AN INTRODUCTION to ABSTRACT ALGEBRA (A One Semester Course)

AN INTRODUCTION TO ABSTRACT ALGEBRA (A One Semester Course)

G. Viglino Ramapo College of New Jersey April 2018

CONTENTS

Part 1 PRELIMINARIES 1.1 Functions 1 1.2 Principle of Mathematical Induction 13 1.3 The Division Algorithm and Beyond 21 1.4 Equivalence Relations 29

Part 2 GROUPS 2.1 Definitions and Examples 41 2.2 Elementary Properties of Groups 54 2.3 Subgroups 62 2.4 Homomorphisms and Isomorphisms 72 2.5 Symmetric Groups 84

2.6 Normal Subgroups and Factor Groups 92 2.7 Direct Products 103

Part 3 FROM RINGS TO FIELDS 3.1 Definitions and Examples 111 3.2 Homomorphisms, and Quotient Rings 121 3.3 Integral Domains and Fields 128

Appendix A: Check Your Understanding Solutions Appendix B: Determinants Appendix C: Answers to Selected Exercises

PREFACE This text is specifically designed to be used in a one-semester undergraduate abstract algebra course. It consists of three parts: PART 1 (25% of text). Lays a foundation for the algebraic construction that follows. PART 2 (50% of text). Focuses exclusively on the primary abstract algebra object: the GROUP. PART 3 (25% of text): Introduces additional algebra objects, including Rings and Fields. For our part, we have made every effort to assist you in the journey you are about to take. We did our very best to write a readable book, without compromising mathematical integrity. Along the way, you will encounter numerous Check Your Understanding boxes designed to challenge your understanding of each newly-introduced concept. Detailed solutions to each of the Check Your Understanding problems appear in Appendix A, but you should only turn to that appendix after making a valiant effort to solve the given problem on your own, or with others. In the words of Desecrates: We never understand a thing so well, and make it our own, when we learn it from another, as when we have discovered it for ourselves. I wish to thank my colleague, Professor Maxim Goldberg-Rugalev, for his invaluable input throughout the development of this text.

1.1 Functions 1

1 Part 1 Preliminaries

§1. FUNCTIONS We begin by recalling a bit of set notation and some definitions involving sets: The symbol “ ” is read “is contained in or is an element of.” In DEFINITION 1.1 Two sets A and B are equal, written AB= if: particular; xA  xB SET EQUALITY xA  xB and xB  xA translates to: If x is in A, then x is in B or: xA  xB SUBSET The set A is said to be a subset of the set B, B written AB , if every element in A is also A an element in B, i.e: xA  xB . AB PROPER SUBSET A is said to be a proper subset of B, written AB , if A is a subset of B and AB .

INTERSECTION The intersection of A and B, written AB , is A B the set consisting of the elements common to AB = xx A and xB both A and B. That is: AB = xx A and xB read such that

A UNION The union of A and B, written AB , is the B set consisting of the elements that are in A or AB = xx A or xB in B (see margin). That is: AB = xx A or xB

A B DISJOINT SETS Two sets A and B are disjoint if AB= the empty set AB= U COMPLEMENT Let A be a subset of the universal set U. The Ac complement of A in U, written Ac , is the set of elements in U that are not contained in A: c A A = x xU and xA (More simply: xx A , if U is understood) While on the topic of notation we call your attention to the following globally understood mathematical symbols: : read “for every” : read “there exists”  read “such that” For example: x yxy +  0 is read: for every x there exists y such that x+y is greater than 0 2 Part 1 Preliminaries

You’ve dealt with functions in one form or another before, but have All “objects” in mathematics you ever been exposed to a definition? If so, it probably started off with are sets, and functions are no something like: exceptions. The function f A function is a rule...... given by fx = x2 , is the or, if you prefer, a rule is a function...... You are now too sophisticated to accept this sort of “circular defini- subset f = xx 2 x   tion.” Alright then, have it your way: of the plane. Pictorially: DEFINITION 1.2 For given sets X and Y, we define the Carte- sian Product of X with Y, denoted by XY , CARTESIAN to be the set of ordered pairs: PRODUCt XY = xy xX and yY A function such as 2 f = xx x   DEFINITION 1.3 A function f from a set X to a set Y is a subset is often simply denoted by fXY  such that for every xX there fx= x2 . Still, in spite of FUNCTION their dominance throughout exists a unique yY . mathematics and the sciences, A function f from a set X to itself is said to be functions that can be described OPERATOR in terms of algebraic expres- an operator on X. sions are truly exceptional. The symbol f: XY is used to indicate that f Scribble a curve in the plane is a function from the set X to the set Y, and for which no vertical line cuts the curve in more than one yfx=  denotes that xy  f . point and you have yourself a DOMAIN function. But what is the “rule” The set X is said to be the domain of f, and for the set g below? yYxy   f for some xX g RANGE is said to be the range of f. gx While the domain of f is all of X, . the range of f need not be all of Y. Moreover, for AX and BY : .x IMAGE OF AX fA= faaA is called the image of Note that the set S below, is not INVERSE IMAGE OF –1 a function: A under f, and f B = xXfx  B . BY is called the inverse image of B. . S COMPOSITION OF FUNCTIONS . x Consider the schematic representation of the functions f: XY and Why not? g: YZ in Figure 1.1, along with a third function gf: XT . f . g fx .gfx x.

X Y Z gf Figure 1.1 As is suggested in the above figure, the function gf: XZ is given by: gf x = gfx first apply f and then apply g 1.1 Functions 3

Formally: DEFINITION 1.4 Let f: XY and g: YZ be such that COMPOSITION the range of f is contained in the domain of g. The composite function gf: XZ is given by: gf x = gfx

Throughout the text the sym- EXAMPLE 1.1 Let f:  and g:  be given by bol  will be used to denote 2 the set of real numbers. fx= x + 1 and gx= 2x – 5 . Find: (a) gf 3 (b) fg x

SOLUTION: 2 (a) gf 3 ==gf3 g3 + 1 ===g10 2105 – 15 2 2 (b) fg x ==fgx f2x – 5 =2x – 5 +41 =x –2620x +

EXAMPLE 1.2 2 2 ab let f: M22   and g:    (see mar- M22 = abcd   cd ab (the set of two-by-two matrices) gin) be given by f = ab cd and 2 = ab ab   cd (The set of two-tuples) gab = ab– . Find: In general: n = x x  x 1 2 n 13 ab (The set of n-tuples) (a) g f  (b) g f   24  cd

SOLUTION: 13 13 (a) g f ==gfg1324    24 24 ===g38 38–5–

ab ab (b) g f ===gfgabcd ab– cd  cd cd

CHECK YOUR UNDERSTANDING 1.1

 ab 2 Let f: M22   be given by f = ad+ and g:   R cd be given by gx= 2xx 2 . Determine: 13 ab Answer: (a) 10 25 (a) gf  (b) gf  (b) 2a + 2d a2 ++2ad d2 24 cd 4 Part 1 Preliminaries

BIJECTIONS AND THEIR INVERSES

DEFINITION 1.5 A function f: XY is: ONE-TO-ONE One-to-one if fa==fb a b ONTO Onto if for every yY there exists xX such that fx= y . BIJECTION A bijection if it is both one-to-one and onto.

EXAMPLE 1.3 4 Let f:   M22 be given by:

fxyzw = –2y x 3wz Show that f is a bijection

SOLUTION: To show that f is one to one, we start with fxyzw = fxyzw and go on to show that xyzw = xyzw : –y = –y  yy=    –2y x –2y x 2x = 2x  xx=  fxyzw = fxyzw  =   3wc 3wc 3w = 3w  ww=    zz=  zz=   xyzw = xyzw

ab To show that f is onto, we take an arbitrary element  M22 cd and set our sights on finding xyzw  R4 such that

fxyzw = ab: cd

–y = a  ya= –    ab –2y x ab 2xb=  xb=  2  fxyzw =  =   cd 3wz cd 3wc=  wc=  3    zd=  zd=  The above argument shows that f will map the element

b c 4 ab ---–a d ---  R to  M22 . Let’s check it out: 2 3 cd

b –2–a --- b c 2 f --- –a d --- ==ab  2 3 c cd 3--- d 3 1.1 Functions 5

CHECK YOUR UNDERSTANDING 1.2

4 (a) Show that the function f: M22  R given by ab f = dc –  3ab is one-to-one and onto. cd

(b) Show that the function f: M22  M22 given by ab ba f = is neither one-to-one nor onto. Answer: See page A-1. cd cd+2b

Consider the bijection f: 0123  abcd depicted in Fig- ure 1.2(a) and the function f –1: abcd  0123 in Figure 1.2(b). The function f –1 , called the inverse of the function f, was obtained from f by “reversing” the direction of the arrows in Figure 1.2(a). .X Y X. Y 3 .d 3 .d 2. .c 2. .c 1. .b 1. .b 0. a 0. a f . f –1 . (a) (b) Figure 1.2 In general: DEFINITION 1.6 The inverse of a bijection f: XY , is the –1 INVERSE FUNCTION function f : YX given by: f –1y ==x where fx y More formally: f –1 = yx xy  f Returning to Figure 1.2, we observe that the inverse of the bijection f f is also a bijection. We also note that if we apply f and then f–1 we will 3 d –1 2 . .c end up where we started, and ditto if we first apply f and then f (see 1 margin). In general: f–1 b 0 a THEOREM 1.1 Let f: XY be a bijection. Then: (a)f –1: YX is also a bijection. (b)f –1fx = x xX and ff–1y = y yY 6 Part 1 Preliminaries

–1 –1 –1 PROOF: (a) f is one-to-one: If f y1 ==f y2 x , then: –1 –1 Recall that to say that y1 x  f and y2 x  f fx= y is to say that xy  f . (see Defini-  xy 1  f and xy 2  f tion 1.3).  y1 = y2 (since f is a function) f –1 is onto: Let xX . Since f is onto, there exists yY such that xy  f . Then: yx  f –1 f –1y = x .

