cos((} + cp)

sin(O + cp)

- sin(O + cp)

- cos(O + cp)

(

)

)

.

)=( )

This establishes closure and associativity is immediate.

To =

cos 0 sin 0 . O cos O - sm

1 0 0 1

Since

T_ 8 T8 = T-8 + 8 = To = (T8) - 1 = T- 8 ·

is

the identity.

(� ) 0 1

'

Thus 0(2) is a group which is infinite and Abelian since

T8T
) is a sub-integral domain of D and that the mapping

x -+ (x, l ) (x E D) embeds

[Intuitively x T .] --+

D in F.

Defi n ition 7 The field F, constructed in Theorem 9, is called the quotient field of D.

D


162

Remark From Z we construct Q as the quotient field of Z. From the polynomial domain Q [x] we construct Q(x) (note round brackets) as the field of 'rational functions in x' over Q.

Exercises 5.2

1 . In Z7 find the roots of the following polynomials and factorize the polynomials as fully as possible: 2 2 x + 2x + 4, x + x + 3, X 6 - 1 . -

2. In Zr find the inverses of 2, 3 and 6.

3. In z47 find the inverses of 3, 23, 24 and 32. 4. Over Z5 solve, if possible, the following pairs of simultaneous equa tions. 2x + 3y = I, (i) 3x + 2y = 4. " " X + Jy = 2, (11) 3x + 2y = 2. 5. Over Z3 1 solve, if possible, the following simultaneous equations. 1 5x + 25y = 1 6, Sx + 21y = 1 8. 6. For the field F, given as Example 9 no. 3 above and consisting of the four elements 0, 1 , a, {3, solve the following simultaneous equations. ax + Y = 1 , X + f3y = {3.

7. Prove that if R is a ring with an identity then the subset invertible elements of R is multiplicatively a group.

U(R)

of

8. Let F be a field and x an indeterminate. What are the invertible elements of F [x] ?

be

the subset { a + b v'3 1 a, b integral domain. Is S a field?

9. Let S

E Z}

of JR. Prove that S is an

163

Rings

1 0. Let S be the subset {a + b v's l a, b E Q} of JR. Prove that S is a field. 1 1 . Let S be the subset { a + b v'6 1 a, b E Q}. Is S a field? 1 2. Let D be an integral domain of prime characteristic p. Prove that, for all a, b E D and n E N , Generalize.

1 3. Let Q

(

)

be the set of 2 x 2 matrices of the form a + ib

c + id -c + id a - ib

(a, b, c, d E IR) .

Prove that Q is a ring which satisfies all the requirements of a field except commutativity of multiplication. (Q is the ring of the quater mons, first discovered by W. R. Hamilton, 1805-65 . )

5 .3 Euclidean Domains We have shown that the integral domain of the integers Z and the polynomial domain Q [x] both have division and Euclidean algorithms. Here we study certain integral domains which are assumed to admit a division algorithm from which a Euclidean algorithm is naturally derivable.

Defi nition 8

--+

Let D be an integral domain. Let there exist a mapping

v : D \ {0} with the following two properties.

{0, 1 , 2, . . . }

1 . For all a, b E D \ {O} , v(a) � ( ab) . 2. For all a , b E D \ {0} , there exist q, r E D such that a = bq + r where either r = 0 or, if r =I 0, then v(r) < v(b) . Then D with the mapping v is called a Euclidean domain.

Exa m ples 12 1.

Z

with the mapping v defined by v( a) = I a I (a E Z, a =I 0) is a Euclidean domain. To see this, we observe


164

(i)

2.

that

v(ab) = l ab I = I a l i b i � I a I = v(a) (a, b E Z, a, b ::/; 0) since v( a) � 1 , and that (ii) becomes simply a restatement of the division algorithm for Z. F [x] , where F is any field, with the mapping v defined by v( J (x)) = deg f(x) ( J (x) E F[x] , f(x) ::/; 0) is a Euclidean domain. To see this, we observe first that for

f(x), g(x) E F[x], f(x) ::/; 0, g(x) ::/; 0, v( J (x)g(x)) = deg ( J (x)g(x)) = deg f(x) + deg g(x) � deg f(x) = v( J (x) ) . Again 2, in Definition 9 , is the division algorithm for F[x] . Strictly we have only discussed the division algorithm for polynomial domains over Q or lR but the arguments previously employed for these particular fields easily adapt to any field. We illustrate some of these arguments in the next example.

Exa m ple 13 Consider Z5 [x] . Let a(x) = 3x4 + 2x3 + 2x 2 + x + 3 and b(a) = x 2 + 2. Suppose we want to find q(x), r(x) E Z5 [x] as in the definition of a Euclidean domain. We perform a long division as follows:

x 2 + 2) 3x4 + 2x3 + 2x2 + x + 3 3x 4 + x2 3 2x + x2 + x 2x 3 + 4x

Thus q(x)

- - = -3x2 + 2x + 1 , r(x) = 2x + 1 .

Before considering other Euclidean domains we prove a convenient lemma. unit if and only if it is invertible.

Recall that an element of a ring is a

Rings

165

Lem ma 3 Let D be a Euclidean domain with identity

1,

as

in Definition 8 .

For all a E D \ {0} , v( 1 ) � v(a) . 2. For a E D \ {0} , v(1) = v(a) if and only if a is a unit of D.

1.

P roof =

1 . v(1 ) � v(1a) v(a) (a E D , a -::j; 0) . 2. Suppose a is a unit. Then there exists b E D such that ab = 1 . Then we have v(a) � v(ab) v(1 ) � v(a) and so v(1 ) v(a) . =

=

Conversely suppose v(a) = v(1 ) . There exist q, r E D such that 1 = aq + r , where if r -::j; 0 we have v(r) < v(a) = v(1) which is false and so r = 0. But, then, aq = 1 and a is a unit. D Our next example of a Euclidean domain occ urs as a subintegral domain of the field of complex numbers. We recall that for complex numbers z1 and z2 we have I z1 z2 l = I z1 l l z2 1 , I Z1

+ z2 l

� I z1 I

+

I z2 l ·

Lem m a 4 Let Z [i] denote the set of complex numbers of the form m + ni ( m, n E Z ) . Then Z [i] is an integral domain.

P roof Z [i] is a non-empty subset of the field C. Now for m1 , m2 , n 1 , � E Z

+ n 1 i) + (m 2 + n 2 i) = ( m 1 + m 2 ) + ( n 1 + n2 )i E Z [i] , - (m l + n 1 i) = ( -m d + ( -n ) i E Z [i] , (m 1 + n 1i ) (m2 + n 2i ) = (m 1 m2 - n 1 n2 ) + (m 1 n2 + n 1 m2 )i E Z [i] .

(m 1

i

The remaining axioms for an integral domain are evidently satisfied.

D

Groups, Rings a nd Fields

166

Defin ition 9 A complex number of the form (after K.F. Gauss) .

m + ni (m, n E Z)

is

called a Gaussian integer

Theorem 10 The Gaussian integers Z [i] form a Euclidean domain under the mapping v given by

v(a) = l a l 2 (a E Z[i] , a ;i O) . Proof 1 . We want to prove that

v(a)

�

v(ab) (a, b, E Z[i] \ {0} ) .

We note first that b = b 1 + b2 i where b 1 , b2 E Z and b� + b� � 1 . Hence v(a) l a l 2 � l a l 2 l b l 2 = l ab l 2 = v(ab) .

so

v(b) = I b I =

=

2. Given a, b

E Z[i] \ {0} , we have to find suitable q, r such that a = bq + r. Now is a complex number which must be of the form a + f3i where a, f3 E Q. We now use a fairly obvious property of Q. Since a must lie between two consecutive integers, we may therefore choose that integer m for which a = m + c:, I c: I < � (for example, a = lf lies between 3 and 4 and we choose m = 4, c: = - �-) Similarly we choose n so that /3 = n + 77, 1 77 I < l Let q = m + ni . Then q E Z[i] . Let r = a - bq. Then r E Z[i] . Thus

�

� = a + /3i = (m + c:) + (n + 77)i = (m + n i) + (c: + 77i) = q + (c: + 77i)

and hence

r = a - bq bq + b(c: + 77 i ) - bq = b(c: + 77 i) . If r = 0 the appropriate q, r have been found. If r # 0 then =

v(r) = l b(e + 77 i) l 2 = l b l 2 l e + 77 i l 2 = l b l 2 (c:2 + 772 )

�

�

l b i2

G + �)

= l b l 2 < l b l 2 = v(b). Hence we have established that

Z [i] is a Euclidean domain.

0

Exa m ple 14 r

5 + 6i, b = 2 - 3i. Suppose we have to find suitable q, E Z[i] such that a = bq + r where either r 0 or if r =F 0 then I r I < I b I = V22 + 32 = m. Now a 5 + 6i (5 + 6i) (2 + 3i) -8 + 27i = 8 27 . = - 13 + 13 1 = b 2 - 3i (2 - 3i) (2 + 3i ) 13 Choosing the integers nearest to - fa and � we obtain, respectively, -1 and 2. Let q = -1 + 2i and r = a - bq = (5 + 6i) - (2 - 3i) (-1 + 2i) = (5 + 6i) (4 + 7i ) = 1 - i. We note that l r l = V1 2 + 1 2 = v'2 < m. Thus we have found an appropriate q , r. However, we may also note that q , r are not uniquely determined, for if q ' = 2i and r' = a - bq' = (5 + 6i) - (2 - 3i) (2i) = - 1 + 2i then l r' l = J5 and so q ' and r' also satisfy the required conditions. Let a =

=

Exercises 5. 3

1. 2.

Find the units of the Euclidean domain Z[i] . In the Euclidean domain Z[i] for given that a = bq + r where 0 � l r l < l b l :

a, b E Z[i]

find

q, r E Z[i]

such

a = 1 + 13i, b = 4 - 3i, (ii) a = 5 + 15i, b = 7 + i, (iii) a = 5 + 6i, b = 2 - 3i. 3. Let D = {a + .Bv'21 a, .B E Z}. Prove that D is a Euclidean domain if v : D \ { 0 } -+ {0, 1 , 2, . . . } is given by v(a + ,B-./2) I (a - .Bv'2)(a + ,B-./2) I = I a2 - 2.82 1 (a, .B E Z) . (i)

=

4. (Hard) Let w be a non-trivial complex cube root of 1 . Then w3 = 1 , w =F 1 and so w2 + w + 1 = 0 and the complex conjugate of w is

= w2 = �. (For the considerations of this example it is best to ignore w -1 ± i v'3.) the fact that w 2 Prove that D = {a + ,B w I a, .B E Z } is a Euclidean domain if v : D \ { 0 } -+ {0, 1 , 2, . . . } is given by v(a + ,B w) = I a + ,B w l 2 = (a + ,B w) (a + ,BW) . w

=

168


5.4 Ideals and Homomorphisms

P. Fermat (1601-65) , in commenting on the 'Arithmetica' of Diophantus (c. 250 AD ) , formulated a statement, of which he claimed to have a proof, that the equation, had no solutions for non-zero integers x, y, z. This celebrated assertion, known as Fermat's Last Theorem, has in fact been established as correct owing to the recent work of A. Wiles (1953- ) . In the 19th century, however, its proof was attempted by many mathematicians among whom was E.E. Kummer (1810-93) who tried heroically to establish a complete result. Arising from his efforts he created a theory of 'ideal numbers' but it fell to Dedekind to introduce 'ideals', albeit in a number-theoretic and so commutative context . For our purposes we shall give a general definition of an 'ideal' appropriate to a possibly non-commutative context.

Defin ition 10 Let R be a ring and let I be an additive subgroup of R. 1 . If for all x E R and for all a E I, it follows that xa E I, then I is said to be a left ideal of R. 2. If for all x E R and for all a E I, it follows that ax E I, then I is said to be a right ideal of R. 3. If for all x E R and for all a E I, it follows that xa E I and ax E I, then I is said to be a tw�sided ideal, or briefly, an ideal of R. R and {0} are immediately ideals of R, and {0} is called the zero ideal of R. If I satisfies 1 , 2 or 3 and I ::/; {0} , I ::/; R, then I is said to be proper . By definition every left or right ideal of R is a subring of R but not every sub ring of R is necessarily a left or right ideal. Every element of R gives rise to a left or right ideal which may not, however, be proper ( see Theorem 1 1) .

Exa m ples 15 1 . Let n E Z. Then nZ = {nx l x E Z} , the subring of those integers divisible by n, is an ideal of Z. (In a commutative ring every left or right ideal is also a two sided ideal. )

169

Rings

2. Let

R be the ring M2 (Q)

of all 2

x

2 matrices over

Q. Let

Then L is easily shown to be an additive subgroup of R and L is also a left ideal since, for x , y, z, t , p, q E Q,

( y) ( ) ( X

z

t

0 p 0 q

=

0 xp + yq (J zp + t q

)E

L.

Lem ma 5 1 . Let R be a ring with an identity 1 and let I be a non-zero left ideal of R. If I contains a unit of R then I = R. 2. A field has no proper left or right ideals.

P roof 1 . Let u be a unit of R such that u E I. Then there exists v E R such that vu = 1 . Let x E R. Then x = x1 = x (va ) = (xv) u E I since I is an ideal. Hence I = R. 2. Every non-zero element of a field is a unit. Hence the result follows. D

Exa m ple 16 We have seen, in the Example above, that M2 (Q) has at least one proper left ideal. We wish to show that M2 ( Q ) has no proper two-sided ideals. Suppose I is a non-zero ideal of M2 (Q) . Let

be a non-zero element of I. Then certainly one of a, b, c, d is non-zero. For the sake of argument we suppose a "I 0. (The reader should consider in turn the cases of non-zero b, c, d and modify the argument to follow.) Since I is an ideal

is an element of I and so also is

Similarly

belong to I. Hence, for

x, y, z, t E Q,

(: :) = (� �) (� �) + (� :) (� �) + G :) G �) + (: �) C �) El

The next theorem, which may be passed over on a first reading, brings together, nevertheless, some useful facts and, at the same time, serves to define some convenient notation.