(b) Let xX . Since xfx  f , fx x  f –1 , which is to say: xf= –1fx . As for the other direction:

CHECK YOUR UNDERSTANDING 1.3 Verify that for any bijection f: XY : Answer: See page A-2. ff–1y = y yY

EXAMPLE 1.4 (a) Find the inverse of the binary function 4 f:   M22 given by:

fxyzw = –2y x 3wz (see Example 1.3) (b) Show, directly, that –1ab ab ff = cd cd

ab SOLUTION: (a) For given we determine xyzw such that cd

fxyzw = ab: cd

–y = a  ya= –    ab –2y x ab 2xb=  xb=  2  fxyzw =  =   cd 3wz cd 3wc=  wc=  3    zd=  zd= 

–1ab b c Conclusion: f = ---–a d --- cd 2 3 1.1 Functions 7

Note: In Example 1.3 we showed that for fxyzw = –2y x : 3wz b c f ---–a d --- = ab 2 3 cd

–1ab b c That being the case: f = ---–a d --- . cd 2 3

b –2–a --- –1 b c 2 (b) ff ab ==f---–a d --- =ab   cd 2 3 c cd 3--- d 3 since fxyzw = –2y x 3wz

CHECK YOUR UNDERSTANDING 1.4

4 Find the inverse of the bijection f: M22  R given by ab f = dc –  3ab and verify, directly, that: Answer: cd –1 z  3 w f xyzw = –1 –1 –y x ffxyzw = xyzw and that f f ab = ab . For the rest: See page A-2. cd cd As it turns out, one-to-one and onto properties are preserved under composition:

THEOREM 1.2 Let f: XY and g: YZ be functions with the range of f contained in the domain of g. Then: (a) If f and g are one-to-one, so is gf . (b) If f and g are onto, so is gf . (c) If f and g are bijections, so is gf .

PROOF: (a) Assume that both f and g are one-to-one, and that:

gf x1 = gf x2

Which is to say: gfx1 = gfx2

Since g is one-to-one: fx1 = fx2

Since f is one-to-one: x1 = x2 8 Part 1 Preliminaries

(b) Assume that both f and g are onto, and let zZ . We are to find xX such that gf x = z . Let’s do it: Since g is onto, there exists yY such that gy= z . Since f is onto, there exists xX such that fx= y . It follows that gf x ===gfx gy z . (c) If f and g are both bijections then, by (a) and (b), so is gf .

Theorem 1.2(c) asserts that the composition gf of two bijections is again a bijection. As such, it has an inverse, and here is how it is related to the inverses of its components: This is an example of a so- called “shoe-sock theorem.” THEOREM 1.3 If f: XY and g: YZ are bijections,

Why the funny name? then:

The shoes come off and then the socks the then and off come shoes The –1 –1 –1

In the reverse process: reverse the In gf = f g One puts on socks then shoes then socks on puts One PROOF: For given zZ , let xX be such that gf x = z ; which is to say, that gf –1z = x . We complete the proof by showing that f –1g–1 z is also equal to x: gf –1z = x zg= f x zgfx=  g–1z = fx f –1g–1z = x f –1g–1 z = x

CHECK YOUR UNDERSTANDING 1.5

4 –2y x The function f:   M22 given by fxyzw = has 3wz

–1ab b c inverse f = ---–a d --- (see Example 1.4), and the func- cd 2 3

ab z  3 w tion g = dc –  3ab has inverse g–1 xyzw = . cd –y x 4 4 Determine the function gf:    and its inverse; and then show, –1 Answer: See page A-2. directly, that gf –1 = f g–1 . 1.1 Functions 9

EXERCISES

Ex cerises 1-19. Let U = 123 O = 135 E = 246 , A = 5nn U B = 3nn U C = 123 15 , D = 24610 F = 11 12 13 14 . Determine: 1.OE 2.OE 3.AB 4. AB 5.BC 6.BC 7.CD 8. CD 9.Oc  Ec 10.Oc  A 11.CO 12. OA c 13.CD  F 14.CDF  15. CF  D 16. CF c  F 17. Bc  C  DO 18.OE c  AB c 19. OE c  OA c 20. Establish the following set identities (all capital letters represent subsets of a universal set U): (a) DeMorgan’s Theorems: (i) AB c = Ac  Bc (ii) AB c = Ac  Bc (b) Associative Theorems: (i) ABC  = AB  C (ii)ABC  = AB  C (c) Distributive Theorems: (i) ABC  = AB  AC (ii) ABC  = AB  AC Exercises 21-23. Prove that: 21.Ac  B c = AB c 22. AB c c  B = Ac  B 23. AB  AB c = A

Exercises 24-26. Give a counterexample to show that each of the following statements is False. 24.AB c = Ac  Bc 25.AB c = Ac  Bc 26. AB c  Cc = Ac  BC c Exercises 27-30. Is f:  (a) One-to-one? (b) Onto? 3x – 7 x2 + 1 27. fx= ------28.fx= x2 – 3 29. fx= ------30. fx= x3 –2x + x + 2 x4 + 1

Exercises 31-33. Is f:  2 (a) One-to-one? (b) Onto? 31. fx= xx 32.fx= x 1 33. fx= x2 + 2x x + 5

Exercises 34-36. Is f: 2  2 (a) One-to-one? (b) Onto? 34.fxy = yx – 35.fxy = xx + y 36. fxy = 2xx + y 10 Part 1 Preliminaries

4 Exercises 37-38. Is f: M22   (a) One-to-one? (b) Onto?

ab ab 37.f = a–2b cc– d 38. f = ab– cdb– a cd cd

4 Exercises 39-40. Is f:   M22 (a) One-to-one? (b) Onto?

ab ba+ a ba+ 39. f abcd = 40. f abcd = cb+ a2b2 cb+ da+

Exercises 41-49. Show that the given function f: XY is a bijection. Determine f –1: YX –1 –1 and show, directly, that f f x = x xX and that ff y = y yY . 41.X ==Y , and fx= 3x – 2 . x + 1 42.X ==–0  0  Y –1  1  , and fx= ------. x 2x 43.X ==–1 –  –1  Y –2  2  , and fx= ------. x + 1 44.XY==2 , and fab = –b a . 45.XY==2 , and fab = 5ab + 3 .

ab bc 46.XYM==22 , and f = . cd da

ab c 2d 47.XYM==22 , and f = . cd ab–

4 2bc+ 1 48.X ==  Y M22 , and fabcd = . da–

 a 3 49.XM= 31 Y = , and f b = 2aa – b bc+ .  c

50. Prove that a function f:  is one-to-one if and only if the function g:  given by gx= –fx is one-to-one.

51. Prove that for any given f: XYg  : YS  and h: ST : hgf = hg f . 1.1 Functions 11

52. Let f: XYg  : XY  and h: YW be given, with h a bijection. (a) Prove that if hfh= g , then fg= . (b) Show, by means of an example, that (a) need not hold when h is not a bijection. 53. Let SX , Y   , and f: SY be given. Prove that there exists a function g: XY such that fx= gx for every xS . (That is, a function g which “extends” f to all of X.) 54. Let SX , Y   , and f: XY be given. Prove that there exists a function g: SY such that fx= gx for every xS . (That is, a function g which is the “restriction” of f to the subset S.)

Exercise. 55-60. (Algebra of Functions) For any set X, and functions f: X   and g: X   , f we define fg+ : X   fg– : X   fg : X   and ---: X   as follows: g fg+ x = fx+ gx fg– x = fx– gx fg x = fx fx f fx --- x = ------ if gx 0 g gx

55. Prove that for any f: X   and g: X   : fg+ = gf+ and fg = gf . 56. Exhibit f:  , g:  , such that fg–  gf– . 57. Exhibit one-to-one functions f:  , g:  , such that fg+ is not one-to-one. 58. Exhibit onto functions f:  , g:  , such that fg+ is not onto. 59. Exhibit one-to-one functions f:  , g:  , such that fg is not one-to-one. 60. Exhibit onto functions f:  , g:  , such that fg+ is not onto.

PROVE OR GIVE A COUNTEREXAMPLE

61. If AB   and BC   , then AC   .

62. If AB= or BC= , then AC= .

63. If ABC  and CBA  , then AC= .

64. If AB = AC , then AC= .

65. If AB = AC , then AC= .

66. If AB = AB , then either A =  or B =  .

67. If ABC  and BCD  , then AC  BD . 12 Part 1 Preliminaries

68.AB  AC = ABC  . 69. If no element of a set A is contained in a set B, then A cannot be a subset of B. 70. Two sets A and B are equal if and only if the set of all subsets of A is equal to the set of all subsets of B.

71. = . 1.2 Principle of Mathematical Induction 13

§2. Principle of Mathematical Induction This section introduces a most powerful mathematical tool, the Prin- A form of the Principle of ciple of Mathematical Induction (PMI). Here is how it works: Mathematical Induction is actually one of Peano’s axioms, PMI which serve to define the posi- Let Pn denote a proposition that is either true or false, depend- tive integers. ing on the value of the integer n. [Giuseppe Peano (1858-1932).] If: I.P1 is True. And if, from the assumption that: II. Pk is True one can show that: III.Pk+ 1 is also True. then the proposition Pn is valid for all integers n  1 Step II of the induction procedure may strike you as being a bit strange. After all, if one can assume that the proposition is valid at nk= , why not just assume that it is valid at nk= + 1 and save a step! Well, you can assume whatever you want in Step II, but if the proposition is not valid for all n you simply are not going to be able to demonstrate, in Step III, that the proposition holds at the next value of n. Just imagine that the propositions P1  P2  P3 Pk Pk+ 1  are lined up, as if they were an infinite set of dominoes:

P(1 P(2 P P P(6) P(7) P(8) P(9) P(10) ) ) (3) (4) P(5) ......

If you knock over the first domino (Step I), and if when a domino falls (Step II) it knocks down the next one (Step III), then all of the dominoes will surely fall. But if the falling kth domino fails to knock over the next one, then all the dominoes need not fall. The Principle of Mathemati- cal Induction might have been To illustrate how the process works, we ask you to consider the sum better labeled the Principle of of the first n odd integers, for n = 1 through n = 5 : Mathematical Deduction, for Sum of the first n odd integers Sum n Sum inductive reasoning is used to 1 1 formulate a hypothesis or con- 11 jecture, while deductive rea- 1 + 3 4 2 4 soning is used to rigorously 1 + 3 + 5 9 3 9 establish whether or not the 1 + 3 + 5 + 7 16 4 16 conjecture is valid. 1 + 3 + 5 + 7 + 9 25 5 25 6 ? Figure 1.3 14 Part 1 Preliminaries

Looking at the pattern of the table on the right in Figure 1.3, you can probably anticipate that the sum of the first 6 odd integers will turn out to be 62 = 36 , which is indeed the case. Indeed, the pattern suggests that: The sum of the first n odd integers is n2 Using the Principle of Mathematical Induction, we now establish the validity of the above conjecture: Let Pn be the proposition that the sum of the first n odd integers equals n2 . I. Since the sum of the first 1 odd integers is 12 , P1 is true. The sum of the first 3 odd II. Assume Pk is true; that is: integers is: 135+++ +2k – 1 = k2 135++ 2  3 – 1 see margin The sum of the first 4 odd III. We show that Pk+ 1 is true, thereby completing the proof: integers is: 1357+++ 2  4 – 1 the sum of the first k + 1 odd integers

Suggesting that the sum of  2 2 the first k odd integers is: 135+++ +2k – 1 + 2k + 1 ==k + 2k + 1 k + 1 13++ +2k – 1 induction hypothesis: Step II (see Exercise 1).