Theorem 1 1 Let R be a ring. 1 . Let

a E R. Then Ra defined by Ra = {xa l x E R}

is a left ideal of R.

3.

+ + • • •

+ a2 +

+ + . .. +

L 1 , L2 , , Ln be left ideals of R. Then L 1 L2 Ln defined by L 1 L2 + an I ai E Li, i = 1 , 2, . . . , n} + Ln = { a 1 is a left ideal of R. Let a l l � , an E R. Then Ra 1 . . + Ran = { x 1 a 1 Xn an I xi E R, i 1 , 2, . . , n}

2 . Let

.

. .

• • . •

+ R� + .

is a left ideal of R.

.

.

.

+ x2� + ... +

= .

171

Rings

P roof 1. 0 =

2.

Oa E R.a , and for all x 1 1 x 2 , y E R we have x 1 a + x 2 a = (x 1 + x 2 )a E R.a , -(x l a) = ( -x l )a E R.a . Thus Ra is a left ideal. We prove the result for two left ideals A and B. The general result is obtained by a simple induction. A + B = {a + b l a E A, b E B} and A + B is an addi tive subgroup of R. Let x E R, a E A, b E B. Then we have x(a + b) = xa + xb E A + B since A, B are left ideals. Thus A + B is a left ideal.

3. This follows from 1 and 2 above.

D

In our considerations of groups we found that normal subgroups of a group and the homomorphisms of a group were important and directly related con cepts. A similar relationship will be seen to exist between ideals of a ring and 'homomorphisms' of a ring.

Defin ition 1 1 Let R and S be rings and let

I : R --+ S be a mapping such that l (x + y) = l (x) + l (y), l (xy) = l (x)I(Y) (for all x, y E R) .

Then I is said to be a homomorphism from R to S. The homomorphism 1, as defined above, preserves the additive and multi plicative structure of R in its image I(R) in T. Thus I is a homomorphism of R considered as an additive Abelian group and considered as a multiplicative semigroup.

Exa m ples 17

be a given integer, n > 0. Then the mapping I : Z --+ Zn given by l(a) = a where a, the equivalence class to which a belongs (a E Z), is a homomorphism

1 . Let

n

since we know that

2.

a + b = a + b, ab = ab. Let F[x] be a polynomial ring over a field F and let a E F. Then let us define I : F [x] --+ F by I : a(x) --+ a(a) (a(x) E F[x]) .


172

In other words, f replaces a polynomial by the result of substituting a for x in the polynomial. But we know that a(x) + b(x) = c(x), a(x)b(x) = d(x)

(a(x) , b(x ), c(x) , d(x)

E F[x] )

implies that a( a ) + b( a ) = c( a ) , a( a )b( a ) = d( a ) and so f is a homomorphism.

Theorem 12 Let R and S be rings and let f : R --+ S be a homomorphism. Then the following statements hold. 1 . If OR and Os are the zero elements of R and S respectively then f(OR) = Os. 2. For all x E R, - f(x) = f( -x) where -f(x) is the inverse with respect to addition of f( x) in S and -x is the inverse with respect to addition of x in R. 3. f(R) is a subring of S. 4. Let K = {x E R l f(x) = Os} , Os defined in 1 . Then K is an ideal of R.

Proof 1 and 2 hold since f is a homomorphism of R into S where both rings are con sidered solely as additive groups. Furthermore, by the same reasoning, f(R) is an additive subgroup of S. To prove that f(R) is indeed a subring of S we have merely to establish multiplicative closure of f(R) in S and this occurs since f is a homomorphism from R to S considered as multiplicative semigroups. Hence, finally, we conclude that f(R) is a subring of S. To prove 4, we observe that K is the kernel of the homomorphism f from the additive group R to the additive group S. Let now a E K, x E R. Then

f(xa) = f(x)f(a) = f(x)Os = Os,

f(ax) = f(a) f(x) = Osf(x) = Os.

Hence xa, ax

E K and so K is an ideal.

D

Defi nition 12 Let R and S be rings and let f : R --+ S be a homomorphism. Then K = {x E R l f(x) = Os} is an ideal of R which is called the kernel of f, written Ker f.

173

Rings

We now come to several definitions and results which correspond to definitions and results previously considered in Chapter 4. As the results and proofs are so similar, those to be provided here will be less detailed than usual. The reader should, however, ensure that he or she fully understands the arguments. We recall that an ideal I of a ring R is, in fact, an additive subgroup of the additive group R and so we have cosets of I in R where the coset containing x E R is

x + I = {x + a i a E I}.

Lemma 6 Let R and S be rings and let f : R --+ S be a homomorphism with kernel K . Let x and y be elements of R. Then f(x) = f(y) if and only if x + K = y + K.

P roof We have shown previously that y + K are equal.

f(x) = f(y) if and only if the cosets x + K and 0

Theorem 13 Let I be an ideal of the ring R. Let R/ I denote the set of cosets of I in R. Two binary operations are defined on R/ I as follows: Let x + I and y + I be cosets of I in R and define

(x + I) + (y + I) = (x + y) + I, (x + I ) (y + I) = xyi. Then the binary operations are well-defined and under these operations R/I is a ring. Further the mapping R --+ R/I given by x --+ x + I (x E R) is a homo morphism of R onto R/ I with kernel I.

P roof Certainly under the definition of addition R/ I is an additive group. We have to show that multiplication is well-defined in R/I. Thus we have to show that if x + I = x' + I (x, x' E R) and y + I = y' + I (y, y' E R) then xy + I = x'y' + I. But certainly x + I = x' + I implies that x' = x + a for some a E I, also y' = y + b for some b E I. Then x'y' = (x + a) (y + b) = xy + ay + xb + ab. But I is an ideal and hence we have ay + xb + ab E I. Thus x'y' + I = xy + (ay + xb + ab) + I = xy + I. Our definition of multiplication is well founded.

174


To prove the associativity of multiplication we have for all

x, y, z E R

[(x + I)(y + I)] (z + I ) = (xy + I) (z + I) = (xy)z + I = x(yz ) + I = (x + I) (yz + I) = (x + I) [(y + I) (z + I)] . The distribution laws may be proved and we conclude that R/ I is a ring. The mapping f : R --+ R/I given by f(x) = x + I (x E R) is a homomorphism of the additive group R onto Rj I. For x, y E R we have f(xy) = xy + I = (x + I ) (y + I ) = f(x)f(y) and so f is a ring homomorphism. Finally Ker f = { x E R I J (x) = I } = {x E R i x + I = I } = I. D Definition 13 Let I be an ideal of the ring R. The ring R/I, as considered above, is said to be a

factor-ring of R.

Exa m ple 18 For each n E Z, nZ = { nx I x E Z} is an ideal of Z and the factor-ring ZjnZ is the ring which we have called Zn ·

Defi nition 14 Let R and S b e rings and let f : R --+ S b e a homomorphism. I f f is surjective then f is said to be an epimorphism. If f is injective then f is said to be a mono morphism. If f is bijective then f is said to be an isomorphism.

Rema rk In Theorem 8 we had an 'embedding', as it was called, of an integral domain D

into its quotient field F. By an embedding we now understand an isomorphism and we have therefore already shown that there exists an isomorphic copy of D contained in F. In this instance we may identify D with its isomorphic copy and so regard D as a sub-integral domain of F, just as we regard Z as embedded in, and a sub-integral domain of, Q.

Rings

175

Theorem 14 Fi rst Isomorph ism Theorem Let R and S be rings and let Rf K and S are isomorphic.

f

:

R -+

S be an epimorphism with kernel K. Then

P roof

K is an ideal of R and so we have an epimorphism R -+ Rf K. We now define h : R/ K -+ S by h(x + K) = f(x) (x E R) . By the proof of the First Isomorphism Theorem for groups h is a well-defined mapping which is, in fact, an isomorphism between the additive groups R/ K and S. We merely have to show that h preserves multiplication. Let x, y E R, then we have h[(x + K) (y + K)] = h[xy + K] = f(xy) = f(x)f(y) = D h(x + K)h(y + K). The proof is finally complete.

Lem ma 7 Let R be a ring. Let S be a subring of R and let I be an ideal of R. Then S + I = { x + a I x E S, a E I } is a subring of R and I n S is an ideal of S.

P roof We leave this proof to the Exercises.

D

Theorem 15 Second Isomorphism Theorem Let R be a ring. Let S be a subring of R and let I be an ideal of R. Then the factor-rings (S + I)/I and Sj (I n S) are isomorphic.

P roof We define p :

S -+ (S + I)/I by p(x) = x + I (x E S). Then p is an epimorphism. If K is the kernel of p then, by the First Isomorphism Theorem, S/ K is isomorphic to ( S + I)/ I. On proving that K = I n S the proof is complete. D


176 Exercises 5. 4

1 . Prove that

A

=

{ ( � �) lx, Q} t E

2. Let L 1 and L2 be left ideals of a ring ideal of R. Generalize. 3. Let

L1

�

L2

�

is a right ideal of M2 { Q).

R.

Prove that

L 1 n L2 is a left

L3 � . . . be a countable ascending chain of left ideals

of a ring R. Prove that

00

U Li is a left ideal of R.

i=l

X be a non-empty subset of a ring R. Let A = {a E R I xa 0 for all x E X} . Prove that A is a right ideal of R.

4 . Let

=

5. Let R be a ring not necessarily possessing an identity. Let a E R. Prove that { na + xa I n E Z, x E R} is a left ideal of R containing a. 6. Let R and S be rings and let f : R --+ S be a homomorphism. Let L be a left ideal of S. Prove that {x E R l f(x) E L } is a left ideal of R.

7. Let R and S be rings and let f : R --+ S be an epimorphism. Let L be a left ideal of R. Prove that f(L ) is a left ideal of S. 8. Let R be a ring and let S be a subring of R. Let I be an ideal of R. Prove that S + I = { x + a l x E S, a E I } is a subring of R and I n S is an ideal of S. 9. Let R, S and T be rings and let f : R --+ S and g : S --+ T be given homomorphisms. Prove that g o f is a homomorphism from R to T. 10. Let

R be the ring of 2 x 2 matrices of the form a b (a, b E R) . -b a

(

)

Prove that the mapping

( -ba

. --+ a + 1b

is an isomorphism of C with R.

(a, b E R)

177

Rings

5.5 Principal Ideal and Unique Factorization Domains

We here continue our study of integral domains with a view to obtaining proper ties very similar to those of the integral domain of the integers z.

Definition 15 Let D be an integral domain. Let a, b E D, b a if there exists c E D such that a = be.

::/; 0. Then b is said to be a divisor of

We recall that u is a unit of a ring R if there exists v

E R such that uv =

vu

= 1.

Defi nition 1 6 Let D be an integral domain. Let a, b exists a unit u in D such that a = bu.

E D.

Then

a is an 8&80Ciate of b if there

Notice that if a is an associate of b then b is an associate of a for if a = bu where uv = vu = 1 then av = buv = b1 = b. An element u is a unit if and only if u is an associate of 1. We further recall that if R is a ring and a E R then Ra = { xa I a E R} and if, in particular, R has an identity 1 then a = 1a E Ra.

Lem ma 8 Let D be Da � Db.

an integral domain. Let

a, b E D.

Then

b

divides

a

if and only if

P roof If b divides a then a = cb for some c E D. Then Da = Deb � Db. Conversely if Da � Db then a E Da � Db and so there exists c E D such that a = cb. D

Lem ma 9 Let D be an integral domain. Let equivalent.

1 . Da = Db. 2 . a divides b and b divides a. 3. a and b are associates. 4. a = bu for some unit u of D.

a, b E D.

The following statements are

178


Proof By Lemma 8, 1 and 2 are equivalent and by remarks above 3 and 4 are equi va lent. We prove the equivalence of 2 and 3. If a divides b and b divides a there exist c, d E D such that b = ac, a = bd. Then a = b d = acd from which, as D is an integral domain, 1 = cd and so d and c are units. Hence a and b are associates. On the other hand if a and b are associates there exist units u, v E D such that a = bu and b = av. Hence b divides a and a divides b. This completes the proof. D

Exa m ples 19 1 . Two integers m , n are associates in Z if and only if m = ±n. In Z Lemma 9 is obviously true. 2. Let F be a field. The units of F[x] , the polynomial ring of F over x, are the non-zero elements of F. Therefore f(x) , g(x) E F[x] are associates if and only if f(x) = cg(x) where c E F, c =I 0.

Defi nition 17 Let D be an integral domain. A non-zero element p of D, which is not a unit, is said to be irreducible if whenever a E D and a divides p then a is an associate of 1 or p.

Defi nition 18 Let D be an integral domain. A non-zero element p of D, which is not a unit, is said to be a prime if whenever p divides ab (a, b E D) then p divides a or p divides b. For our convenience we make the following definition for a restricted class of rings; we content ourselves with remarking that, with appropriate modification, the restrictions in the definition may be removed.

Defi n ition 19 Let R be a commutative ring with an identity. 1 . An ideal P of R, P =I R, is said to be prime if whenever a , b E R and ab E P, then either a E P or b E P. 2. An ideal M of R, M =I R, is said to be maximal if whenever M � I and I � R for any ideal I, then either M = I or I = R.

179

Rings

Theorem 16 Let

1.