EXAMPLE 1.5 Use the Principle of Mathematical Induction to establish the following formula for the sum of the first n integers: nn+ 1 123+++ +n = ------ 2

SOLUTION: Let Pn be the proposition: nn+ 1 123+++ +n = ------(*) 2 11+ 1 I.P1 is true: 1 =------ Check! 2 kk+ 1 II. Assume Pk is true: 123+++ +k = ------2 III. We are to show that Pk+ 1 is true; which is to say, that (*) holds when nk= + 1 : k + 1 k + 1 + 1 k + 1 k + 2 123+++ ++kk+ 1 ==------2 2 Let’s do it: 123+++ ++kk+ 1 = 123+++ +k + k + 1 kk+ 1 induction hypothesis: = ------+ k + 1 2 ==------kk + 1------+ 2k------+ 1------k + 1 ------k + 2 2 2 1.2 Principle of Mathematical Induction 15

CHECK YOUR UNDERSTANDING 1.6 (a) Use the formula for the sum of the first n odd integers, along with that for the sum of the first n integers, to derive a formula for the sum of the first n even integers. (b)Use the Principle of Mathematical Induction directly to establish Answer: See page A-3. the formula you obtained in (a). We pause momentarily to recall three number theory definitions. In the present discussion, Z denotes the set of integers. DEFINITION 1.7 nZ is even if kZn  = 2k . EVEN AND ODD nZ is odd if kZn  = 2k + 1 . DIVISIBILITY A nonzero integer a divides bZ , written ab, if bak= for some kZ .

THEOREM 1.4 Let b and c be nonzero integers. Then: (a) If ab and bc , then ac . (b) If ab and ac , then ab+ c . (c) If ab , then abc for every c.

PROOF: (a) If ab and bc , then, by Definition 1.7: bak== and cbh for some h and k . Consequently: cbhak==hakh = =at (where tkh= ). It follow, from Definition 1.7, that ac . Note how Definition 1.7 is used in both directions in the above proof. (b) If ab and ac , then bah= and cak= for some h and k. Consequently: bc+ ==ah+ ak ah+ k =at (where thk= + ). It follows that ab+ c . (c) If ab , then bak= for some k. Consequently, for any c: bc=== ak cakc at (where tkc= ). It follows that abc .

CHECK YOUR UNDERSTANDING 1.7 Prove or give a counterexample. (a) If ab+ c , then ab or ac . Answer: See page A-3. (b) If ab and ab+ c , then ac . 16 Part 1 Preliminaries

The “domino effect” of the Principle of Mathematical Induction need not start by knocking down the first domino P1 . Consider the following example where domino P0 is the first to fall.

EXAMPLE 1.6 Use the Principle of Mathematical Induction to show that 45n – 1 for all integers n  0 .

SOLUTION: Let Pn be the proposition 45n – 1 .

I.P0 is true: 450 – 1 , since 50 –111 ==–0 . II. Assume Pk is true: 4 5k – 1 .

k + 1 What motivated us to III. We show Pk+ 1 is true; namely, that 45– 1 : write –1 in the form k + 1 k k – 5 + 4 ? Necessity did: 5 –551 ==–551 – 5 + 4 (see margin) We had to do something = 55k – 1 + 4 to get “5k – 1 ” into the picture (see II). The desired conclusion now follows from Theorem 1.4: Clever, to be sure; but such Theorem 1.4 (c): k k a clever move stems from 45– 1  455– 1 and then: stubbornly focusing on k k what is given and on what Theorem 1.4(b): 455– 1 and 4 4 4 5 5 – 1 + 4 needs to be established.

CHECK YOUR UNDERSTANDING 1.8 (a) Use the Principle of Mathematical Induction to show that n!  n2 for all integers n  4 . (b) Use the Principle of Mathematical Induction to show that Answer: See page A-4. 6 n3 + 5n for all integers n  1 .

ALTERNATE FORMS OF MATHEMATICAL INDUCTION We complete this section by introducing two equivalent forms of the Principle of Mathematical Induction — equivalent in that any one of API is often called the them can be used to establish the remaining two. Strong Principle of Induc- One version, which we will call the Alternate Principle of Induction tion. A bit of a misnomer, (API), is displayed in Figure 1.3(b). As you can see, the only difference since it is, in fact, equiva- between PMI and API surfaces in (*) and (**). Specifically, the propo- lent to PMI. sition “Pk True” in (a) is replaced, in (b), with the proposition “Pm True for all integers m up to and including k”. Let Pn denote a proposition that is either true or false, depending on the value of the integer n. PMI API If P1 is True, and if: If P1 is True, and if (*) Pk True  Pk+ 1 True (**): Pm True for 1 mk  Pk+ 1 True then Pn is True for all integers n  1 then Pn is True for all integers n  1 (a) (b) Figure 1.4 1.2 Principle of Mathematical Induction 17

We establish the equivalence of PMI and API by showing that (*) holds if and only (**) holds. Clearly, if (*) holds then (**) must also hold. As for the other way around: Assume that (**) holds and that (*) does not. (we will arrive at a contradiction)

If (*) does not hold, then there must exist some k0 for which Pk0 is True and Pk0 + 1 is False. Since Pk0 + 1 is False, and since (**) holds, we know that Pk1 is False for some 1 k1 k0 . But we are assuming that Pk0 is True. Hence Pk1 is False for some 1  k1  k0 . Repeating the above procedure with k1 playing the role of k0 we arrive at Pk2 is False for some 1  k2  k1 .

Continuing in this fashion we shall, after at most k0 – 1 steps, be forced to conclude that P1 is False — contradicting the assumption that P1 is True.

EXAMPLE 1.7 Use API to show that for any given integer n  12 there exist integers a  0 b  0 such that n = 3a + 7b . SOLUTION: I. Claim holds for n = 12 : 12= 3 4 + 70 II. Assume claim holds for all m such that 12 mk . III. To show that the claim holds for nk= + 1 we first show, directly, that it does indeed hold if k +131 = or if k +141 = : 13== 3 2 +30727 1 and 14  +  Now consider any k + 115 . If k + 115 , then 12  k + 1 – 3  k . Appealing to the induction hypothesis, we choose a  0 b  0 such that: k + 1 –33 = a + 7b It follows that k +31 = a + 1 + 7b , and the proof is complete. 18 Part 1 Preliminaries

Here is another important property which turns out to be equivalent to the Principle of Mathematical Induction:

+ Z denotes the set of + positive integers. THE WELL-ORDERING PRINCIPLE FOR Z Every nonempty subset of Z+ has a smallest (or least, or first) element. Note that subsets of Z need not have first elements. A case in We show that the Alternate Principle of Mathematical Induction point implies the Well-Ordering Principle:  –42024 –  Let S be a NONEMPTY subset of Z+ . Note also that the bounded set If 1  S , then it is certainly the smallest element in S, and we are done. x   5 x 9 does not contain a smallest Assume 1  S , and suppose that S does not have a smallest element element (5 is not in the set). (we will arrive at a contradiction): Let Pn be the proposition that nS for nZ + . Since, 1  S , P1 is True. Suppose that Pm is True for all 1 mk , can Pk+ 1 be False? No: To say that Pk+ 1 is False is to say that k + 1  S . But that would make k + 1 the smallest element in S, since none of its predecessors are in S. This cannot be, since S was assumed not to have a smallest element. Since P1 is True (1  S ) and since the validity of Pm for all 1 mk implies the validity of Pk+ 1 , Pn must be True for all nZ + ; which is the same as saying that no element of Z+ is in S — contradicting the assumption that S is NONEMPTY.

CHECK YOUR UNDERSTANDING 1.9 Show that the Well-Ordering Principle implies the Principle of Math- Answer: See page A-4. ematical Induction. 1.2 Principle of Mathematical Induction 19

EXERCISES

Exercises 1-29. Establish the validity of the given statement.

1. For every integer n  1 , 2n – 1 is the nth odd integer.

3n2 – n 2. For every integer n  1 , 147+++ +3n – 2 = ------. 2 n 2n – 1 2n + 1 3. For every integer n  1 , 12 ++++32 52  2n – 1 2 = ------------------. 3

4. For every integer n  1 , 12 ++++22 32  n2 = nn------+ 1------2n------+ 1 . 6

44n – 1 5. For every integer n  1 , 44++++2 43  4n = ------------. 3 1 1 1 1 1 6. For every integer n  1 , --- +++------ +----- = 1– ----- . 2 4 8 2n 2n

1 1 1 1 7. For every integer n  1 , 1 + --- 1 + --- 1 + --- 1 + --- = n + 1 . 1 2 3 n

1 – xn + 1 8. For every integer n  1 and any real number x  1 , x0 ++++x1 x2  xn = ------. 1 – x

n a1 – rn + 1 9. For every integer n  1 , and any real number r  1 , ari = ------.  1 – r i = 0

10. For every integer n  0 : 524n + 2 + 1 .

11. For every integer n  1 : 943n – 1 .

12. For every integer n  1 : 35n – 2n .

13. For every integer n  1 , 52n + 7 is divisible by 8.

14. For every integer n  1 , 33n + 1 + 2n + 1 is divisible by 5.

15. For every integer n  1 , 4n + 1 + 52n – 1 is divisible by 21.

16. For every integer n  1 , 32n + 2 –98n – is divisible by 64.

17. For every integer n  0 , 2n  n . 20 Part 1 Preliminaries

18. For every integer n  5 , 2n – 4  n .

19. For every integer n  5 , 2n  n2 .

20. For every integer n  4 , 3n  2n + 10 . 2n ! 21. For every integer n  1 , ------is an odd integer. 2nn! 22. For every integer n  4 , 2nn ! . 23. (Calculus Dependent) Show that the sum of n differentiable functions is again differentiable. d 24. (Calculus Dependent) Show that for every integer n  1 , xn = nxn – 1 . dx Suggestion: Use the product Theorem: If f and g are differentiable functions, then so is fg d d d differentiable, and fxgx = fx gx+ gx fx . dx dx dx 1 25. Let a1 = 1 and an + 1 = 3 – ----- . Show that an + 1  an . an 1 26. Let a1 = 2 and an + 1 = ------. Show that an + 1  an . 3 – an 1 1 1 27. For every integer n  1 , 1 ++++------ ------ 2n +11 – . 2 3 n

28. For any positive number x, 1 + x n  1 + nx for every n  1 . 29. For every integer n  8 , there exist integers a  0 b  0 such that n = 3a + 5b .