R be a commutative ring with an identity.

An ideal P of R is prime if and only if Rj P is an integral domain.

2. An ideal M of R is maximal if and only if R/ M is a field.

P roof We note that the factor-ring in identity 1 .

1.

1 and 2 is, at least, a commutative ring with an

P be prime. Let a, b E R be such that (a + P) (b + P) = P. Then we have ab + P = P and so ab E P. AB P is prime a E P or b E P which implies a + P = P or b + P = P. Hence, Rj P has no proper divisors of zero and so Let

is an integral domain. Let now Rj P be an integral domain. Let a, b E R be such that ab E P. Then (a + P ) (b + P) = ab + P = P and so we have divisors of zero in R/P. Since R/ P is an integral domain a + P = P or b + P = P from which a E P or b E P. Hence P is prime. 2 . Let M be maximal. Let a E R be such that a + M ::1 M. Then a ¢ M and so the ideal Ra + M contains a and M and so M C Ra + M. Since M is a maximal ideal Ra + M = R. Then there exist x E R and m E M such that

xa + m = 1 . Hence

1 + M = (xa + m) + M = xa + m + M = xa + M = (x + M)(a + M) . Thus x + M is the inverse of a + M. Hence R/ M is a field. Let now Rj M be a field. Let I be an ideal such that M s; I s; R. If M ::p I there exists a E I, a ¢ M. Then a + M ::1 M and so, as R/ M is a field, a + M is invertible in R/ M and there exists b E R such that (a + M)(b + M) = 1 + M. Therefore ab = 1 + m for some m E M. Hence, as a E I, m E M and M k I we have 1 E I. But then I = R. D

Theorem 17 Let

D be an integral domain. Then a prime is also an irreducible element.


180

Proof Let p be a prime in D. Let a E D and suppose a divides p. We wish to show t hat a is an associate of 1 or p. We have p = ab for some b E D. Since p is prime p divides ab and so either p divides a or p divides b. If p divides a then a = pc for some c E D and so p = ab = pcb and 1 = cb. Thus b is a unit and p and a are associates. On the other hand, if p divides b then b = dp for some d E D and so we have p = ab = adp and 1 = ad. Thus a is an associate of 1. This completes the proof. D In the integral domain Z every irreducible element is also prime. In analogy with Z we wish to investigate those integral domains in which every irreducible element is prime. To facilitate our investigation it is convenient to make a defi nition which is obviously modelled on the Fundamental Theorem of Arithmetic in z.

Defi nition 20 Let D be an integral domain in which every non-zero element, which is not a unit , is expressible as a finite product of irreducible elements. Furthermore, whenever a non-zero element x of D which is not a unit , is written as X = P1P2

·

· ·

Pm

= Ql Q2

·

·

·

Qn

where p1 , P2 , . . . , pm ; q1 , q2 , . . . , qn are irreducible elements then m = n and, with a suitable reordering if necessary, Pi and qi are associates ( i = 1 , 2, . . . , n) . Then D is said to be a unique factorization domain (U.F.D.). By the Fundamental Theorem of Arithmetic Z is a U.F.D.

Theorem 18 In a unique factorization domain every irreducible element is prime.

Proof

D be a U.F.D. Let x be an irreducible element of D. Suppose x divides ab (a, b E D) . We have to show that x divides a or x divides b. By supposition there exists y E D such that xy = ab. Let

181

Rings

Now y, a and b are products of irreducible elements, say,

a = P1P2 . Pm , b = q1 q2 qm Y r 1 r2 re , where p1 , P2 , . . . , Pm ; q1 , (J2 , . . . , qn ; r 1 , r2 , . . . , rt are irreducible elements. Then xr 1 r2 Tt = P1P2 Pm q1 q2 qn . But these are two factorizations into irreducible elements and so, as D is a U.F.D . , we must have 1 + t = m + n and, more importantly, every irreducible · ·

· ·

·

·

·

·

·

·

·

=

·

· ·

·

· ·

element on the left-hand side must be an associate of an irreducible element on the right-hand side. Thus x is an associate of some Pi or some qi . But this implies that x divides a or x divides b. D

Defi nition 2 1 Let D be an integral domain. Let a, b, c be non-zero elements of D. Then c is said to be a common divisor of a and b if c divides a and c divides b. A common divisor d, which is itself divisible by any other common divisor, is called a greatest common divisor ( G .C.D. ) . The terminology of this definition corresponds to the terminology of Defini itions 3 and 4 of Chapter 2 in the case of the integers. As in that case, we may show that in a U.F.D . any two non-zero elements have a G.C.D. We defer the discussion of a G.C.D. until after we have discussed 'principal ideal rings'.

Definition 22 Let D be an integral domain. Let a E D. Then Da { xa I x E D } is said to be a principal ideal. An integral domain in which every ideal is principal is called a principal ideal domain (P.I.D. ) . =

Lem ma 10 Let D be a principal ideal domain. Any two non-zero elements a and b of D have a G. C.D. d given by

Da + Db = Dd. P roof

Da and Db are ideals of D and so Da + Db is also an ideal. Hence as D is a P.I.D . there exists d E D such that Da + Db = Dd. We claim that d is indeed a G.C.D. of a and b. Certainly Da � Dd and so d divides a and similarly d divides b. Thus


182

d is a common divisor of a and b. Suppose c E D is also a common divisor of a and b. Now d E Da + Db and so there exist x, y E D such that d = xa + yb . But now if c divides a and c divides b we must have that c divides d. Thus d is a G.C.D. of a and b. D We now relate our earlier discussion of Euclidean domains to principal ideal domains.

Theorem 19 A Euclidean domain is a principal ideal domain.

P roof Let D be a Euclidean domain with mapping v : D \ {0} --+ { 0, 1 , 2, }. Let I be an ideal of D. We wish to prove that I is a principal ideal. H I = { 0} the result is true so suppose I ::F {0}. Choose an element b of I for which, amongst all the non-zero elements of I, v(b) is as small as possible. As I is an ideal Db � I. We shall show that Db = I. Let a E I. As D is a Euclidean domain there exist q, r E D such that

a = bq + r where either r 0 or, if r ::F 0, then v(r) < v(b). But, as I is an ideal, r = a - bq E I. But if r ::F 0 we have v(r) < v(b) and this contradicts the choice of b as being such that v(b) is as small as possible. Thus r 0 and a = bq E Db. Hence, finally, lli = L D =

=

Exa m ples 20

1 . Z is a Euclidean domain and we now know that every ideal of Z is of the form nZ = {nx l x E Z} for some n. 2. The polynomial ring F[x] , where F is a field, is a Euclidean domain and so is a P.I.D .

Theorem 20 In a principal ideal domain every irreducible element is prime.

183

Rings

P roof

D be a P.I.D . Let p be an irreducible element of D. Let p divide ab where a, b E D. We wish to show that p divides a or p divides b. Suppose p does not divide a. Let c E D be such that c divides p and c divides a.

Let

Since p is irreducible c is a unit or an associate of p. If c is an associate of p then as c divides a so also does p divide a which is false. Hence c is a unit. Hence, by Lemma 10,

Da + Dp = Dd where d is necessarily a unit and so Dd = D, giving Da + Dp = D. Hence there exist x, y E D such that xa + YP = 1. But then

xab + ypb = b from which p divides

b. Hence p is a prime.

0

Corollary

D be a principal ideal domain. Let p be an irreducible element of D. a 1 , � , , an E D and suppose p divides a 1 � an . Then for some i (1 � i � n) , p divides ai.

Let Let

•

.

•

.

.

.

Proof Either p divides result.

a1

or p divides

a 2 a3 . . . an .

A simple induction proves the 0

Our aim is to prove that a P.I.D . is necessarily a U.F.D. We are thereby enabled to deduce that a Euclidean domain is a U.F.D . At a first reading the reader may wish to confine himself or herself to the fact of the deduction and to avoid the necess ary proofs. We have a lengthy lemma before the main theorem.

Lem ma 1 1 Let D be a principal ideal domain. Let a be a non-zero element of D which is not a unit of D. Then a is the product of a finite number of irreducible elements.


184

Proof We argue by contradiction. Suppose therefore that a is not the finite product of irreducible elements. Since a cann ot itself be irreducible, a is divisible by a 1 , say, a 1 E D, where a 1 is neither a unit nor an associate of a. Hence there exists b1 E D which is also neither a unit nor an associate of a, such that

a = a 1 b1 . But this implies that and since a 1 is not an associate of a we have

Da C Da 1 . Now either a 1 or b1 is not a finite product of irreducible elements. Suppose a 1 is not a product of irreducible elements. We apply the same argument to a 1 which is not a product of irreducible elements and we obtain a 2 dividing a 1 where

Da 1

C

Da 2

and where a2 is not a product ofirreducible elements. We continue this process to obtain a strictly ascending chain of ideals Dan (n E N) such that

Da c Da 1 where each an

c

D� c . . .

(n E N) is not a product of irreducible elements. Let I=

00

U

n= l

Dan ·

Then I is an ideal of D ( see Exercises 5 .4, no. 3) and so I = De for some e E D. But e must belong to some subset of the infinite union and so there must exist k such that e E Dek. But then

I = De � Dak � I and so I = Dak. But this implies that Dak = Dak + 1 = . . .

and this contradicts the fact that the ideals in the chain are distinct. Hence our orig inal supposition is false and a is indeed a finite product of irreducible elements. D

Theorem 2 1 A principal ideal domain is also a unique factorization domain.

185

Rings

P roof Let D be a P .I.D . By Lemma 1 1 we have shown that every non-zero element of D which is not a unit is a finite product of irreducible elements. We have to show that such a product is essentially unique. Let a E D and let a be expressed as a = P1P2

· · ·

Pm

=

ql q2

·

·

·

qn

where p1 , p2 , . . . , Pm i q1 , q2 , . . . , qn are irreducible elements of D. Then p1 divides q1 q2 . . . qn and so, by the Corollary above, p1 divides one of q1 , q2 , . . . , qn . By renumbering, if necessary, we may suppose p1 divides q1 . Then p1 and q1 are irreducible elements which are associates and so q1 = up1 where u is a unit of D. Then

which implies that P2P3

· · ·

Pm = uq2 q3

· ·

·

qn

=

q I2 q3

·

·

·

qn

where q� = uq2 is an irreducible element. A simple induction now proves the desired result. D

Corol l a ry A Euclidean domain i s also a unique factorization domain.

Exercises 5. 5

1.

Let R be a commutative ring with an identity. Prove that a maximal ideal of R is a prime ideal. Give an example of a ring R and a prime ideal of R which is not maximal in R.

2. Z [i] is the ring of Gaussian integers. Find the G . C.D. d in Z [i] of 4 + 2i and 1 -3i and write d in the form (4 + 2 i ) z1 + (1 - 3 i ) z2 = d

for appropriate z1 , z2

E Z [i] .

3. Let D be a principal ideal domain. Prove that any n non-zero elements a 1 , a2 , . . . , an of D have a G.C.D. d given by Da 1 + Da 2 + . . . + Dan = Dd.


186

5.6 Factorization in

Q[x]

Given a non-zero polynomial f(x) E Q[x] we want to devise a criterion by which the polynomial is a prime, or is equivalently irreducible, in Q[x] . We remark first that any such polynomial f(x) may be written as

f(x) where g(x)

=

1

-

c

g(x)

E Z[x] and c E Z.

Exa m ple 21

It suffices therefore to consider polynomials which are in Z[x] but to attempt their factorization in Q[x] . We aim to show that if a polynomial in Z[x] factors non-trivially in Q[x] then it already has a non-trivial factorization in Z[x] . From this fact we infer that if a polynomial in Z[x] does not factor non-trivially in Z[x] then it does not factor non-trivially in Q[x].

Defi n ition 23 Let f(x) be a polynomial in Z [x] . Then the greatest common divisor of the coefficients of f(x) is called the content of f(x). A polynomial of content 1 is said to be primitive. Let f(x) E Q[x], f(x) ::/; 0. Then we may write d

f(x) = c h(x) -

where

c,

d are integers,

h(x) E Z[x]

and h(x) has content 1 .

We illustrate this assertion by means o f Examples.

Exa m ples 22 1.

14 28 21 1 f(x) = 5 x 2 + 3 x + 2 = 2 . 3 . 5 [84x 2 + 280x + 3 1 5 ] =

7 7 2 30 [ 12x + 40x + 45] = 30 h(x) ,

where h(x) has content

1. 1 1 x +· 1 2. f(x) = 5 xa + g1 x 2 + 6 14 12 1 3 21x2 + 308x + 12] = 8.J. 7 [70x + 1 h(x), = 168 where h(x) has content 1. Lem ma 12

f(x) and g(x) be primitive polynomials in Z[x] . c 1 , c2 E Z, c1 =F 0, � =F 0, such that cd(x) = �g(x) . f(x) = ±g(x) . Let

Suppose there exist Then c 1 = ±� and

P roof Let f(x) = a0 + a 1 x + . . . + anx n (an =F 0) . Then the G.C.D . of a0 , a 1 , . . . , an is and so there exist t0, t 1 , . . . , tn E Z such that

t0 a0 + t 1 a 1 + . . . + tn an = 1. Since cd (x) = �g(x) , c2 divides c1 ao , c 1 a 1 , . . . , cl an and so c2 divides toc1 ao + t 1 c 1 a 1 + + tn cnan = cl (to ao + t 1 a 1 + + tn an ) = c1 . Similarly c 1 divides � · Thus c 1 = ±� and so f(x) = ±g(x) . · · ·

1

· · ·

D

The next lemma holds the key to the use of primitive polynomials.

Lem ma 13 Gauss's Lem ma Let

f(x) and g(x) be primitive polynomials in Z[x]. Then f(x)g(x) is primitive.

P roof

f(x) = ao + a 1 x + . . . + am xm ( am =I 0), g(x) = bo + b1 x + + bn xn (bn =F 0), f(x)g(x) = h(x) = Co + C1 X + . . . + Cm+ n xm + n (Cm + n =f; 0) . If h(x) is not primitive there exists a prime p E Z such that p divides each of eo , e ll . . . , Cm +n· Now p cannot divide all of the coefficients of f(x) or all of the Let

· .

·


188

coefficients of g(x) since f(x) and g(x) are primitive. Suppose therefore that p divides a0, a1 1 , 0-r _ 1 but p does not divide ar where 0 � r < m and that p divides bo, b1 1 , bs _ 1 but p does not divide bs where 0 � s < n. Consider Cr + s· We have • • •

•

•

•

Cr+s = ao br + s + a 1 br + s - 1 + . . . + ar - 1 bs+ 1 + ar bs + ar + 1 bs - 1 + . . . + ar +s bo· Now p divides Cr+s i ao , a 1 , ar _ 1 ; bo , b1 , . . . , bs _ 1 and hence it follows that p divides ar bs and so divides ar or bs . But this is a contradiction and so f(x)g(x) is primitive. D Exa m ple 23

2x 2 + 3x + 1, x 2 + 2 are primitive and so is (2x2 + 3x + 1)(x 2 + 2) = 2x4 + 3x3 + 5x 2 + 6x + 2. We use Lemma 13 to show that if factorization does not occur in Z[x] then it does not occur in Q[x].

Lem ma 14 Let f(x) E Z[x] and let f(x ) have degree n, n > 0. Suppose that f(x) does not factorize in Z[x] into the product of two polynomials of degrees r and s where 0 < r < n, 0 < s < n ( r + s = n ) . Then f(x) does not factorize in Q [x] into the product of two polynomials of degrees r and s.

P roof We argue by contradiction and suppose

f(x)

= g(x)h(x)

where g(x) and h(x) are polynomials in Q[x] of degrees r and s respectively. Now

f(x) = eofo (x), g(x) = C1 9o (x) , h(x) = � ho(x) , d1 d2 where fo (x), 9o (x) and ho (x) are primitive polynomials in Z[x] and eo, c1 1 c2 , d1 1 d2 are in Z. Then eo fo (x) = dC1 9o (x) d� ho (x) 1 2 and

so

Now, by Gauss's Lemma , g0(x)h0 (x) is primitive and so, by Lemma 12, eod1 � = ±c1 c2 and fo (x) = ±go (x)ho (x) . But g0 (x) and ho(x) have degrees r and s respectively and so we have derived a contradiction since f(x) = eo fo (x) = ±g0 (x)h0(x) . Thus the lemma is proved. D We now obtain our main result which is named after F.G.M. Eisenstein

(1823- 52) .

Theorem 22 Eisenstei n 's Criterion Let f(x) E Z[x] and let f(x) exists a prime p such that:

= ao + a 1 x + . . . + anxn (� =I 0) .

Suppose there

p divides a0 , a 1 , . . . , an- 1 1 p does not divide an and p2 does not divide ao . Then f(x) is irreducible as a polynomial in Q[x] . 1. 2. 3.

P roof We argue by contradiction. If f(x) is not irreducible in Q[x] then f(x) is not prime in Q[x] and so f(x) may be factorized in Q[x] into two polynomials of degrees r and s where 0 < r < n , 0 < s < n , r + s = n. There is necessarily a corresponding factorization of f(x) in Z[x] . Hence we may suppose that

f(x) = g(x)h(x) where g(x) and h(x) are polynomials in Z[x] of degrees r and s respectively. Let g(x) = bo + b1 x + . . . + br xr (br =I 0) , h(x) = Co + C1 X + . . . + C8X8 (c8 =I 0). Now a0 = b0eo and since p divides a0 but p2 does not divide a0 , either bo or eo , but not both bo and eo , is divisible by p. Suppose p divides bo but p does not divide Co · If p were to divide b0, b 1 , . . . , br then all the coefficients of f(x) would also be divisible by p and that is false. We suppose therefore that p divides b0 , b 1 , . . . , bk 1 but p does not divide bk for some k where 0 < k :$ r < n. Since ak = bkCo + bk- 1 c1 + . . · + bo ck we have that p divides ak; bo , b1 , . . . , bk 1 and so p divides bkCo. But p does not divide Co and so p divides bk which is false. Hence our initial assumption was wrong and consequently f(x) is irreducible in Q [x] . D _

_

We give some examples of the use of Eisenstein's Criterion.


190

Exam ples 24 1. Let p be a prime. Then xn - p E Q[x] is irreducible by the criterion. 2. 21 6x 2 9x3 4x4 is irreducible in Q[x] on applying the criterion with

+ + +

3.

p = 3.

+ + +...+

71' - 1 is irreducible where p We may also prove that f(x) = 1 x x 2 is a prime. We cannot apply the criterion directly but we may transform f(x) into a polynomial to which the criterion may be applied. Introduce a new indeterminate y where x = 1 + y. Then, temporarily using the quotient field Q(x) = Q(y) of Q[x] , we have 1 /(1 + y) = f(x) = 1 x x 2 + . . . + xP -

+ +

= xP - 1 =

x-1 (1 + y)P - 1 y

[ + ( �) ( � ) � . . . + ( : ) ] 1

=

= Now,

y+

yP - 1

+

y

(�) Y + ( � ) y2 + . . . + ( : ) yP.

we have seen in the proof of Theorem

8,

all of the coefficients in

+ y) are divisible by p except for ( :) 1 , also ( �) = p and so is not divisible by p2 • Thus /( 1 + Y ) is irreducible and so, evidently, is f(x) . as

/(1

=

Exercises 5. 6

+ + + + + +

+ + + +

+

1 . Prove that the following polynomials are irreducible in Q[x] . 5 25x 10x 2 7x3 , 14 21x 49x 2 6x3 , 26 39x 65x 2 10x3 . 2. Prove that 19 + 24x 9x 2 x 3 is irreducible in Q[x] . (Hint: put X = y - 1 .) 3. Prove that x3 + 9 is irreducible in Q[x] .

6 Topics in Group Theory

In this final chapter we extend our knowledge of group theory. Among other aspects of finite groups, we investigate permutation groups and obtain two results of cardinal importance in the theory of finite groups. In the first of these, we establish the structure of Abelian groups and, in the second, we estab lish the existence of the so-called 'Sylow p-subgroups' of a finite group.

6 . 1 Permutation Groups We are here concerned with bijective mappings of a non-empty set into itself. Such mappings may be multiplied under the circle-composition of mappings. From our earlier work we easily obtain the following theorem.

Theorem 1 Let X be a non-empty set and let S(X) be the set of bijective mappings of X onto X. Under the circle-composition of mappings S(X) is a group.

Proof From the Examples in Section 4. 1 (Semigroups) , we know that S(X ) is a monoid. Let f E S(X). Then the inverse f - 1 of f is given as follows. f - 1 is that mapping 191


192

for which f - 1 (a) S(X) is a group. Our interest

= b if and only if f(b) = a (a, b E X) . It is then immediate that D

in this section concerns bijective mappings on finite sets.

Defi n itio n 1 A bijective mapping of a non-empty set X onto itself is called a permutation. Any set of permutations of X forming a group is called a permutation group. The permutation group of all permutations on X is called the symmetric group on X and is frequently denoted by S(X) . If X consists of the n symbols or elements x 1 , x 2 , . . . , xn and more particularly if X = {1, 2, . . . , n} then the symmetric group on X is designated as Sn· A permutation p E Sn is written as

p=

( i1

i2

in

)

· · · in where the notation means that p(i 1 ) = i1 , p(i 2 ) = i2 , , p(in ) = in· but not invariably, i 1 , i 2 , . . . , i n are in the natural order 1, 2, . . . , n.) i1

i2

.

•

.

(Usually

Exa m ple 1 The permutation p such that represented by any of

p(1) = 2, p(2) = 1, p(3) = 4, p(4) = 3

)(

(1

)(

2 3 4 4 2 1 3 2 1 4 3 ' ' 2 1 4 3 1 2 3 4 3 1 2 4

may be

)

Defi n ition 2 A permutation p E Sn is said to fix i E {1, 2, . . . , n} if p(i) = i and to move i E {1, 2, . . . , n} if p(i) =I i . The identity permutation, which is the identity of the group Sn , is that permutation fixing all i E { 1 , 2, . . . , n} .

Lem ma 1

Sn is a group of order n(n - 1) . . . 2. 1 = n ! Proof Let p =

(�

J1

2

n

i2

· · · in

).

193


For }I we have n choices of symbols from {1, 2, . . . , n} . For h we have n - 1 choices of symbols from {1, 2, . . . , n } \ {}I }. For }a we have n - 2 choices of sym bols from {1, 2, . . . , n } \ {}I , ]2 } . Continuing we have only one choice for Jn · The total number of choices is n ( n - 1) (n - 2) . . . 1 giving the result. D

Exa m ple 2 Let X = { 1, 2, 3}. Then the symmetric group S3 has order 3! the six permutations,

= 6 and consists of

( : � :) , is the identity permutation. ( � � �) is the permutation p = 2, p(2) = 3 and p(3) = 1. 1 2 3 If q = , then the product pq is obtained from the composition of 3 1 2 the mappings p and q, namely by applying q and then p as follows: such that p(1)

(

)

1 ___!_. 3 -�-d 2 __!.. 1 � 2 3 __!.. 2 � 3 which we also write

(�

as

�) (�

2 3

2 1

� ) = (�

1 2

�) (�

2 1

( Jl�

2

�) = ( :

2

2

:) .

Convention Let

p= Then

p=

( p;1)

( t�l

2 i2 . . .

�).

tn

�)

q=

(;

h

...

:).

�)

2 2 q= p(2) . . . p ) ' q 1) q( 2) . . . q )


194

and the product pq is given by

( t ; � ) ( t ; �J = ( : \ ! �\ ! (��)) ( : � ) =( � � (pq�(n)) . p 1) p 2 ) . . . p ) 1 2 q 1) ( 2 )

q 1) q 2)

q

1 q( 1) q ( ) . . . q )

(pq (1) (pq ( 2)

(The reader is advised that another convention for the multiplication o f permutations is also to be found in some textbooks. ) We give further examples exhibiting our convention for the multiplication of permutations.

Exa m ples 3

1.

(:

2.

(�

:)(�

2 3 4 5 2 4

2 3 4 2 1 3

: ) (: 52 11 32 : ) ( � (: 52 31 42 : ) =

=

2 3 4 5 6 3 7 1 2 4 2 3 2 3

= (: = (�

Lem ma 2 Let p =

( '1�

2 i2

...

�) ( �

2 2 6 4 5 4

n)

. Then p- 1

tn

)

;) (�

3 7 4 7

.

.

3 4 5 6 7 7 3 6 1 4 1 2 3 4 5 6 2 7 3 6 1 6 6 6

7 5 3 5

:) .

= cl1

2 3 4 2 1 3

i2

2

�).

:)

:)


195

P roof D

Defi nition 3 A permutation p of { 1 , 2, . . . , n } is called a cycle of length r ( 1 < r � n ) if for some subset {i 1 , i 2 , , in } from {1, 2, . . . , n } we have p(i1 ) i 2 , p(i 2 ) = ia , . . . p(ir - d = ir , P (ir ) = i 1 and for j E {1, 2, . . . , n } \ {i l , i 2 , . . . , i r }, p(j) = j. To indicate that the cycle p permutes the integers i 1 , i 2 , , i r cyclically and that the remaining integers are fixed by p we write p, in an abbreviated notation, as •

•

=

.

. • •

p = (i 1 i 2 · · · ir) · A cycle of length 2 is often called a transposition. Conventionally the identity permutation is regarded as a cycle of length 1 and is written as (1).

Exa m ple 4

:)

2 3 4 5 6 7 is the cycle (1 6 2 3 5) , which may also 3 5 4 1 2 7 be written as (6 2 3 5 1) , (2 3 5 1 6 ) , (3 5 1 6 2 ) or (5 1 6 2 3). We note that in cycle notation we write down only the symbols from 1, 2, . . . , n

(�

which are moved. Consequently only the context enables us to determine

whether the cycle ( 1

(1

2 3 4 5 6 6 3 5 4 1 2

6 2 3 5) represents

(1

2 3 4 5 6 7 8 6 3 5 4 1 2 7 8

) etc. Fortunately ambiguity does not usually arise.

) or ,

Permutations expressed in cycle notation are easy to multiply. We illustrate their multiplication in the following example.

Exa m ple 5 Suppose we wish to prove that in S6 we have

(1

3 4 5 2) (6 2 3 4) (1 3 5)

=

(1

5 3 2 4 6) .


196

Letting r = (1 3 4 5 produc t r st,

2) , s = (6 2 3 4) , t = (1 3 5) we have, for the 1 �3�4 �5 2�2�3 �4 3�5�5�2 4�4�6 �6 5�1�1�3 6�6�2�1

and so rst

=

(1

3 4 5 2) (6 2 3 4)(1 3 5)

=

(1

5 3

2 4 6) .

Exam ple 6 The cycles ( 1 7 integers, { 1 , 2, 4,

8 2 4) and (3 7, 8} and {3, 5 ,

6 5) permute cyclically the disjoint sets of 6}. The cycles are easily seen to commute.

Defi n ition 4 Two permutations from Sn are said to be disjoint if the two subsets of integers moved by the permutations are disjoint.

Lem ma 3 Disjoint permutations commute.