30. Let m be any nonnegative integer. Use the Well-Ordering Principle to show that every nonempty subset of the set nZnm  – contains a smallest element. 31. Use the Principle of Mathematical Induction to show that there are n! different ways of ordering n objects, where n!123=  n . 32. What is wrong with the following “Proof” that any two positive integers are equal: Let Pn be the proposition: If a and b are any two positive integers such that maxab = n , then ab= . I.P1 is true: If maxab = 1 , then both a and b must equal 1. II. Assume Pk is true: If maxab = k , then ab= . III. We show Pk+ 1 is true: If maxab = k + 1 then maxa – 1 b – 1 = k . · By II, a – 1 ==b – 1  a b . 1.3 The Division Algorithm and Beyond 21

§3. The Division Algorithm and Beyond ALL LETTERS IN THIS SECTION WILL BE UNDERSTOOD TO REPRESENT INTEGERS. In elementary school you learned how to divide one integer into another to arrive at a quotient and a remainder, and could then check 5 q your answer (see margin). That checking process reveals an important 3 17 a d 15 result: 2 r THEOREM 1.5 For any given aZ and dZ + , there exist Check: 17= 3 5+ 2 THE DIVISION unique integers q and r, with 0  r  d , such that: adqr= + ALGORITHM ad= qr+ ROOF Here is a “convincing argu- P : We begin by establishing the existence of q and r such that: ment” for your consideration: adqr= + with 0  rd Mark off multiples of d on the Consider the set: number line: SadnnZ= –  and adn–  0 (*) | | | | | -2d -d 0 d 2d We first show that S is not empty: Case 1. If adq= , then let If a  0 , then aad= –  00 , and therefore aS . r = 0 . [0 is playing the role of n in (*)] Case 2. If a is not a multiple If a  0 , then ada–  0 , and therefore adaS–  . of d, then let dq be such that [a is playing the role of n in (*) and remember that dZ + ] dq a d + 1 q . We then have ad= qr+ , where: Since S is a nonempty subset of 0  Z+ , it has a least element d (Exercise 30, page 20); let’s call it r. Since r is in S, there exists q  Z r such that: r = ad– q (**) .a dq dq+ d To complete the existence part of the proof, we show that r  d . In either case 0  rd . Assume, to the contrary, that r  d . From: r – d ==ad– q – d ad– q + 1 (**) we see that r – d is of the form adn– (with n = q + 1 ). Moreover, our assumption that r  d implies that r – d  0 . It follows that r – dS , contradicting the minimality of r. To establish uniqueness, assume that: adqr= + with 0  rd and adq=  + r with 0  r  d This is a common mathematical theme: [We will show that qq=  and rr=  (see margin)] To establish that something is unique, consider Since r  0 and r  d (or –r  –d ): rr–   0 – r  0 – d = –d . two such “somethings” Since rd and r  0 (or –r  0 ): rr–   dr–   d – 0 = d and then go on to show that the two “some- Thus: –d  rr–   d , or rr–   d things” are, in fact, one and the same. From dq+ r = dq + r we have: rr–  = dq – q (a multiple of d) But if rr–   d and if rr–  is a multiple of d, then rr–0 = (or rr=  ). Returning to dq+ r = dq + r we now have: dq+ r = dq + r  dq= dq  dq– q ==0  q q d  0 22 Part 1 Preliminaries

. EXAMPLE 1.8 Show that for any odd integer n, 8 n2 – 1 .

SOLUTION: There are, at times, more than one way to stroke a cat: Using Induction Using the Division Algorithm We show that the proposition: We know that for any n there exists q such that: 82m + 1 2 – 1 n ==2q or n 2q + 1 (*) holds for all m  0 (thereby covering all n ==3q or n 3q +31 or n =q + 2 (**) odd integers n). n ====4q or n 4q +41 or n q +42 or n q + 3 I. Valid at m = 0 : 20 + 1 2 –01 = . II. Assume valid at mk= ; that is: While (*) and (**) may not lead us to a fruitful conclusion, the bottom line does. Specifically: 2 2 2k + 1 –81 = t or4k +84k = t For any n: for some integer t. n ====4q or n 4q +41 or n q +42 or n q + 3 III. We are to establish validity at If n is odd, then there are but the two possibilities: mk= + 1 ; that is, that: n ==4q +41 or n q + 3 2 2k + 1 + 1 –81 = s We now show that, in either case 8 n2 – 1 . for some integer s. Let’s do it: If n = 4q + 1 , then: 2k + 1 + 1 2 – 1 n2 –41 ==q + 1 2 –161 q2 ++8q 11– = 8k = 2k + 3 2 – 1 with k = 2q2 + q 2 = 4k ++12k 8 If n = 4q + 3 , then: = 4k2 + 4k + 8k + 8 n2 –41 ==q + 3 2 –161 q2 ++24q 91– = 8h ===8t +88k + 1 tk++1 8s with h = 2q2 ++3q 1 II

CHECK YOUR UNDERSTANDING 1.10

Answer: See page A-5 Prove that for any integer n, n2 = 3q or n2 = 3q + 1 for some integer q.

DEFINITION 1.8 For given a and b not both zero, the greatest GREATEST COMMON common divisor of a and b, denoted by DIVISOR gcd(a,b), is the largest positive integer that divides both a and b.

THEOREM 1.6 If a and b are not both 0, then there exist s and t such that: gcd a b = sa+ tb 1.3 The Division Algorithm and Beyond 23

PROOF: Let Gx=  0 xmanb= + for some m and n Assume, without loss of generality that a  0 . Since both a and –a are of the form ma+ nb : a = 1a + 0b while –1a = – a + 0b ; and since either a or –a is positive: G   . That being the case, the Well Ordering Principle (page 18) assures us that G has a smallest element gsatb= + . We show that ggcdab=  by showing that (1): g divides both a and b, and that (2): every divisor of a and b also divides g. (1) Applying the Division Algorithm we have: aqgr= + with 0  rg . (*) (**) Substituting gsatb= + in (*) brings us to: aqsatb= + + r r = 1 – qs atb– Since r is of the form ma+ nb with rg , it cannot be in G, and must therefore be 0 [see (**)]. Consequently aqg= , and ga. The same argument can be used to show that gb . (2) If da and db , then, by Theorem 1.4(b) and (c), page 15: dg .

CHECK YOUR UNDERSTANDING 1.11 Show that for any a and b not both zero: Answer: See page A-5 gcd a b = gcd a b .

DEFINITION 1.9 Two integers a and b, not both zero, are rela- RELATIVELY PRIME tively prime if: gcd a b = 1 For example: Since gcd15 8 = 1 , 15 and 8 are relatively prime. Since gcd15 9 = 31 , 15 and 9 are not relatively prime.

THEOREM 1.7 Two integers, a and b, are relatively prime if and only if there exist st  Z such that 1 = sa+ tb

PROOF: To say that a and b are relatively prime is to say that gcd a b = 1 . The existence of integers s and t such that 1 = sa+ tb follows from Theorem 1.6. For the converse, assume that there exist integers s and t such that 1 = sa+ tb. Since gcd a b divides both a and b, it divides 1 [Theorem 1.4(b) and (c), page 15]; and, being positive, must equal 1. 24 Part 1 Preliminaries

THEOREM 1.8 Let abc  Z . If abc , and if gcd a b = 1 , then ac.

PROOF: Let s and t be such that: 1 = sa+ tb Multiplying both sides of the above equation by c: csactbc= + Clearly asac . Moreover, since abc : atbc . The result now follows from Theorem 1.4(b), page 15. CHECK YOUR UNDERSTANDING 1.12 Let abc  Z . Show that if abc and ab , then a and c can not be Answer: See page A-5. relatively prime.

PRIME NUMBERS Chances are that you are already familiar with the important concept of a prime number; but just in case:

DEFINITION 1.10 An integer p  1 is prime if 1 and p are its PRIME only divisors.

For example: 2, 5, 7, and 11 are all prime, while 9 and 25 are not. Moreover, since any even number is divisible by 2, no even number So, 2 is the oddest prime (sorry). greater than 2 is prime.

THEOREM 1.9 If p is prime and if pab , then pa or pb .

PROOF: If pa , we are done. We complete the proof by showing that if pa, then pb: Since the greatest common divisor of p and a divides p, it is either 1 or p. As it must also divide a, and since we are assuming pa , it must be that gcd p a = 1 . The result now follows from Theorem 1.8.

CHECK YOUR UNDERSTANDING 1.13 Let p be prime. Use the Principle of Mathematical Induction to show Answer: See page A-5.  that if pa1a2 an , then pai for some 1 in . The following result is important enough to be called the Fundamen- tal Theorem of Arithmetic. THEOREM 1.10 Every integer n greater than 1 can be expressed uniquely (up to order) as a product of primes. 1.3 The Division Algorithm and Beyond 25

PROOF: We use API of page 16 (starting at n = 2 ) to establish the existence part of the theorem: I. Being prime, 2 itself is already expressed as a product of primes. II. Suppose a prime factorization exists for all m with 2 mk . III.We complete the proof by showing that k + 1 can be expressed as a product of primes: If k + 1 is prime, then we are done. If k + 1 is not prime, then k + 1 = ab , with 2 abk . By our induction hypothesis, both a and b can be expressed as a product of primes. But then, so can k + 1 = ab . For uniqueness, consider the set: SnZ=  + n has two different prime decompositions Assume that S   (we will arrive at a contradiction). The Well-Ordering Principle of page 17 assures us that S has a smallest element, let’s call it m. Being in S, m has two distinct prime factorizations, say:

m ==p1p2ps q1q2qt

Since p1 p1p2ps and since p1p2ps = q1q2qt we

have p1 q1q2qt . By CYU 1.13, p1 qj for some 1 jt .

Without loss of generality, let us assume that p1 q1 . Since q1 is prime, its only divisors are 1 and itself. It follows, since p1  1 , that p1 = q1 . Consequently:

p1p2ps = q1q2qt  p1p2ps = p1q2qt

 p1p2ps – p1q2qt = 0

 p1p2ps – q2qt = 0

p1  0:  p2ps – q2qt = 0

 p2ps = q2qt two distinct prime decompositions for an integer smaller than m — contradicting the minimality on m in S

THEOREM 1.11 There are infinitely many primes. PROOF: Assume that there are but a finite number of primes, say Sp= 1p2  pn , and consider the number:

mp= 1p2pn + 1 26 Part 1 Preliminaries

Since mS , it is not prime. By Theorem 1.10, some prime must

divide m. Let us assume, without loss of generality, that p1 m . Since

p1 divides both m and p1p2pn : p1 mp– 1p2pn [Theorem

1.4(b), page 15]. A contradiction, since mp–11p2pn = .