P roof Let p and q be disjoint permutations from Sn · Then we know that we have three subsets A, B, C, of which C may be empty, such that { 1 , 2, . . . , n}

=

Au BUG

is a disjoint union and such that p permutes the integers of A among themselves but fixes the integers of B U C and q permutes the integers of B among them selves but fixes the integers of A U C. Let a E A, b E B, c E C. Then =

p(a) = q(p(a) )

(pq) (a)

=

p(q(a) )

(pq ) (b )

=

p(q( b ) ) = q ( b )

(pq ) (c )

=

p( q(c) ) = p(c )

Thus pq = qp .

= =

q(p( b) )

( qp ) (a) ,

= =

c = q(c)

( qp ) ( b ) ,

=

q (p(c) )

=

( qp ) ( c) . D


Exa m ple 7

197

(1

9

g)

2 3 4 5 6 7 8 8 3 6 5 7 4 1 2

is a product of the disjoint, and so commuting, cycles

( 4 6 7) .

(1

9

2 8)

and

We obtain easily the following result.

Theorem 2 Every permutation is a product of disjoint cycles. We have broken down any given permutation into disjoint cycles. We may now break down each cycle into transpositions.

Theorem 3 Every permutation is a product of transpositions.

P roof It is enough to show that every cycle is a product of transpositions. Trivially ( 1 ) = ( 12) (12) . Let now (i 1 , i 2 , . . . , i r) be a cycle of length Then, by direct calculation,

r, r

>

1. D

Exa m ple 8

(1

2 3 4 5 6 7 8 7 1 2 3 4 8 5 6

)

=

(1 7 5 4 3 2) (6 8)

(1 2)(1 3) (1 4) (1 5)(1 7) (6 8) = (5 7) (5 1 ) (6 8) (5 2) (1 2) (3 4) (5 3) (1 2) =

We may express a permutation as a product of transpositions in many, indeed too many, ways but whether an even number or an odd number of transpositions is involved in any particular way turns out to be solely determined by the permu tation. In order to prove this mysterious fact we make a digression into rings of polynomials of several commuting indeterminates and we define the 'action' of a permutation upon such polynomials.


198

Defi n ition 5

Z[x 1 , x2 , . . . , Xn] be the polynomial ring of n commuting indeterminates . x Xt 2 , . . . , xn over Z. Let p E Sn and let f(x 1 , x2 , . . . , xn ) E Z[x 1 , x2 , . . . , xn ] · The action of p on f(x 1 , x 2 , . . . , xn ) is then defined to be the polynomial (pf) (x t , x2 , . . . , Xn ) given by (pf) (xt , x 2 , · · · , Xn ) = f(xp( l ) • Xp(2) • . · . , Xp(n) ) · Let

Exam ple 9

f(x 1 , x2 , xa) = xix� + 5x� + 4x�xg E Z [x t , x 2 , xa] . 1 2 3 Let p = E S3 . Then 3 1 2 (pj ) (x 1 , x 2 , x3) = xh� + 5xi + 4xix�. If, in particular, we act upon f(x 1 , x 2 , . . . , xn) E Z[x 1 , x 2 , . . . , xn] , as above, by

(

Let

)

a transposition (ij) (i ::/: j), say, then in effect we interchange the indeterminates xi and xi in the expression for f(x t . x 2 , . . . , xn ) ·

Example 10

(1 2) act on various polynomials. (x� + x� + xa) = x� + xi + x 3 , (x 1 - x 2 ) = x 2 - Xt = - (x t - x 2 ) , (x3 - x 4 ) = xa - x 4 , (x1 - x 2 ) (x 1 - x3 ) (x 2 - x3 ) = (x2 - x1 ) (x 2 - xa) (x l - x3 ) = -(xt - x 2 ) (x 1 - x ) (x 2 - x ) . 3 3

We let the transposition

(1 (1 (1 (1

2) 2) 2) 2)

Defi n ition 6 Let Z[x 1 , x 2 , . . . , xn ] be the polynomial ring of n commuting indeterminates over Z. Let f(x 1 , x 2 , . . . , xn ) be a polynomial from Z[x 1 , x2 , . . . , xnl · If for all trans positions t E Sn

(tf)(x t . x 2 , . . . , Xn ) = f(x t . x 2 , . . . , Xn) then f(x 1 , x 2 , , Xn ) i s said to be symmetric. I f for all transpositions t E Sn (tf) (x 1 , x 2 , · · · • xn) -f(x t , x 2 , . . . , xn ) then f(x 1 , x 2 , . . . , xn ) is said to be skew-symmetric. •

•

•

=

A polynomial may be neither symmetric nor skew-symmetric.

199


Exa m ple 1 1 In Z[x 1 , x2 , x3 ] , x 1 + x2 + x 3 , x� + x� + x�, x 1 x2 + x 1 xa + x2xa and x 1 x2x3 are typical symmetric polynomials. (x 1 - x 2 ) (x 1 - x3) (x2 - x3 ) is a skew-symmetric polynomial. x 1 - x 2 + x3 is neither symmetric nor skew-symmetric. We recall that

denotes the product a 1 a2 . . . an . Let w(x 1 , x 2 , . . . , xn ) be the particular poly nomial from Z[x 1 , x 2 , . . . , xn ] given by

w(x l , x 2 , · · · • x n ) =

n

II r

(xr - X8) ,sr H2 , , Hn be normal subgroups of G such that H; n H1 H2 H; - l Hi+ l . . . Hn = {e} (i = 1 , 2, . . . , n ) and G = H1 H2 Hw • • •

•

•

•

• •

•

Then G is said to be the (internal) direct product of H1 , H2 , . . . , Hn. In this definition, since H1 , H2 , . . . , Hn are normal subgroups of G, it follows that, for each i, 1 :=:;; i :=:;; n, H1 H2 H; _ 1 Hi+ 1 . . . Hn is a normal subgroup of G. . • •

Defi nition 14

Let G be the Cartesian product of the groups A 1 , A2 , . . . , An. A multiplication is defined on G = A 1 x A2 x . . . x An by ( a l l � ' . . . , a n) (a � , a ; , . . . , a �) = (a 1 a � , a2 a; , . . . , ana � ) (a; , a: E A; , i = 1 , 2, . . . , n )

Then G is a group called the (external} direct product of A 1 , A2 , Similarly,

as

• . •

, Aw

above, A; = { (e 1 , e2 , . . . , ei - 1 > a, ei + l • . . . en) I a E A ; } ,

where Ai h as the identity ei (j = 1 , 2, . . . , n } , is isomorphic to A; (i = 1 , 2, . . . , n } and G is the (internal} direct product of A 1 , A2 , , Aw On account of these obvious isomorphisms we sometimes do not always draw a precise distinction between internal and external direct products. • • .

Definition 15 Let p be a prime. A group the order of which is a power of p is called a p-group.

Defi nition 16 Let p be a prime. Let H1 , H2 , . . . , Hn be cyclic groups of order p. Any group iso morphic to the direct product H1 x H2 x . . . x H n is said to be an elementary p-group.


212

Exa m ple 18

The Klein four-group is an elementary Abelian 2-group. We have given the definitions above for groups having a binary operation of multiplication. If we are concerned with Abelian groups then, as we know, an additive notation is commonly employed. The changes of notation and nomen clature are fairly obvious. We may still form the Cartesian product of Abelian groups to obtain an additive group in which addition is performed component wise, yielding an (external} direct sum. We find it to be convenient now to rephrase the definition above for (internal) direct s ums of Abelian groups.

Defi n ition 17 (following Definition 13)

Let A be an Abelian group. Let A 1 , A2 , . . . , An be subgroups of A such that A; n (A1 + A2 + . . . + A ; - 1 + A i+l + . . . + A n) = {0} (i = 1, 2, . . . , n )

and A = A1 + A2 + . . . + An- Then A is the (internal) direct sum of A� > A2 , · . . , An .

Exa m ples 19

1. Let A, B and C be groups of orders 9, 14 and 18 respectively. Then the group G, given by G = A x B x C, is a group of order 9 . 14.18 = 2268. 2. Let A, B, C be cyclic groups of orders 7, 8 and 17 respectively. Then the group G, given by G = A x B x C is a group of order 952. Moreover if A = (a l a7 = eA ) , B = (b l b8 = e B ) and C = (c l c1 7 = e c )

then G is cyclic with generator ( a, b, c) and identity element ( e A , e B, e 0 ) . Exercises 6. 3

1 . Let A, B and C be normal subgroups of a group G. Prove that ABC is a normal subgroup of G.

2. Let the group G be the direct product of the groups H and K. Prove that G is Abelian if and only if H and K are Abelian. 3 . Let H and K be normal subgroups of a group G such that G = HK. Prove that G/( H n K) is the direct product of H f( H n K) and Kf( H n K) . 4. Let A, B and C be subgroups of an Abelian group G such that

G = A + B + C, A n B = {0} , A n C = {0} and B n C = {0} .

Does it follow that G is the direct sum of A , B and C?

213


6.4 Abelian Groups The structure of an Abelian group may be very complicated and, without some conditions on the group, may be difficult to determine. If the group has a finite number of generators then a structure theorem is possible. Here we shall further restrict the considerations to finite Abelian groups for which we obtain a struc ture theorem. We have shown that a cyclic group has subgroups of all possible orders and that the order of an element in a group is the order of the cyclic subgroup generated by that element . If the order of every element of a group is a power of a prime p then it is not immediate, at present , that the group is a p-group since the order might be divisible by some prime other than p and yet have no elements of that order. One of our aims is to eliminate this hypothetical situation. We begin with a lemma which resolves the situation for an Abelian group. As is customary we shall employ additive notation when discussing Abelian groups.

Lem ma 5

Let A be a finite Abelian group. Let p be a prime such that every element of A has an order which is a power of p. Then A is a p-group.

P roof If A is cyclic then, by the remarks above, A is a p-group since if I A I is divisible by a prime q, q f. p, we would have an element of order q, which is false. We may now argue by induction and assume the result is true for all groups of orders strictly less than I A I · Suppose A is not cyclic and let a E A, a =/:- 0. Then the subgroup (a) generated by a has an order which is a power of p. Further, A/ (a) has the property that every element has an order which is a power of p since if x E A and x has order p 0 then

p0 (x + (a)) = p 0 x + (a) = 0 + (a) = (a) and so x + (a) has order dividing p 0 • By the induction assumption Aj(a) is a p-group and so, as I A I = l (a) l = l (a) I I A/(a) l, A is a p-group. This completes the proof.

0

Theorem 8 Let A be a group of order where p� 1 p�2 p�n where p1 , P2 , . . . , Pn are distinct primes and ai > O (i = 1, 2, . . . , n ) . Let Ai = {x E A i pf;x = O } •

•

•


214

(i = 1 , 2, . . . , n ) . Then Ai is a subgroup of order pf' direct sum of At , A2 , . . . , An .

(i = 1 , 2,

.

. . , n ) and A is the

Proof , n .

Ai is a Pi-subgroup of A (i = 1, 2, . . . ) We want to show that A = A t + A2 + . . . + An, Ai n ( A t + A2 + . . . + Ai - l + Ai + l + . . . + A n) = {0} (i = 1 , 2, . . ) ) and finally that I Ai I = pf' (i 1, 2, f ) Since p1 , p2 , . . . , Pn are distinct Let qi be given by I A I = p ' qi (i = 1, 2, primes, qt , q2 , . . . , qn have 1 as their greatest common divisor. Hence there exist t t , t2 , . . . , tn E Z such that t tqt + t2 q2 + + tnqn = 1 . Then, if x E A, X = t tqt X + t 2 q2 x + . . . + tnqnx E A t + A 2 + . . . + A n ) since pf' q ix = I A l x = 0 implies that q ix E Ai (i = 1 , 2,

By Lemma 5,

. , n

. . . , n .

=

. . . , n .

·

· ·

. . . , n .

Let now

y

E Ai n (A t + A2 + . . . + Ai- t + Ai + l + . . . + An)

for some j E { 1 , 2, . . . , n } . Since y E Ai we have p;i Yi = 0. On the other hand, since y E A t + A2 + . . . + Aj t + Aj + t + . . . + An we have

+ Y2 +

+ Yi - 1 + Yi + l + + Yn where Yk E A k (k = 1, 2, . . . , j - 1, j + 1, ) Then + %Yi - 1 + %Yi+ t + + %Y n = 0 + 0 + %Y = %Y 1 + qj Y2 + Y = Y1

·

· ·

·

· ·

· ·

·

. . . n .

· · ·

· ·

.

+ 0 = 0.

Thus the order of y divides the coprime integers pj and

qj and so y = 0. Thus Ai n (A t + A2 + . . . + Ai - t + Ai + t + . . . + An) = {0} (j = 1 , 2, . . . , } . Hence A is the direct sum of A t . A2 , . . . , Am and so I Ad (i = 1, 2,

Since

is a power of Pi (i = 1, 2, . . . , n ) we deduce that ) and our proof is now complete.

. . . , n

n

I Ad

= pf' 0

By virtue of this last theorem we need only consider the structure of a finite Abelian p-group. The following rather obvious lemma is useful.

215


Lem ma 6

Let G be a group. Let H1 , H2 and K be subgroups of G such that K is a normal subgroup of G and K k H1 n H2 . If Ht f K n H2 / K is the trivial subgroup of G/K then K H1 n H2 . =

P roof Let x E H1 n H2 . Then Kx E Ht f K n H2 / K and so, by assumption, Thus x E K giving the result.

Kx = K. D

Lem ma 7

Let A be a finite Abelian p-group. Let a be an element of A of greatest possible order. Then there exists a subgroup H of A such that A is the direct sum of H and (a).

P roof If A is cyclic then A = (a) and the result is true with H = { 0 } . Suppose A is not cyclic. We shall argue by induction on the order of A, assuming that the result is true for groups of orders strictly less than I A 1 Then a is of greatest possible order p m , say, in A and A f. (a) . We claim that there exists in G \ (a) an element of order p. We choose b E G \ (a) to have least possible order amongst the elements of G \ (a). If pb = 0 then our claim is proved. If pb f. 0 then b has order p r where p < p r < p m . But then pb has order p r - l and so, by the choice of b, pb E (a). Thus pb = na for some n E Z and hence

0 = pr b = pr - l (pb) = p r - \na) = (pr - l n)a. l Since a has order p r it must be that pr divides p r - n and so p divides n. Thus n = pq for some q E Z. Let c = b - qa. Since b ¢ (a), we necessarily have c ¢ (a). But pc = pb - pqa = pb - na 0 and so c E A \ (a) and c has order p. =

This proves our claim. We may now suppose that b has order p. Then (b) n (a) = {0 } since (b) n (a) f. 0 implies (b) k (a) which is false. Then a + (b) has order pm in Af (b) and so a + (b) is an element of greatest possible order in A/ (b). By the induction assumption there exists a subgroup such that A/ (b) is the direct sum of this subgroup and ( (a) + (b))/(b). We may write the subgroup in the form H/ (b) for some subgroup H of A.


216

(b) � H, A = H + ((a) + (b)) = H + (a) . In addition we have (( a) + (b)) = (b) . To complete the proof we have to show that H n (a) = { 0} . Let , therefore, X E H n (a) . Then X E H n ((a) + (b)) = (b) . Thus X E (a) n (b) = {0} and we Then, as

Hn

have established the lemma.

D

Corollary A finite Abelian p-group is a direct sum o f cyclic p-subgroups.

P roof Let A be a finite Abelian p-group. By Lemma 7 there exists a cyclic subgroup A 1 and a subgroup H such that

A = A1 + H, A1 n H = { 0}. Now H has strictly lower order than A and so a simple induction ensures that H is a direct sum of cyclic p-groups A2 , A , . . . , A n , say. 3 Thus

and and so, finally,

A is the direct sum of A 1 , A2 , . . . , A n .

D

We are now able to prove our main theorem on the structure of finite Abelian groups.

Theorem 9 F u nda menta l Theorem of Finite Abelia n Groups

A finite Abelian p-group A is direct sum of cyclic p-subgroups. If A is a direct sum of the cyclic p-subgroups A1 , A2 , , A m and is also the direct sum of the cyclic p-subgroups B1 , B2 , . . . , B n then m = n and, with a suitable renumbering if necessary, A1 is isomorphic to B; (i = 1 , 2, . . . , n ) . .

.

•

Proof The Corollary above yields the first statement of the theorem. We now prove the second statement. First of all, let A be an elementary Abelian p-group. Then all non-zero elements of A have order p and so any cyclic subgroup of A has order p.


217

Thus since A is the direct sum of cyclic subgroups A1 , A2 , . . . , A m and of cyclic subgroups B1 , B2 , . . . , B n each of these subgroups has order p and so

A m i = I A I = I B1 B2 p m = I A 1 A2 Thus m = n and the result holds. X

X . . . X

X

X . . . X

Bn l = p n .

We shall argue by induction and we make the assumption that the result is true for all Abelian p-groups of orders strictly less than I A 1 . Let now A be not an elementary Abelian p-group. Suppose the notation is chosen so that A 1 , A2 , . . . , A r are cyclic groups of orders � p2 and A r+ l • A r+ 2 , . . . , A m are cyclic p-groups of orders p ( 1 < r � m ) . Similarly suppose the notation is chosen so that B1 , B2 , . . . , B. are cyclic subgroups of orders � p2 and Bs + 1 , Bs + 2 , . . . , Bn are cyclic p-groups of orders p ( 1 < s � n ) . Letting pA be the subgroup {pa l a E A} (that this is a p-subgroup is easily verified ) we have, in an obvious adaptation of this notation, X

X . . . X

X

X . . . X

pA = pA l pA2 pA l pA2 =

X

X . . . X

X

X . . . X

pA r pA r+ l pA r ,

pA m

and X

X . . . X

pBn pB8 pB8 + 1 pA = pBl pB2 = pBl pB2 pB By induction r = s and, with a suitable renumbering if necessary, pA; is iso morphic to pBi (i = 1 , 2, . . . , r ) . Hence Ai is isomorphic to B; (i = 1 , 2, . . . , r ) . Thus the direct products A 1 x A2 x . . . x A r and B1 x B2 x . . . x Br are X

X . . . X

•.

isomorphic. But then

I A r+ l x A r+ 2 x . . . x A m i = I A : A1 x A 2 x . . . x A r l Br I = I A : B1 B2 = I Br + l Br + 2 · Bn l · n r r Hence p m - = p - and so m = n and this completes the proof. X

X

X . . . X X

·

·

X

D

Exercises 6. 4

A be an Abelian group. Let M be the set of elements of A of finite order. Prove that M is a subgroup of A and that Aj M contains no non-trivial elements of finite order.

1 . Let

A be an Abelian group. Let is a subgroup of A.

2. Let

n

E N. Prove that n A = { na I a E A}


218

3. Up to isomorphism find all Abelian groups of order 20 . 4. Up to isomorphism find all Abelian groups of order 72. 5. Up to isomorphism find all Abelian groups of order 2250 .

6 . 5 p-Groups and Sylow Subgroups From the Fundamental Theorem of Abelian Groups every Abelian group of order p� 1 p�2 p�n , where p1 , P2 , . . . , P n are distinct primes and O:; > 0 (i = 1 , 2, . . . , n ) , has a subgroup of order p �; (i = 1 , 2, . . . , n ) . We have here a number-theoretic relationship between the order of an Abelian group and of the orders of certain subgroups of the group. We develop this relationship for non-Abelian groups but first we must investigate p-groups. We begin with a useful counting result. •

•

•

Theorem 10

Let G be a finite group. Let Z(G ) be the centre of G and let C1 , C2 , . . . , C N be the conjugacy classes of G, each of which contains at least two elements of G. Let X r E cr and let Cc(x r ) be the centralizer of X r in G (r = 1 , 2, . . . , N) . Then I G I = I Z(G) I

N

+ L I G : Cc(x r ) l . r= l

P roof Z(G ) consists of those elements of G which are self-conjugate. Thus G = Z(G )

U C1 U C2 U . . . U CN

is a disjoint union. By the Corollary to Theorem 20, Chapter 4, the number of elements in cr is given by I G : Cc(x r ) I and so, from the disjoint union above, I G I = I Z(G ) I

Theorem 1 1

n

N

+ L I G : Cc(x r ) l . r= l

D

Let G be a finite group of order p (n � 1) where p is a prime. Then the centre Z(G ) of G is non-trivial.

219


P roof In the notation of Theorem 10, we have I G : Cc (xr) I > 1 and, necessarily, I G : Cc(x r ) I divides I G I . Thus since I G I = p n we deduce that p divides I G : Cc(x r ) I , r = 1 , 2, . . . , N. Hence p divides E;:'= l I G : Cc(x r ) I · Thus as N

I G I = I Z ( G ) I + L IG : Cc(x r ) l r= l we conclude that p divides I Z ( G) I and so we obtain the result . pn =

D

vVe give some examples showing the importance of Theorem 1 1 .

Exa m ples 2 0

1 . Let G be a group of order p2 where p is a prime. Then we may prove that G is

Abelian as follows. Certainly we now know that the centre Z ( G ) of G is non trivial and so must be of order p or p2 • But if Z ( G) is of order p then GI Z ( G) is also of order p and so is cyclic. By Example 2 5 no. 3 of Section 4 . 4 G is then Abelian. A group of order p3 need not be Abelian. Both the quaternion and dihedral groups of order 8 are non-Abelian. 2. Let G be a group of order p n (n :;::: 1) where p is a prime. Then we assert that G contains a normal subgroup of index p. We know that Z ( G) =/:. {e}. If Z ( G) = G then G is Abelian and the reader should verify that the assertion follows from the Fundamental Theorem of Abelian Groups. We suppose therefore that Z ( G) =/:. G and we intend to argue by induction on I G I · Thus by the induction assumption GI Z ( G ) has a normal subgroup of index p. By Exercises 5.4, no. 4 , the normal subgroup of index p may be written as H I Z ( G ) where H is a normal subgroup of G. Then I GI Z ( G ) : HI Z ( G ) I = p implies that

and so

IG : HI

1 =1

I Z ( G) I = p IGI I Z (G ) I IHI X

�: = p, proving the assertion.

We come now to the first of two major and fascinating theorems on the struc ture of finite groups. Insofar as this text is concerned they will represent the cul mination of our work in group theory. A.-L. Cauchy (1 789- 1857) proved that every group of order pa m, where p is a prime not dividing m, contains a subgroup of order p, but it was P .L. Sylow (1832-1918) who established the existence of a subgroup of order pa and which was subsequently named after him.


220

Theorem 12 (Sylow)

Let G be a finite group of order p a m where p is a prime not dividing m. Then G has a subgroup of order p a .

Proof If G is Abelian the result has already been shown. Suppose G is non-Abelian and make the induction assumption that the result is true for all groups of orders strictly less than I G I · In the notation of Theorem 10 we have N

I G I = I Z(G ) I + L I G : Cc(x r ) I · r= l We consider two cases. If p divides I G : Cc(x r ) I for r = 1 , 2 , . . . , N then p divides I Z(G ) I · But then Z(G ) has a subgroup Q of order p b where 1 : 0 then m must divide 7 which is impossible. Thus r = 0 and m = 1 . Also n divides 5 2 . 7 and n = 1 + 7s for some s E { 0 , 1 , 2, . . . } . If s > 0 then n must divide 5 2


223

which is impossible. Hence the Sylow 5-subgroup P and the Sylow 7-subgroup Q are both unique. Thus G is isomorphic to the direct product P x Q and is therefore Abelian as P and Q are Abelian. 2 . Let G be a finite group and let P be a Sylow p-subgroup of G. Let H be a sub group of G such that Nc (P) 2N. Then there exists q, 1 � q � N , such that a! = a9 where p > 2N � 2q. AlSo b2 = (aP - 9 ) 2 = a2P - 29 = aPaP - 29 = a9 a!' - 29 = ap - q = b.

4.2 1 . No. 1 E N, -1

ft {0} U N.

2. x = ex = ( x - 1 x)x 3.

=

x - 1 ( xx)

=

x- 1 x = e.

(ab2 ) - 1 (c2 a - 1 r 1 (c2 b2 d) (ad) - 2 a = b - 2 a - 1 ac - 2 c2 b2 da 2 a - 2 a = d- 1 a - 1 • (abc) - 1 (ab) 2 d( d- 1 b- 1 ) 2 bdc = c- 1 b- 1 a- 1 ababda 1 b- 1 d- 1 b- 1 bdc = c- 1 ac.

aa = c "# b = de = (ab)c. -x (1 x) - 1 5 . (ac) (x) = a(c(x) ) = a(1 - x) = = = � = d( x) . 1-X X-1 -X

4. No. a(bc)

=

�

Thus ac = d, etc.

7.

6. Use de Moivre's Theorem. Use de Moivre's Theorem.

8. Let J, g be strictly monotonic. Then 0 � x 1 < x 2 � 1 implies 0 � g(xJ ) < g(x 2 ) � 1 and so 0 � ( f o g) (x 1 ) < ( f o g) (x 2 ) � 1 . Thu.<J f o g is strictly monotonic. Circle-composition is associative and t(x ) x gives identity. Inverse / - 1 of f is given by /- 1 ( x) = y if and only if f(y) = x. Prove r 1 is strictly monotonic! =

Hints to Sol utions

9·

( -ba ab ) (

-yx( xy) -) (-bx -( �

is associative.

G is Abelian.

ax - by ay a o E c, 1 -b

) _(

235

ay + bx E G. Matrix multiplication -by + ax b -1 a -b 1 = 2 a + u- b a E G. a

)

)

L2

y x-vt t-? x )(x t) t) = t -?x), t-?x-�( x (Jy-u1 - � Js -1 -�y)� (x-vJ1 t--u(� A t v ) ) J1 - � A t - �x) (x- wt t-?x) ( J1,-)x�-(Ju1 - �v)t , J1 ,-)�A

= �' s = � y1 -� y1 -� (Tu o Tv , = Tu (Tv (x, Tu ( Y, s )

(ii) L et

=

=

s

,

=

(1 +

+

where X �

(1 +

=

X

F!�� Jl-� + cr

,

X

c

Circle-composition is associative , T0 is the identity and (Tv) -1 = T- v·

4.3

1.

Let G

=

{e, a, a2 , a3 , a\ a5 } , a6 = e. Cayley table is: e a a2 a3 a4 as

e e a a2 a3 a4 as

a a a2 a3 a4 as e

From Cayley table { e, a } , { e, a2 ,

a2 a a3 a4 as e a a4 }

a3

a4

as

as e a a2

as e a a2 a3

e a a2 a3 a4

are the proper subgroups.


236

2. ALetn Ba, bn ECA. . n. B n C n . . .

Then

ab, a-1 E A, ab, a-1 E B, . . .

3. { e, a, b} is a normal subgroup. { e, c }, { e , d}, { e, !}

Thus

ab, a -1 E

are conjugate subgroups.

No other proper subgroups.

4. 5.