CHECK YOUR UNDERSTANDING 1.14 Answer: See page A-5. Let a and b be relatively prime. Prove that if an and bn , then ab n . 1.3 The Division Algorithm and Beyond 27

EXERCISES

Exercises 1-3. For given a and d, determine integers q and r, with 0  r  d , such that ad= qr+ . 1.a ==0 d 1 2.a ==–5 d 133 3. a ==–134 d 5 Exercises 4-6. Find the greatest common divisor of a and b. 4.a ==120 b 880 5.a ==–10 b 55 6. a ==–134 b 5

Exercises 7-10. The least common multiple of nonzero integers a1a2  an , written lcma1a2  an , is the smallest positive integer that is a multiple of each ai ; i.e. is divisible by each ai . Find: 7.lcm 12 20 8.lcm359 9.lcm23915 10. lcm– 3 2421

e1 e2 en f1 f2 fn 11. Let ap= 1  p2 pn and ap= 1  p2 pn , where the pi s are distinct primes and where ei  0 and fi  0 for all i. Let mi = minei f1 (the smaller of the two numbers), and Mi = maxei f1 (the larger of the two numbers). Prove that:

m1 m2 mn M1 M2 Mn (a) gcdab = p1  p2 pn (b) lcdab = p1  p2 pn (see Exercise 7-10)

12. Prove that if 3 does not divide n, then n = 3k + 1 or n = 3k + 2 for some kZ .

13. Let n be such that 3 n2 – 1 . Show that 3 n .

14. Show that if n is not divisible by 3, then n2 = 3m + 1 for some integer m.

15. Show that an odd prime p divides 2n if and only if p divides n.

16. Prove that if a = 6n + 5 for some n, then a = 3m + 2 for some m.

17. Show that 2 n4 – 3 if and only if 4 n2 + 3 .

18. Prove that any two consecutive odd positive integers are relatively prime. 19. Let a and b not both be zero. Prove that there exist integers s and t such that nsatb= + if and only if n is a multiple of gcd a b .

20. Prove that the only three consecutive odd numbers that are prime are 3, 5, and 7.

21. Show that a prime p divides n2 if and only if p divides n.

22. Prove that every odd prime p is of the form 4n + 1 or of the form 4n + 3 for some n.

23. Prove that every prime p  3 is of the form 6n + 1 or of the form 6n + 5 for some n. 28 Part 1 Preliminaries

24. Prove that every prime p  5 is of the form 10n + 1 , 10n + 3 , 10n + 7 , or 10n + 9 for some n.

25. Prove that a prime p divides n2 – 1 if and only if pn– 1 or pn+ 1 .

26. Prove that every prime of the form 3n + 1 is also of the form 6k + 1 .

27. Prove that if n is a positive integer of the form 3k + 2 , then n has a prime factor of this form as well.

28. Prove that a  1 and b  1 are relatively prime if and only if no prime in the prime decomposition of a appears in the prime decomposition of b.

29. Prove that if the integer n  1 satisfies the property that if nab , then na or nb for every pair of integers a and b, then n is prime.

30. Prove that n  1 is prime if and only if n is not divisible by any prime p with pn .

PROVE OR GIVE A COUNTEREXAMPLE

31. There exists an integer n such that n2 = 3m – 1 for some m. 32. If a = 3m + 2 for some m, then a = 6n + 5 for some n. 33. If m and n are odd integers, then either mn+ or mn– is divisible by 4. 34. For any a, and b not both 0, there exist a unique pair of integers s and t such that gcd a b = sa + tb . 35. For every n, 34n – 1 . 36. For every nZ + , 34n + 1 . 1.4 Equivalence Relations 29

§4. EQUIVALENCE RELATIONS

Recall that XY , called In Section 2 we defined a function from a set X to a set Y to be a sub- the Cartesian Product set fXY  such that: of X with Y, is the set of all ordered pairs xy , For every xX there exists a unique yY with xy  f . with xX and yY . Removing all restrictions, we arrive at a far more general concept than that of a function: DEFINITION 1.11 A relation E from a set X to a set Y is any RELATION subset EXY  . A relation from a set X to X is said to be a relation on X. Each and every subset of  , including the chaotic one in the margin, is a relation on  , suggesting that Definition 1.11 is a tad too general. Some restrictions are in order:

DEFINITION 1.12 A relation E on a set X is a subset EXX  and is said to be: REFLEXIVE Reflexive: xx  E for every xX . (Every element of X is related to itself) SYMMETRIC Symmetric: If xy  E then yx  E . (If x is related to y, then y is related to x) TRANSITIVE Transitive: If xy  E and yz  E then xz  E . (If x is related to y, and y is related to z, then x is related to z) EQUIVALENCE An equivalence relation on a set X is a relation RELATION that is reflexive, symmetric and transitive. The notation x~y is often used to indicate that x is related to y with respect to some understood relation E. Utilizing that option, we can rephrase Definition 1.12 as follows: An equivalence relation ~ on a set X is a relation which is Reflexive: if x~x for every xX . Symmetric: if x~y , then y~x . and Transitive: if x~y and y~z , then x~z .

a c EXAMPLE 1.9 Show that the relation ---~--- if ad= bc is an b d equivalence relation on the set of rational numbers. 30 Part 1 Preliminaries

SOLUTION: a a As you know, when it Reflexive: ---~--- since ab= ba. comes to rational num- b b bers, one simply writes a c c a 2 4 2 4 Symmetric: ---~---  ad ==bc cb da  ---~--- . --- = --- rather than ---~--- . b d d b 3 6 3 6 a c c e Transitive: ---~--- and ---~--  ad ==bc and cf de b d d f (*) (**) a e We establish the fact that ---~-- by showing that af= be : b f bc bc de af ==------ f ------ ------=be d d c see (*) see (**)

Recall that ab means EXAMPLE 1.10 Show that the relation a~b if 23ab– is that a divides b (see Defi- an equivalence relation on Z. nition 1.7, page 15). SOLUTION: The relation a~b if 23ab– is: Reflexive. a~a , since: 3aa–2= a here, a is playing the role of b Symmetric. Assume that a~b , which is to say, that: 3ab–2= h for some hZ (*) We are to show that b~a , which is to say, that: An expression of the form 3ba–2= n for some nZ 2hb+ a = ------is unaccept- Lets do it. From (*) b = 3a – 2h . 3 able in the solution pro- Hence: 3ba–33===a – 2h –2a 4a – 3h 2n cess, since we are involved with the set Z of integers and not “fractions.” TRANSITIVE: Assume that a~b and bc ; which is to say, that: (1) 3ab–2= h and (2) 3bc–2= k for hk  Z We are to show that a~c ; which is to say, that: 3ac–2= n . Let’s do it. From (2): c = 3b – 2k . Hence: 3ac–3= a – 3b – 2k From (1): ===2hb+2– 3b – 2k hkb+ – 2n

CHECK YOUR UNDERSTANDING 1.15 Two sets A and B are said to have the same cardinality, written CardA = CardB , if there exists a bijection f: AB . Show that the relation A~B if CardA = CardB is an equivalence relation on any collection S of sets. Note: In a sense, the term “same cardinality” can be interpreted to mean “same number of elements.” The classier terminology is used since the expression “same Answer: See page A-6 number of elements” suggests that we have associated a number to each set, even those that are infinite. A further discussion on cardinality if offered in the exercises. 1.4 Equivalence Relations 31

DEFINITION 1.13 Let ~ be an equivalence relation on X. For each x  X , the equivalence class of x , EQUIVALENCE 0 0 CLASS denoted by x0 , is the set: x0 = xXx ~x0

In words: The equivalence class of x0 consists of all elements of X that are related to x0 . We now show that any element in x0 will generate the same equivalence class: THEOREM 1.12 Let ~ be an equivalence relation on X. For any x1 x2  X : x1  x2  x1 = x2

PROOF: Assume that x1  x2 . We show that x1  x2 (a similar argument can be used to show that x2  x1 and that therefore x1 = x2 ): xx 1  xx 1

By transitivity, since x1~x2: xx 2  xx 2

Conversely, if x1 = x2 , then, since x1  x2 : x1  x2 .

EXAMPLE 1.11 Determine the set n nZ of equivalence classes corresponding to the equivalence relation a~b if 23ab– of Example 1.10.

SOLUTION: Let’s start off with a = 0 . By definition: 0 ==bZ 2 –b 2nn Z  the even integers Since 1 is not in [0], 1 will differ from 0 (Theorem 1.12). Specifically: 1 ==bZ 21– b 2n + 1 nZ (the odd integers)

In the above example the give equivalence relation decomposed Z into disjoint equivalence classes; namely: Z ==0  1 even integers  odd integers To put it another way: the equivalence classes in Example 1.11 effected a partition of Z, where: 32 Part 1 Preliminaries

DEFINITION 1.14 A set of nonempty subsets S   A of a To put it roughly: PARTITION set X is said to be a partition of X if: A partition of a set S chops S up into disjoint pieces. (i) XS=     A

(ii) If S  S   then S = S

In the above, S is being indexed by the set A, as is the case with the union    A S A = 12   . In particular: If , then:   A S ==S S  S and S ==S S  S    A    12 1 2     1 2   A   12

And, if AZ==+ 123 , then:   S ==S S and S =S    A i iZ + i i = 1    i iZ + i = 1

Figure 1.5(a) displays a 5-subset partition S1S2 S3 S4 S5 of the indicated set. An infinite partition of 0  is represented in Figure  1.5(b): nn + 1 n = 0

S5  S2 [ )[ )[ )[ )[ )[ )[ S 0 1 2 3 4 5 6 4 S3 (a) (b) Figure 1.5 CHECK YOUR UNDERSTANDING 1.16 Determine if the given collection of subsets of  is a partition of  ?

(a) nn + 1 nZ

(a): No (b): Yes   (b) n nZ ii + 1 i = 0  –1i –  –i i = 0 There is an important connection between the equivalence relations on a set X and the partitions of X, and here it is: THEOREM 1.13 (a) If ~ is an equivalence relation on X, then the set of its equivalence classes, x xX , is a partition of X.