E G, a E A n B n c . . . Then x- 1 ax E A, x -1 ax E B, etc. Let u, v E a-1Ha. u = a-1 ha, v = a-1 ka(h, k E H ). uv = a -1 haa-1 ka = a-1 hka E a-1 Ha and u-1 = a-1 h-1 a E H. Thus a- 1 Ha is a subgroup. Hence n x-1 Hx is a subgroup. Let c = n x-1 Hx Then xEG xEG c E x -1 Hx for all x E G . Let y E G . y-1 cy E y- I x -1 Hxy = (xy) -1 H(xy) for all x E G. As x runs over the elements of G so does xy and So y -1 cy E x -1 Hx for all x E G. Thus y-1 cy E n x -1 Hx X

xeG

which is therefore normal.

6.

Let A , B be non-singular matrices. Then AB, A- 1 are non-singular as (AB) -1 = B-1 A -1 and ( A -1 ) -1 = A, etc. 1 a -I 1 a 1 b 1 a+b 1 -a = 0 1 ' 0 1 0 1 0 1 0 1 · Thus H is a subgroup which is obviously Abelian. Let c E Cc(A) , x E Nc(A) . We need to prove x-1cx E Cc(A) . Let a E A. Then xax-1 E A and so (x -1 cx) -1 a(x -1cx) = x -1 c-1 xax -1 cx = x -1 xax -1 x = a. Hence we have x-1cx E Cc(A) . Suppose NH is a subgroup. Then by closure HN � NH. Let x E NH. Then x- 1 E NH, x- 1 = nh (n E N, h E H ). Then x = h -1 n -1 E HN and So NH � HN. Thus HN = NH. Suppose now NH = HN. Let x, y E NH, x = nh, y = mk where m, n E N, h, k E H. Since hm E HN = NH, let hm = m'h' (m' E N, h' E H ). Then xy = nhmk = nm'h'k E NH and x-1 = h -1 n -1 E HN NH. Hence NH is

( )( ) (

7.

8.

)( ) ( =

)

=

a subgroup.

9. Each Hi contains I Hi I - 1 = I H I - 1 elements other than e. Thus H1 U H2 U . . . U Hn contains at most n( I H I - 1) elements other than e and so I H1 U H2 U . . . U Hn l :$ n(IH I - 1) + 1 . If G = H1 U H2 U . . . u Hn , then I G I :$ n( I H I - 1) + 1 :$ I G : H I (I H I - 1) + 1 = I G I - I G : H I + 1 which is impossible unless I G : H I = 1 and H = G.

Hints to S o lutio ns

237

4.4 1 . Let G = Nc(H)a 1 U Nc(H)a2 U . . . U Nc(H)an be a coset decomposition. Then the conjugate subgroups a11 Ha 1 , a21 H� , . . . , a;:;-1 Han are distinct and there are no other conjugate subgroups. 2.

x E G implies x - 1 E Hai (same i ) . Thus x E ai 1 H and so G = a1 1 H U a2- l H U . . . U an- l H . a;- l H = ai- l H 1' f and on1y 1·f a;- l = ai-l h- 1 (h E H ) , th at is a; = hai or Hai = Hai.

3. 1 2Z, 1 2Z + 1 , . . . , 12Z + 1 1 .

4.5 1. (g o f) (xy) 2. 3. 4.

= g( f (xy)) = g( f (x) f (y)) = g( f (x))g( f (y)) = (g o f) (x) (g o f) (y)

(x, y E G ) . aabb = f(a)f(b) = f(ab) = abab. Cancelling a and b gives ab = ba. f(x) = x -1 a -1 xa E Z(G ). f (x) f (y) = x -1 a -1 xay-1 a-1 ya = y-1 (x -1 a-1 xa) a-1ya = (xy)- 1 a- 1 (xy)a = f (xy) (x, y E G ). H/N = {Nh l h E H } . Let x, y E H. (Nx) (Ny) = Nxy, (Nx)- 1 = Nx- 1 . Thus H/N is a subgroup of GjN. Let W be a subgroup ofGjN. Let H be the subset of G given by {x E G I Nx E W} . Then prove that H is a subgroup of G, certainly HjN = W. x- 1 Hx H if and only if (Nx) -1 (H/ N) (Nx) = HjN. =

5. Proper subgroups are {e , a , b , c} , {e , a , d, f } , {e, a , g , h } and centre Z(G ) {e, a } . Classes are {e} , { a } , { b, c } , { d, f } , {e , g} . G = Z(G ) u Z(G )b u Z(G )du Z(G )g. Gj Z(G ) is a Klein four-group.

5.1 1. z2

Sum 0 I

z3

Sum 0 1 2

0 0 I

1 1 0

0 0 1 2

1 1 2 0

Product 0 0 0

I 0 I

Product 0

1 0 I 2

1

2 2 0 I

0 1 2

0

0 0 0

2

0 2 I

=


238 z4

2

Sum 0

0

I

1

2

3

I 2 3

I 2 3

2 3 0

3 0 I

0 I 2

=

0

3

I

Product 0 0 0 I 2 3

0 0 0

0

0

2

3

I 2 3

2 0 2

3 2 I

0

Z

X = 6, y 3, = 9. 3. x = a2 u 1m1 + a 1 u2m2 = a2 ( 1 - �m2 ) + a 1 �m2 = � - a2 u2m2 + a1 �m2 = � (mod �) . Similarly x a1 (mod mi } . 4. proper Ring axioms are easily verified. (ln, OE)(On, 1E) = (On, OE) implies that divisors of zero exist. 5. x(x - 1) = 0. Hence x = 0 or x - 1 = 0. 6. Verify assertions. 2.

=

5.2

1. x2 + x + 2 = (x - 3) 2 , x2 + 2x + 4 = (x - I) (x - 4) , x2 + x + 3 does not factorize, x - 1 = (x - 1 ) (x - 2) (x - 3) (x - 4) (x - 5 ) (x - 6). 2. 2.4 = I, = I, = I. 3. 3.16 = I, 23.45 = I, 24. 2 = I, 32.25 = I. 4. (i) �x + �y = ! } thus x + �y = �}x + 4y = 3 3x + 2y = 4 Solutions: X = y = 2; X = I, y = 3; X = 2, y = X = 3, y = X = 4, y = I. (ii) x + 3y = 2 } thus 33xx ++ 42yy == I2 } ' 3x + 2y = 2 Solutions: 3 . 2y = 3. 4 , giving y = 2, x = I. 5. 1 �x + 25y = 16 27x + 1 �y = 4 } } thus 8x + 21y = 1 8 27x + 5y = 22 ' Solutions: 7. 9y = 7 . 1 3 , giving y = 29 , x = 23. 6. ax + y = I } t us ax + y = -I } . x + f3y = {3 ax + y = 1 Solutions: x = y = I; x = I, y = {3; x = a, y = a ; x = {3, y = 6

-

-

-

-

-

-

-

6.6

3.5

4;

0,

0;

h

0,

0.

239

Hints to Solutions 7.

u, v E U(R) , (uv) - 1 = v-1u-1 , so uv E U(R) etc.

8. F \ {0} .

a, b, c E Z, (a + bv'3) + (c + dv'3) = (a + c) + (b + d) v'3 E S, (a + bv'3) (c + dv'3) = ( ac + 3bd) + - (a + bv'3) = (-a) + ( -b) v'3 E S, (ad + be ) v'3 E S. Now S � lR and so S is a subring of JR. S is an integral domain but is not a field. 10. Certainly S is an integral domain. But 0 f. a + bv'5 E S implies 1 = a - bv'5 E S and so S IS. a field. a + bv�o a + 5b 1 1 . S is a field. Prove! 12 . Induction: (a + b) P = ((a + b) P ) P = (a P + bP ) P = (a P ) P + (b P ) P = a P + b p . 9. For all

2

2

n

13.

n+l

n

n+1

n+ 1

(t.·r� t.·t( ; )

Write elements

z _ w

as

n

n

where z,

w

n

are complex conjugates of z ,

Hence prove result.

5.3

1 . a + bi is a unit if and only if a2 + b2 = 1 . Thus units are ± 1 , ± i. 2. (i) 1 + 13i = (- 1 + 2 . ) + - 2 + 1 . 1 5 51 4 - 3i r = (1 + 13i) - (4 - 3i) (- 1 + 2i) = -1 + 2i. q = - 1 + 2i , (ii) 5 + 1 5i = 1 + 2I" . 7+T r q = 1 + 2i , 0. (iii) 5 + 6� = (-1 + 2i) + � + _!_ i . 1 - 31 13 13 q = - 1 + 2i, r = (5 + 6i) - (2 - 3i) ( -1 + 2i) = 1 - i. 3. For a , "(, 8 E Z,

(

=

(

·

)

)

Now follow the proof for the Gaussian integers.

w.


240

4.

=

w2 = w = -1 - w, w'W = 1 . D is a ring since (a + f3w) + (-y + 6w) (a + -y) + ({3 + 6)w, (a + {3w) (-y + 6w) = (a-y - {36) + ( -{36 + fYy + a6)w and .

D 5; C

�::: = (m + c) + (n + ry)w (m, n E Z, l e i � � . 1 77 1 � D· r

= (a + {3w) - (-y + 6w) (m + nw) = (-y + 6w)(c + ryw) . Now prove I c + rywl 2 = c2 + 772 - cry � �

and follow proof for Gaussian

integers.

2. The intersection of a collection of left ideals is a left ideal. 3.

00

U Li . For some N, a1 , � E LN. Thus immediately we have i=I a1 + a2 , -a� , xa1 E LN. Hence a1 + a2 , -at . xa1 E U L; which is therefore

x E R. a1 , a2 E

oo

i=l

a left ideal.

4. 0 E A since xO = 0 0 + 0 = 0, x( -a) = right ideal.

(x E X ). y E R, a1 , � E A. x(a 1 + a2 ) = xa 1 + xa2 = -(xai ) = 0, x(a1y) = (xai )y = Oy = 0. Hence A is a

la + Oa, a is in subset. Now prove subset is a left ideal. 6. M = {x E Ri f (x) E L}. y E R, u, v E M. f(u + v) = f(u) + f(v) E L, f(-u) = -f(u) E L, f (yu) f(y) f (u) E L. Thus we have u + v, -u, yu E M. Hence M is a left ideal. 7. f (L) is certainly a subring of S. Let a E f (L), x E S. Then a = f (b) (b E L) , x = f(y) (y E R). xa f(y)f(b) = f(yb) E f(L) . Thus f (L) is a left ideal of S. 8. x, y E S, a, b E I. (x + a) + (y + b) = (x + y) + (y + b) E S + I, -(x + a) = (-x) + (-a) E S + I, (x + a) (y + b) = xy + (ay + xb + ab) E S + I ( l is an ideal ) . Thus S + I is a subring. Alcro I n S is a subring of S. If x E S, a E I n S, then xa E I (I left ideal ) and xa E S (S subring ) . Thus xa E I n S which is a left ideal of S. 9. R, S, T are additive and multiplicative semigroups. Hence result. 5. Since a =

=

=

Hints to Solutions

241

10. ( a + i b) + (c + id) = (a + c) + i(b + d) -+ c - �b b d) (a + ib) (c + id) = (ac - bd) + i(bc + ad) --+ -b � d- c �b

( � : ::�) ( :) + ( �d �)

( : : ::� :�) ( :) ( �d :)

.

As mapping is bij ective result follows.

5.5 1 . M maximal implies R/ M is a field and so R/ M is an integral domain. Hence M is prime. {0} is prime , but not maximal, in Z. 2. 4 + 2i = (1 - 3i)i + (1 + i) , 1 - 3i = (1 + i) ( -1 (4 + 2i) 1 + (1 - 3i) (-i) = 1 + i (G.C.D.).

- 2i ) .

Hence it follows that

3. Certainly Da1 + Da2 + . . . + Dan = Dd for some d. Then x1 a1 +xn an = d for some x1 . x 2 , , xn E D . Prove now d is G.C.D.

+ x 2 a2 + . . .

• • •

5.6 1 . Three polynomials are irreducible by Criterion for p = 5, 7, 13 respectively. 2.

19 + 24x + 9x2 + x 3 = 3 + 9y + 6y2 + y3 which is irreducible by Criterion for

p

3.

= 3.

x 3 + 9 not irreducible in Q[x] implies that x 3 + 9 ha.s a linear factor x + a (a E Q) . Then a3 + 9 = 0 (a E Q) which is impossible.

2. Direct verification. 3. (1 (6

2) (1 4 ) (6

6) ( 1 7) ( 1 8) (6 5) (6

5) (1 3) odd , (1 3 ) ( 1 2 ) (6 1 ) (6 7) even.

4)(1

2) odd ,

4. ( 1 ) , (1 2 ) (3 4) , (1 3 ) (2 4) , (1 4 ) (2 3), (1 2 3) , (1 3 2) , (1 4 2), (1 2 4) , (2 4 3) , (2 3 4) , (1 4 3) , (1 3 4) . 5. Skew-symmetric.


242

6. Ker f = 7.

8. 9.

{a E G la.x = X for all X E G } =

n

xeX

Stabc(x) .

G = a 1 Stabc (x) U a2 Stabc (x) U . . . U CLnStabc(x ) . Prove orbit of x consists precisely of a .x, a2 .x, . . . , an .x and that a;.x = ai .x if and only if 1 aj 1 a; E Stabc(x) . a, b E G. (ab) .Hi = (ab)H; (ab) -1 = abH;b-1a-1 = a.(b.H;). Stabc(Hi) = Nc(Hi) · Hence n = I G : Nc(H ) I . e -+ (1 ), a -+ (1 2 3), b -+ (1 3 2), c -+ (1 2) , d -+ (1 3), f -+ (2 3).