(b) If S   A is a partition of X, then the relation x1~x2 if x1 x2  S for some   A is an equivalence relation on X. 1.4 Equivalence Relations 33

PROOF: (a) We Show that: (i) Xx=   xX

and (ii) If x1  x2  , then x1 = x2 . (i): Since  is an equivalence relation, x~x for every xX . It follows that xx  for every xX , and that therefore Xx=  . xX

(ii): If x1  x2  , then there exists x0  x1  x2 .

Since x0  x1 and x0  x2 : x0~x1 and x0~x2 .

By symmetry and transitivity: x1~x2

By Theorem 1.12: x1 = x2

(b) Let S   A be a partition of X. We show that the relation:

x1~x2 if there exists   A such that x1 x2  S is an equivalence relation on X: Reflexive: To say that x~x , is to say that x belongs to the same S as itself, and it certainly does.

Symmetric: x~y    Ax  yS   yx  S  y~x Transitive: Assume x~y and y~z . We show that x~z :

Since x~y : xy  S for some   A . Since z~y : yz  S for some   A .  Since S S  (y is contained in both sets): S = S .     It follows that both x and z are in S (or in S if you prefer),   and that, consequently: x~z .

CONGRUENCE MODULO n Here is a particularly important equivalence relation of the set of integers:

THEOREM 1.14 Let nZ + . The relation a~b if na– b is an equivalence relation on Z.

PROOF: Reflexive: a~a since na– a . Symmetric: a~bnab –  nb– a b~a . 34 Part 1 Preliminaries

Transitive: a~b and b~c  na– b and nb– c

Theorem 1.4(b), page 15:  nab– + bc–  na– c ac

In the event that na– b , we say that: a is congruent to b modulo n and write ab mod n

THEOREM 1.15 Let nZ + . If aa mod n and bb mod n , then: (a) ab+  ab+ mod n (b) ab ab mod n

PROOF: (a) If na– a and nb– b , then: naa– + bb–  nab+ – ab+ (a) If na– a and nb– b , then: (1) aa– = hn and (2) bb– = kn for hk  Z We are to show that nabab– ; which is to say that ab– ab = ns Lets do it: ab– ab = ab– ab + ab– ab = aa– babb+ – ==hnb+ akn nhb+ ak =ns

CHECK YOUR UNDERSTANDING 1.17

+ Let nZ . Let ad= an + ra and bd= bn + rb with 0  ra  n and 0  rb  n (see Theorem 1.5, page 21). Prove that:

ab mod n if and only if ra = rb Answer: See page A-6 (same remainder when dividing by n) Theorem 1.13 assures is that the equivalent classes associated with the equivalence relation of Theorem 1.15 partition the set of integers. Focusing on n = 5 , we see that the equivalence class containing 0 consists of all multiples of 5, as the remainder of any multiple of 5 when divided by 5, is the same as that obtained by dividing 0 by 5 (see CYU 1.17). Specifically:

0 5 =  –20 – 15 – 10 – 5 0 5 10 15 20  Note that the above equivalence class has many “names”. It can also, be called the equivalent class containing 235, among infinitely many other choices: 1.4 Equivalence Relations 35

 0 5 ===125 5 –15 5 The same can be said about the four remaining equivalence classes:

1 5 = –14 – 9 – 4161116

2 5 = –13 – 8 – 3271217

3 5 = –12 – 7 – 2381218

4 5 = –11 – 6 – 1491319 Note that 5 = 0 . Can we define a sum on the above five equivalence classes? Yes:

a 5+ b 5 = ab+ 5 The above sum is well defined, in that it is independent of the chosen representatives in the two equivalence classes. Indeed:

THEOREM 1.16 + For given nZ , let Z n denote the set of equivalence classes associated with the equivalence relation a~b if na– b ; i.e:

Z n = 0 n1 n  n – 1 n Then:

(a) For any a n b n  Z n , the operation

a n+ b n = ab+ n is well defined.

(b) For any a n b n c n  Z n : a n+ b n + c n = a n+ b n+ c n (associative property)

PROOF: (a) We show that if a n = a n and b n = b n , then

ab+ n = ab+ n (i.e the sum is independent of the chosen representatives for the equivalence classes a n and b n ): a n ==a n  na– a aa– hn , for hZ

and: b n ==b n  nb– b bb– kn , for kZ . Since ab+ – ab+ ==aa– – bb– hk– n :

ab+ n = ab+ n

(b) a n+ b n + c n ==ab+ n+ c n ab+ + c n

= abc+ + n

= a n+ b n+ c n 36 Part 1 Preliminaries

CHECK YOUR UNDERSTANDING 1.18

(a) Verify that the product a nb n = ab n in Zn is well

defined. That is: if a n = a n and b n = b n , then:

ab n = ab n .

(b) Prove that a nb nc n = a nb n c n

EXERCISES

Exercises 1-3. Show that the given relation is an equivalence relation on Z. 1.a~b if ab= . 2.a~b if 2 a – 3b . 3.a~b if 5 ab– . Exercises 4-7. Show that the given relation is an equivalence relation on Q, the set of rational numbers. a c a c a c 4.---~--- if --- – ---  Z . 5.---~--- if 2 bd+ . b d b d b d a c a c 6.---~--- if ad– bc b2 + d2 = 0 . 7.---~--- if ad 2 –0bc 2 = . b d b d Exercises 8-13. Show that the given relation is an equivalence relation on  . 8.x~y if x2 = y2 . 9.x~y if xy= . 10.x~y if x + 1 = y + 1 . 11.x~y if xy–  Z . 12.x~y if sinx = siny + 2 . 13. x~y if x2 –0y2 = . Exercises 14-17. Show that the given relation is an equivalence relation on 2 . . 14.x0 y0 ~x1 y1 if x0 + y0 = x1 + y1 . 15.x0 y0 ~x1 y1 if x0y0 = x1y1 . 2 2 2 2 16.x0 y0 ~x1 y1 if x0 + y0 = x1 + y1 . 17.x0 y0 ~x1 y1 if x0 = x1 . Exercises 18-21. Show that the given relation is not an equivalence relation on 2 .

18.x0 y0 ~x1 y1 if x0 = y1 . 19.x0 y0 ~x1 y1 if x0 – y1 = y0 – x1 .

20.x0 y0 ~x1 y1 if x0x1 = y0y1 . 21.x0 y0 ~x1 y1 if x0x1  0 and y0y1  0 . Exercises 22-30. Determine whether or not the given relation is an equivalence relation on 3 .

22.x0y0 z0 ~x1y1 z1 if y0 = y1 .

23.x0y0 z0 ~x1y1 z1 if x0 ++y0 z0 = x1 ++y1 z1 .

24.x0y0 z0 ~x1y1 z1 if x0 = y1 + z1 .

25. x0y0 z0 ~x1y1 z1 if x0z0 + 2y0  x1z1 + 2y1 .

26.x0y0 z0 ~x1y1 z1 if x0 + 2y0 – 3z0 = x1 + 2y1 – 3z1 . 2 2 27. x0y0 z0 ~x1y1 z1 if x0 ++y0 z0 = x1 ++y1 z1 . 2 2 2 2 2 2 28.x0y0 z0 ~x1y1 z1 if x0 ++y0 z0 = x1 ++y1 z1 .

29. x0y0 z0 ~x1y1 z1 if x0 +++++y0 z0 x1 y1 z1  0.

30.x0y0 z0 ~x1y1 z1 if y0z0 = y1z1 . 38 Part 1 Preliminaries

Exercises 31-34. Determine whether or not the given relation is an equivalence relation on M22 .

31.ab~ ab if ad= . cd cd

32.ab~ ab if abc= abc . cd cd

33.ab~ ab if ad– bc = ad– bc . cd cd

34.ab~ ab if ad– bc = ad– bc . cd cd

Exercises 35-41. Show that the given relation is an equivalence relation on FZ= f: ZZ (the set of functions from Z to Z). 35.f ~ g if f1 = g1 . 36.f ~ g if fn= gn for every nZ . 37.f ~ g if fn= gn for every nZ . 38.f ~ g if 2 fn+ gn for every nZ . 39.f ~ g if fn+ m = gn+ m for every nm  Z . 40.f ~ g if 32fn+ gn for every nZ . 41.f ~ g if 32gf n + fn for every nZ . Exercises 42-47. Describe the set of equivalence classes for the equivalence relation of: 42. Exercise 1 43. Exercise 3 44. Exercise 5 45. Exercise 9 46. Exercise 15 47. Exercise 17 Exercises 48-52. Show that the given collection S of subsets of the set X is a partition of X. 48.X =  , S = –0 0 0  . 49.XZ= , S = 3nn Z 3n + 1 nZ 3n + 2 nZ . 50.XZ= +  Z+ , SS=  whereS = ab gcd a b = n . n nZ + n

51.X =  , SS= b b   where Sb = xy yxb= + .

2 2 2 52.X =  , Sxy=  x + y = r r   . 2 2 2 53.X =  , SS= r r   where Sr = xy x + y = r .

Exercises 53-54. (Congruences) Let nZ + . Use the Principle of Mathematical Induction to show that:   54. If ai  ai mod n for 1 im , then a1 +++a2 am  a1 +++a2 am mod n .  55. If ai  ai mod n for 1 im , then a1a2am  a1a2 am mod n . 1.4 Equivalence Relations 39

56. Show that the relation A~B if CardA = CardB is an equivalence relation on PX for any set X. Suggestion: Consider Theorem 1.1, page 5.

PROVE OR GIVE A COUNTEREXAMPLE

57. The union of any two equivalence relations on any given nonempty set X is again an equivalence relation on X. 58. The intersection of any two equivalence relations on any given nonempty set X is again an equivalence relation on X. + 59. For abnm   Z , let Sn and Sm denote the set of equivalence classes associated with the equivalence relations a~b if na– b and a~b if ma– b , respectively. If nm , then

Sn  Sm . 60. If n  2 , then every integer is congruent modulo n to exactly one of the integers 0  mn . 61. If CX , A~B if AC = BC is an equivalence relation on PX . 62. There exists an equivalence relation on the set 12345 for which each equivalence class contains an even number of elements. 40 Part 1 Preliminaries 2.1 Definitions and Examples 41

2Defloration - a professional takes Mirella's virginity Part 2

Groups §1. DEFINITIONS AND EXAMPLES The following properties reside in the familiar set Z of integers: Property Example:

Closure ab+  Z a bZ 57+  Z

Associative 1.abc+ + = ab+ + c a bZ 541+54+ = + + 1

Identity 2.a + 0 = a a  Z 40+4=

Inverse 3.aa+0– = a  Z 55+0– =

A generalization of the above properties bring us to the definition of a group — an abstract structure upon which rests a rich theory, with numerous applications throughout mathematics, the sciences, architecture, music, the visual arts, and elsewhere: A binary operator on a set X is a function that assigns to DEFINITION 2.1 A group G * , or simply G, is a nonempty any two elements in X an element in X. Since the function GROUP set G together with a binary operator, *, (see value resides back in X, one margin) such that: says that the operator is closed. Associative Axiom: 1.a*b*c = a*b *c for every abc   G .