6.2

1. b = a - \ a3 = e, b2 = e implies a = b = e, I G I = 1 . 2. From aba? = b, b- 1 ab = a- 2 = a2 • Hence b- 1 a2 b = a4 = e and so a2 = e. Thus ab = b and a = e. G = (b l b2 = e). 3. The elements of G are of the form an or ban (n = 0, 1, . . . , 7) . Thus I G I = 1 6 . a 4 h as order 2 but (ban ) 2 = ban ban = b2 b-1an ban = b2 a - n an = b2 i= e (n i= 0) . Hence a4 generates only subgroup of order 2. 4. G is the dihedral group D3 . 5. Label vertices consecutively as 1 , 2, 3, 4. Group of transformations is { (1 ) , (1 4) ( 2 3) , (1 2 ) ( 3 4 ) , (1 3) (2 4) }. 6. As in text for Dn . 6.3

1. AB is normal. Hence ABC = (AB)C is normal. 2. G = H x K. a, b E G, a = xy, b = zt (x, z E H; y, t E K). ab = (xz) (yt), ba = (zx) (ty). Hence result. 3. Obvious as Hj(H n K) n Kj(H n K) = { H n K}. 4. No. G = (a, b l a + b = b + a, 2a = 2b = 0}, A = (a}, B = (b), C = (c), where c = a + b. Then G = A + B + C, A n B = A n C = B n C = { 0 } .

243

Hints to Solutions

6.4

1 . a, b E M implies rna = 0, nb = 0 (rn, n E N) . rnn(a + b) = n(rna) + rn(nb) = 0 + 0 = 0, rn( -n) = - (rna) 0, thus M is a subgroup. Suppose x E A, r E N is such that r (x + M) M. Then rx + M = M and so rx E M. Then for some s E N, s ( rx ) = 0 and so ( sr ) x 0 and x E M. 2. na + nb = n(a + b) (a E A) , - (na) = ( -n)a. Hence nA is a subgroup. In solutions 3-5 let Cn be cyclic of order n. 3. 20 = 22 . 5 . c2 X c2 X Cs , c4 X Cs . 4. 12 = 23 .3 2 . c2 x c2 x c2 x c3 x C3 , c4 x c2 x c3 x C3 , Cs x C3 x C3 , =

x

=

=

C2 x C9 , C4 x c2 x Cg , Cs x Cg . 2 5. 2250 = 2.3 . 5 3 . c2 X c3 X c3 X Cs X Cs X Cs , c2 X Cg X Cs X Cs X Cs , c2 X c3 X c3 X Cs X Cs X C2s , c2 X Cg X Cs X c25 • c2 X c3 X c3 X cl 25 • c2 x c9 x C1 2s · c2

6.5

1.

2.

C2

x

Z(G ) i= { e } . If Z(G ) i� not a subgroup of H then H C HZ( G ) � Nc (H ) . If Z (G ) � H then argue by induction on I G I . Result is true for G/Z(G ) . Then HjZ(G ) is not the normalizer of H/Z(G ) in GjZ(G ) from which H is not the normalizer of H in G. Deduction is clear. r G I = 4 5 32 .5. Let rn be the number of Sylow 3-subgroups. Then we have rn = 1 + 3r and 1 + 3r divides 3 2 . 5 . Hence r = 0, rn = 1 . Let P be the unique Sylow 3-subgroup. Then P has order 9 and so is Abelian. Let n be the number of Sylow 5-subgroups. Then n = 1 + 3 s and 1 + 3 s divides 32 . 5 . Hence s = 0, n = 1 . Let Q be the unique Sylow 5-subgroup. Then Q has order 3 and so is Centre

=

cyclic. Since

G = P X G result follows. I G I = 207 = 3 2 . 23. Let rn, n be the numbers of Sylow 3-subgroups and 23subgroups respectively. rn = 1 + 3r, n = 1 + 23 s and rn and n divide 32 .23. Hence rn = 1 , n = 1 . G is the direct product of its (Abelian ) Sylow subgroups

of orders 9 and 23 and so is Abelian.

3. P is a Sylow �subgroup of G and of PN. Let l PN I = I P irn, I G I = I P irnn where rn, n are integers not divisible by p. But PN / N is isomorphic to P/(P n N) and so

I PN I INI

I P I g1Vlng . . N = I P n N I I PN I = I P n N I I P i rn P n N rn. =I I I l I PI IPn Nl I PI

244


Thus P n N is a Sylow p-subgroup of N. Hence a Sylow p-subgroup of Gf N has order I P : P n N l = l PN : Nl. Thus PN/N is a Sylow p-subgroup of

GfN.

4. If p = q the group is Abelian and result is true. If p > q let n be the number of Sylow p-subgroups. Then n = 1 + rp and n divides pq. Hence r = 0, n = 1 and the Sylow p-subgroup is unique. 5. If 1 + 5k divides 60 then 1 + 5k is 1 or 6. 6. 351 = 3 3 . 1 3. Let n be the number of Sylow 13-subgroups. Then we have n = 1 + 13k and 1 + 13k divides 3 3 .13. Hence 1 + 13k = 1 or 27. If 1 + 13k = 1 there is a unique Sylow 13-subgroup. If 1 + 1 3k = 27 the group contains 27 Sylow 13-subgroups. Then the number of elements in Sylow 1 3subgroups is given by 27(13 - 1 ) + 1 = 325. Thus there are 351 - 325 = 26 remaining elements. But a Sylow 3-subgroup h88 27 elements including the identity element. Thus the remaining elements must be from a Sylow 3subgroup which is unique. Hence there is a normal Sylow 1 3-subgroup or a normal Sylow 3-subgroup.

Suggestions for Further Study

1. Beginning Texts R.A . Dean , Elements of A bstract A_lgebra (Wiley, 1967) . J.B. Fraleigh, A First Course in A bstract A lge bra (Addison-Wesley, 1 989) . J.A. Galli an , Cont emporary A bstract Algebra (D.C. Heath, 1 994) .

2 . Advanced Texts P.M. Cohn, Algebra, vols. 1-3 (Wiley, 1 991 ) - has an extensive bibliography. N. Jacobson , Basic A lgebra, vols. I and II (W.H. Freeman, 1 985) . B.L. van der Waerden , Modem A lgebra, vols. I and II (F. Ungar , 1 948) - the classic text and still worth a look.

3. Hic;torical Texts V.J. Katz, A History of Mathematics (HarperCollins, 1 993) . B.L. van der Waerden , A History of Algebra (Springer-Verlag, 1 985).

245

Index

Abel, N . H . 93 Abelian 93, 107 Abelian group 93, 107 absolute value 54 act 198, 201 action 1 98 addition 82 , 94 algorithm 50 Al-Khwarizmi 50 alternating group 2 0 1 arithmetic modulo n 145 Arithmetica 168 associate 1 7 7 associativity 24, 82 , 94, 1 0 0 Associativity a n d Index Law 9 5 axioms 82 Axioms of Addition 82 Axioms of Distributivity 83 Axioms of Multiplication 83 Bertrand, J . 69 bijective 20 binary 82 BODMAS 84 cancellation 100, 1 0 1 cardinality 2 Cartesian product 7 Cauchy, A.-L. 2 1 9 Cayley table 102 Cayley, A. 93, 203 central 1 1 3 centralizer 1 19, 1 2 6 centre 1 1 3 characteristic 155 Characterization of a Field 1 5 1

Chebyshev, P.L. 69 China 1 49 Chinese Remainder Theorem 149 circle-composition 2 1 , 24 closure 82, 94, 100 codomain 18 coefficient 73 common divisor 50, 1 8 1 commutative 8 8 , 107 commute 88, 107 complement 6 complex numbers 1 9 composite 49 congruence 145 congruent 146 conjugacy 1 1 3 conjugacy class 1 13, 1 32 conjugate 1 1 2 , 120 constant polynomial 73 content 186 coprime 62 corps 151 coset 1 2 8 coset decomposition 129 countable 41 cubic 73 cycle 1 95 cyclic 1 1 1 D e Moivre, A . 1 1 1 D e Morgan Laws 1 2 D e Morgan, A . 1 2 Dedekind, R 7 1 , 1 5 1 , 168 degree 73 Descartes, R 7

246

Index dihedral group 1 1 8, 209 Diophantus 168 direct product 2 1 0 direct product (external) 2 1 0 direct product (internal) 2 1 1 direct sum 2 1 2 direct sum (external) 2 1 2 direct sum (internal) 2 1 2 disj oint 5, 1 96 disj oint union 5 Distributive Laws 1 2 divides 48, 7 4 Division Algorithm 52 , 54, 76 divisor 48 , 7 4, 75, 1 77 divisor of zero 88 domain 1 8 Dyck, W. von 93 Eisenstein, F.G. M. 189 Eisenstein's Criterion 189 element 1 , 3 elementary p-group 2 1 1 empty set 2 Epimenides of Crete 45 epimorphism 1 39, 1 74 equal 4, 73 equivalence class 30 equivalence relation 28, 58 Eratosthenes 68 Euclid 56 Euclidean Algorithm 56 , 79 Euclidean domain 163 even 198 factor-group 1 39 factor-ring 17 4 Fermat, P. 1 68 Fermat's Last Theorem 1 68 Fibonacci 40 field 151 finite characteristic 154 finite group 1 02 finite index 1 2 9 finite order 1 0 2 , 1 1 1 finite set 2 First Isomorphism Theorem 1 4 1 , 1 75 fix 192 Fraenkel , A.H. 7 1 function 1 8 Fundamental Theorem o f Arithmetic 66 Galilei , G. 1 1 0 Galois, E. 9 3 Gauss, K.F. 6 9 , 166 Gaussian integer 1 66 Gauss's Lemma 187 general linear group 124 generate 1 1 1 generator 1 1 1 generators 206 Goldbach, C . 70, 84

247 greatest common divisor 5 1 , 75, 8 1 group 99 Hadamard, J. 70 Haimes, P.R. 150 Hamilton, W. R. 163 Hasse diagrams 143 Hasse , H . 143 Hilbert , D. 71 Hindu-Arabic numerals 40 homomorphism 1 34, 1 7 1 ideal 1 68 identity 87, 95, 100 identity element 87, 95 identity permutation 192 image 18 indeterminate 73 index 129 Index Law for Groups 1 06 induce 202 infinite cardinality 2 infinite characteristic 154 infinite dihedral group 207 infinite group 1 02 infinite index 129 infinite order 102, 1 1 1 infinite set 2 injective 20 integers 19 integral domain 82, 88 International Standard Book Number 1 50 intersection 5 inverse 100, 150 inverse of addition 84 invertible 1 50 irrational 19 irreducible 1 78 , 1 79 , 1 86 isomorphism 139, 1 74 kernel 136, 172 Klein four-group 105 Klein, F. 105 Kerper, 1 5 1 Kronecker, L. 4 7 , 93 Kummer, E.E. 168 Lagrange, J .-L. 1 30 Lagrange's Theorem 130 left coset 1 28 left ideal 168 Leonardo of Pisa 40 Liber Abbaci 40 linear 73 linear fractional transformations 206 Lorentz, H.A. 1 15 map 18 mapping 18 matrix 42, 91 maximal l 78 modular 145

248 modulus 54 monic 73 monoid 95 monomorphism 139, 1 7 4 move 192 multiplication 82, 94 Napoleon 41 natural numbers 1 9 non-Abelian 1 07 non-commutative 107 non-singular 124 normal 120 normalizer 1 2 2 , 126 not congruent 146 not equal 4 null set 2 odd 1 98 one-one 20 one-one and onto 20 onto 20 orbit 202 order 1 1 1 , 1 1 2 Orwell, G . 4 1 partition 3 2 permutation 1 92 permutation group 192 p-group 2 1 1 P lato 63 polynomial 72, 73 prime 49, 178, 1 79, 186 prime characteristic 1 55 prime number 49 primitive 186 principal ideal 1 8 1 principal ideal domain 1 8 1 Principle of Induction 3 5 Principle of Well-ordering 3 4 product 82 proper 4, 1 1 6 proper divisor o f zero 88 property 4 p-subgroup 2 16 quadratic 73 quaternion group 144 quaternions 163 quotient field 159, 161 range 18 rational functions 1 62 rational numbers 19 real numbers 1 9 reflexive 28 related 27 relations 2 06 relative complement 6 Remainder Theorem 77

I ndex right coset 128 right ideal 1 68 ring 82 Ruffini, P. 93 Russell, B . 45 Second Isomorphism Theorem 142, 175 self-conjugate 113 • semigroup 94 set 1 sieve 68 simple 143 skew-symmetric 198 St. Paul's Epistle 45 stabilizer 202 Steinitz, E. 1 5 1 strictly positive integers 1 9 subfield 1 56 Subfield Criterion 156 subgroup 1 1 6 Subgroup Criterion 1 1 6 subring 90 Subring Criterion 90 subset 3 sum 82 surjective 2 0 Sylow subgroup 2 2 0 Sylow, P.L. 2 19 , 2 2 1 symbols 1 92 symmetric 2 8 , 1 98 symmetric group 1 92 The Elements 56, 63, 66, 67 Theaetetus 63 Theodorus 63 transitive 28 transposition 195 trivial divisors 48, 74 two-sided ideal 168 uncountable 41 union 5 unique factorization domain 1 80 Unique Solution of Equation 1 02 unity 87 universal set 4 Valllle-Poussin, C . J . de Ia 70 Venn diagrams 8 Venn, J . 8 Waerden , B.L. van der 150 Weber, H. 93, 1 5 1 Weyl, H. 4 7 Wiles, A. 1 6 8 Wussing, H. 9 3 zero 83 zero characteristic 154 zero ideal 168 zero polynomia 73

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