Evariste Galois defined Identity Axiom: 2. There exists an element in G, which we will the concept of a group in label e, such that a*ea= for every aG . 1831 at the age of 20. He was killed in a duel one Inverse Axiom: 3. For every aG there exists an element, year later, while attempt- a  G such that a a = e . ing to defend the honor of * a prostitute. In particular, Z + is a group; with “+, 0, and –a ” playing the role of We show, in the next section, “*, e, and a ” in the above definition. that both the identity element e and the inverse element a Is the set of integers under multiplication a group? No: of Axioms 2 and 3 are, in fact, both unique and “ambidex- While “regular” multiplications is an associative trous:” binary operator on Z, with 1 as identity, no integer a*ee==*aa  other than 1 has a multiplicative inverse in Z. a*a ==a*ae Yes, there is a number whose product with 2 is 1: 1 1 2  --- = 1 , but ---  Z . 2 2 Bottom line: The set of integers under multiplication is not a group. 42 Part 2 Groups

CHECK YOUR UNDERSTANDING 2.1 Determine if the given set is a group under the given operation. If not, specify which of the axioms of Definition 2.1 do not hold. (a) The set Q of rational numbers under addition. (b)The set  of real numbers under addition. (c)The set  of real numbers under multiplication. (a), (b), and (d) are groups. (d)The set + = r   r  0 of positive real numbers under (c) is not a group. multiplication. We now move Theorem 1.16 of page 35 up a notch: THEOREM 2.1 + For given nZ , let Z n denote the set of equivalence classes associated with the equivalence relation a~b if na– b ; i.e:

Z n = 0 n1 n  n – 1 n

Then: Z n + with a n+ b n = ab+ n is a group

PROOF: We already know that + is a well defined associative operator. The identity and inverse axioms of Definition 2.1 are also met:

Identity: For any a n  Z n : a n+ 0 n ==a + 0 n a n .

Inverses: For any a n  Z n : a n+ –a n ==aa– n 0 n Molding Theorem 2.1 into a more compact form by replacing each equivalent class a n with the smallest nonnegative integer in that class, we come to: THEOREM 2.2 + For given nZ , let Zn = 012 n – 1 ,

You are invited to for- and let a +n br= , where ab+ = dn+ r . mally establish this result in Exercise 51. Then Zn +n is a group. The above sum is called addition modulo n. Note that a+n 0 = a for every aZ n , and that for any aZ n : ana+0– =

For example, if n = 5 then Z5 = 01234 , and:

1+5 2== 3 4+5 4= 3 and 3+5 2 0 344 + mod 5 2.1 Definitions and Examples 43

CHECK YOUR UNDERSTANDING 2.2

Complete the following (self-explanatory) group table for Z4 +4 .

+4 0 1 2 3

0 3 since 0+433= 1 2

21since 2+431=

3 02since 3+432=

Answer: See page A-7. since 1+412= since 3+410=

Groups containing infinitely many elements, like Z + and  + , are said to be infinite groups. Those containing finite may elements,

like Zn +n which contains n elements, are said to be finite groups.

DEFINITION 2.2 Let G be a finite group. The number of ele- ORDER OF A GROUP ments in G is called the order of G, and is denoted by G .

GROUP TABLES AND BEYOND

The group Z4 , with table depicted in Figure 2.1(a), has order 4. Another group of order 4, the so-called Klein 4-group, appears in Figure 2.1(b).

+4 0123 * eabc Z4: K: 00123 eeabc 11230 aaecb 22301 bbcea 33012 ccbae (a) (b) Figure 2.1 Is K really a group? Well, the above table leaves no doubt that the closure and identity axioms are satisfied (e is the identity element). Moreover, each element has an inverse, namely itself: ee=== e aa ebb e and cc =e . Finally, though a bit tedious, you can check directly that the associative property holds [for example: ab acab== and aba==ac b ]. You can also see that K is an abelian group; where:

Abelian groups are also said DEFINITION 2.3 A group G * is abelian if to be commutative groups. ABELIAN GROUP a*bb= *a for every ab  G 44 Part 2 Groups

We will soon show that Z4 and K are the only groups of order 4, but first: THEOREM 2.3 Every element of a finite group G must appear once and only once in each row and each column of its group table.

th PROOF: Let Gea= 1 a2  an – 1 . By construction, the i row

of G’s group table is precisely aieaia1 aia2  aian – 1 . The fact that every element of G appears exactly one time in that row is a consequence of Exercise 50, which asserts that the function f : GG ai given by f g = a g is a bijection. As for the columns: ai i

CHECK YOUR UNDERSTANDING 2.3 Answer: See page A-8. Complete the proof of Theorem 2.3. We now show that the two groups in Figure 2.1 represent all groups An alternative proof is offered in Appendix B, of order four. To begin with, we note that any group table featuring the page B-1. four elements eabc must “start off” as in T in Figure 2.2, for e represents the identity element.

* eabc eeabc T: aa bb cc

* eabc * eabc * eabc eeabc eeabc eeabc E: aae B: aab C: aac bb bb bb cc cc cc

Figure 2.2 Since no element of a group can occur more than once in any row or column of the table, the -box in T can only be inhabited by e, b or c, with each of those possibilities displayed as E, B, and C in Figure 2.2. Repeatedly reemploying Theorem 2.2, we observe that while E leads to two possible group tables, both B and C can only be completed in one way (see Figure 2.3) 2.1 Definitions and Examples 45

* eabc eeabc aaecbE bbcae 1 * eabc * eabc * eabc two options eeabc eeabc eeabc ccbea E: aae aaecb aaecb bb bb bbc * eabc cc cc ccb eeabc aaecb only option only option E bbcea 2 ccbae

* eabc * eabc e a b c * eabc eeabc eeabc e e a b c eeabc B: aab aabce a a b c e aabceB bb bb b b c bbcea cc cc c c e cceab only option only option only option

* eabc * eabc * eabc * eabc eeabc eeabc eeabc eeabc C: aac aaceb aaceb aacebC bb bb bbe bbeca cc cc ccb ccbae only option only option only option Figure 2.3 At this point we know that there can be at most four groups of order 4, and their corresponding group tables appear in Figure 2.4. The group tables for Z4 and K of Figure 2.1 are also displayed at the bottom Fig- ure 2.4.

E : eabc eabc eabc 1 * B: * C: * E2: * eabc eeabc eeabc eeabc eeabc aaecb aabce aaceb aaecb bbcae bbcea bbeca bbcea ccbea cceab ccbae ccbae

+ 0123 Z4: 4 K: * eabc 00123 eeabc 11230 aaecb 22301 bbcea 33012 ccbae Figure 2.4

While table E2 and the Klein group table K are identical, those of the remaining four tables in Figure 2.4 look different. But looks can be deceiving: 46 Part 2 Groups

To show, for example, that Z4 and E1 only differ superficially, we begin by reordering the elements in the first row and first column of Z4 in Figure 2.5(a) from “0, 1, 2, 3” to “0, 2, 1, 3” [see Figure 2.5 (b)]. We then transform Figure 2.5(b) to E1 in (c) by replacing the symbols “0, 2, 1, 3” with the symbols “e, a, b, c,” respectively, and the operator

symbol “+4 ” with “*”. + 0123 Z4: 4 +4 0213 E1: * eabc 00123 00213 eeabc 11230 22031 aaecb 22301 11320 bbcae 33012 33102 ccbea (a) (b) (c) Figure 2.5 So, appearances aside, the group structure of E coincides with that This “appearances aside” 1 concept is formalized in of Z4 . In a similar fashion you can verify that tables B and C of Figure Section 4. 2.4 only differ from table Z4 syntactically.

PERMUTATIONS AND SYMMETRIC GROUPS

For any non-empty set X, let SX = f: XXf is a bijection . We then have: The composition operator “ ” is defined on page 3. THEOREM 2.4 For any non-empty set X, SX  is a group.

PROOF: Turning to Definition 2.1:

Operator. fg  SX : gfS X [Theorem 1.2(c), page 7].

Associative. fgh   SX : hgf = hg f [Exercise 51, page 10].

Identity. fS X : fIX ==IX ff , where IX: XX is the identity function: IX x = x for every xX . –1 Inverse. fS X : ff = IX [Theorem 1.1(b), page 5].

The elements (functions) in SX are said to be permutations (on X), and SX  is said to be the symmetric group on X.

IN PARTICULAR:

For X = 12 n , SX  is called the symmetric group of degree n, and will be denoted by Sn .

Let’s get our feet wet by considering the symmetric croup S3 , the set of permutations on X = 123 . Since there are n! ways of ordering n objects (Exercise 31, page 20), the group S3 consists of 3!== 1 2 3 6 elements: 2.1 Definitions and Examples 47

e  1 2 3 4 5 11 12 13 11 13 12 22 23 21 23 22 21 33 31 32 32 31 33 In a more compact (and more standard) form (see margin), we write: Directly below each ele- 1 2 3 1 2 3 1 2 3 ments of the first row e = , 1 = , 2 = appears its image under the 1 2 3 2 3 1 3 1 2 permutations. The fact that 3 lies below 1 in  , for 1 2 3 1 2 3 1 2 3 4  = ,  = ,  =  example, simply indicates 3 1 3 2 4 3 2 1 5 2 1 3 that the permutation 4 maps 1 to 3: 13 . Note that 0 is the identity function e: 01 = 1 02 =2 and 03 = 3

The symmetric group S3 Figure 2.6 Generalizing the above observation we have:

THEOREM 2.5 The symmetric group Sn of degree n contains n! elements. Juxtaposition may also be used to denote the composition operation in Sn . For EXAMPLE 2.1 Referring to the group S3 featured in Figure 2.6, example, for   Sn : Determine:  represents  –1 (a) 2 4 (b) 42 (c) 2 4  and  =  see margin

SOLUTION: (a) To find 24 we first perform 4 and then apply 2 to the resulting function values: From Figure 2.6: 4 2   2 4 132 1  2 13 13 21 22 221  24 ==2  1 5 32 31 313 3  3

(b) Using the standard form we show that 42 = 3 :

1 2 3 1 2 3 first  :  2  1 2 3 3 1 2 3 1 2    4 2 ==3 1 2 3  1 3 2 then 4: 1 3 2 3 2 1

1 2 3 (c) Tor arrive at the inverse of the permutation  =  , 2 3 1 2 simply reverse its action: 1 2 3 –1 3 1 2 1 2 3 –1 ==== 2 3 1 2 1 2 3 2 3 1 1 48 Part 2 Groups

CHECK YOUR UNDERSTANDING 2.4 Answer: 1 2 3 4 5 With reference to the symmetric group S5 , determine  and  ,  :   5 4 3 2 1 where: 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 :  = and  = 4 3 2 5 1 1 5 2 3 4 5 3 2 1 4

Adhering to convention, we will start using ab (rather than a*b ) to denote the binary operation in a generic group. Under this notation, the symbol a–1 (rather than a ) is used to denote the inverse of a, while e continues to represent the identity element. In a generic abelian group, however, the symbol “+” is typically used to represent the binary operator, with 0 denoting the identity element, and –a denoting the inverse of a. To summarize: In Summery: Original Form Product Form Sum Form (Reserved only for abelian groups) 1.a*bG ab  G 1.ab G 1.ab+  G 2.a*b*c = a*b * c 2.abc= ab c 2.abc+ + = ab+ + c 3. a*ea= 3.ae= a 3.a + 0 = a

4.a*a = e 4. aa–1 = e 4. aa+0– =

SOME ADDITIONAL NOTATION: Referring to the product form, For any positive integer n: For any positive integer n: –n do not express a in the form an represents aaaa na represents aaa+++ +a ----1- (there is no “division” in an n a’s n a’s –n –1 n the group). and a = a and –n ana= – From its very definition we We also define a0 to be e. We also define 0a to be 0. find that the following exponent rules hold in any group G: Utilizing the above notation: For anynm  Z : anam = anm+ DEFINITION 2.4 (Product form) A group G is cyclic if there an m = anm n CYCLIC GROUP exists aG such that Ga= nZ . In the sum form, it is accept- able utilize the notation ab– . (Sum form) An abelian group G is cyclic if By definition: ab– = ab+ – . there exists aG such that GnanZ=  .

GENERATOR In either case we say that the element a is a generator of G, and write Ga=  .

EXAMPLE 2.2 Show that: (a) Z6 is cyclic (b) S3 is not cyclic. 2.1 Definitions and Examples 49

SOLUTION: (a) Clearly Z6 = 1 . In fact, as we now show, 5 is also a generator of Z6 (don’t forget that we are summing modulo 6): Note that: 15 = 5 505=  + 5

25 ==5+65 4 10= 1 6 + 4

35 ==5+65+65 3 15= 2 6 + 3

45 ==5+65+65+65 2 20= 3 6 + 2

55 ==5+65+65+65+65 1 25= 4 6 + 1

65 ==5+65+65+65+655+ 6 0 30= 5 6 + 0

Since every element of Z6 = 012345 is a multiple of 5, we conclude that Z6 = 5 . (b) We could use a brute-force method to verify, directly, that no element of S3 generates all of S3 . Instead, we appeal to the following theorem [and Example 2.2(b)] to draw the desired conclusion. THEOREM 2.6 Every cyclic group is abelian.

PROOF: Let Ga== an nZ . For any two elements as and at in G (not necessarily distinct) we have: asat ===ast+ ats+ atas

CHECK YOUR UNDERSTANDING 2.5

(a) Show that 1 and 5 are the only generators of Z6 .

(b) Show that S2 is cyclic.

Answer: See page A-8. (c) Show that Sn is not cyclic for any n  2 . At this point we have two groups of order 6 at our disposal:

Z6 +n and S3 Do these groups differ only superficially, or are they really different in some algebraic sense? They do differ algebraically in that one is cyclic while the other is not, and also in that one is abelian while the other is not. 50 Part 2 Groups

EXERCISES

Exercise 1-11. Determine if the given set is a group under the given operator. If not, specify why not. If it is, indicate whether or not the group is abelian, and whether or not it is cyclic. If it is cyclic, find a generator for the group. 1. The set 2nn Z of even integers under addition. 2. The set 2n + 1 nZ of odd integers under addition.

3. The set of integers Z, with a*bc= , where c is the smaller of the two integers a and b (the common value if ab= ). ab 4. The set Q+ of positive rational numbers, with a b = ------. * 2 a2 5. The set x   x  0 , with a b = ----- . * b 6. The set 02468 under the operation of addition modulo 10. 7. The set 0123 under multiplication modulo 4. (For example: 2*3 = 2 , since 23 ==614 + 2 ; and 3*3 = 1 , since 33 ==924 + 1 .) 8. The set 01234 under multiplication modulo 5. (See Exercise 7.) 9. The set ab+ 2 ab  Z under addition.

10. The set ab+ 2 ab  Q with not both a and b equal to 0 under the usual multiplication of real numbers. 11. The set ZZ = ab ab  Z , with ab + cd = ac+  bd+ .

Exercise 12-23. Referring to the group S3 : 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 e = ,  = ,  =  = ,  = , =  1 2 3 1 2 3 1 2 3 1 2 3 1 3 2 4 3 2 1 5 2 1 3 determine: 12.  and   2 3 n + 2 4 4 2 13.3 and 3 14.3 for nZ .

n + –2 –3 –n + 15. 1 for nZ . 16.3 and 3 17.3 for nZ .

–n + 2 3 n + 18.3 for nZ . 19.2 and 2 20.2 for nZ .

n + –2 –3 –n + 21. 2 for nZ . 22.2 and 2 23.2 for nZ . 2.1 Definitions and Examples 51

Exercise 24-33. For 123456 123456 123456  =   =   =  234561 214365 654321 Determine: 24.  25.  26.  27.  28.  29. 5 30. 100 31. 101 32. 100 33. 101

34. Let S = 1 . Show that S * with 1*11= is a group. Is the group abelian? Cyclic?

ab ab aa+ bb+ 35. Is M22  + with + = a group? If so, is it abelian? Cyclic? cd cd cc+ dd+

ab ab aa bb 36. Is M22  * with * = a group? If so, is it abelian? Cyclic? cd cd cc dd

ab ab aa+ bc ab+ bd 37. Is M22  * with * = a group? If so, is it abelian? Cyclic? cd cd ca+ dc cb+ dd

* abc 38. Let S=  abc along with the binary operator: aabc . Is S * a group? bbbc cccc

*201 39. Let S = 012 along with the binary operator: 2201 . Is S * a group? 0012 1120

40. Let Sxy=  xy   . Show that S * with xy *xy = xx+ – 1 yy++1 is a group. Is the group abelian? Cyclic?

41. For n  0 , let Pn denote the set of polynomials of degree less than or equal to n. Show that n n n i i i Pn * with a x * b x = a + b x is a group. Is the group abelian?  i  i  i i i = 0 i = 0 i = 0

42. Let Sxy=  xy   . Show that S * with xy *xy = xx++2 yy+ is a group. Is the group abelian?

43. Let Q denote the set of rational numbers. Show that Q * with a*b= abab++ is not a group. 52 Part 2 Groups

44. Let QaQa=   –1 . Show that Q * with a*b= abab++ is a group. Is the group abelian?

45. Let Gea= a  a . Show that the function f : GG given by f g = ga is 1 2 n – 1 ai ai i a bijection

46. (a) Give an example of a group G in which the exponent law ab n = anbn does not hold in a group G, for nZ "" (b) Prove that the exponential law ab n = anbn does hold if the group G is abelian. (c) Express the property ab n = anbn in sum-notation form.

47. Let G be a group and abc  G . Show that if ba= ca , then bc= .

48. Let a be an element in a group G. Show that if a = 2 , then ab= ba for ever bG .

49. Let p and q be distinct primes numbers. Find the number of generators of Zpq .

+ 50. (a) Show that the group Zn of Theorem 2.1 is cyclic for any nZ .

(b) Prove that mZ n is a generator of Zn if and only if m and n are relatively prime. Suggestion: Consider Theorem 1.7, page 23.

51. Let F denote the set of all real-valued functions. For f and g in F , let fg+ be given by fg+ x = fx+ gx . Show that F  + is a group. Is the group abelian? 52. Prove Theorem 2.2.

53. Let G and H be groups. Let GH = gh gGhH   with:

gh *gh = gg hh (a) Show that GH  * is a group. (b) Prove that GH  * is abelian if and only both G and H are abelian.

54. Let X be a set and let PX be the set of all subsets of X. Is PX * a group if: (a) A*BAB=  (b) A*BAB=  2.1 Definitions and Examples 53

PROVE OR GIVE A COUNTEREXAMPLE

55. The set  of real numbers under multiplication is a group

56. The set + = r   r  0 of positive real numbers under multiplication.

57. Let G be a group and abc  G . If bc , then ba ca .

2 58. The group S3 contains four elements  such that  = e and three elements  such that 3 = e .

59. The group S  is abelian.

60. Let G be a group and ab  G . If ab= b , then ac= c for every cG .

61. Let G be a group and ab  G . If ab= ba , then ac= ca for every cG .

62. The cyclic group Z + has exactly two distinct generators. 54 Part 2 Groups

§2. ELEMENTARY PROPERTIES OF GROUPS We begin by recalling the group axioms, featuring both the product and sum notations: Product Form Sum Form (Typically reserved for abelian groups) Closure ab G ab+  G Axiom 1.abc= ab c 1.abc+ + = ab+ + c Identity Axiom 2.ae= a 2.a + 0 = a Inverse Axiom 3. aa–1 = e 3. aa+0– =

For aesthetic reasons, a set of Actually, as we show below, both the identity element of Axiom 2 and axioms should be independent, in that no axiom or part of the inverse elements of Axiom 3 work on both sides; but first: an axiom is a consequence of the rest. One should not, for LEMMA 2.1 Let G be a group. If aG is such that a2 = a , example, replace Axiom 2 in Definition 2.1, page 41: then ae= . eGaeaaG  =   with: PROOF: –1 –1 –1 e  Gaeea = = aa  G aa= a  aa a = aa  aaa= eaeea =  = e Axiom 1 Axiom 3 Axiom 2

In sum form: THEOREM 2.7 Let G be a group. ForaG : aa+0– == –a + a 0 a + 0 ==a  0 + a a (a) aa–1 = ea –1ae= (b) ae= a ea = a

PROOF:

(a) a–1a a–1a ===a–1aa–1 aa–1e aa–1a

Axiom 1 Axiom 3 Axiom 2 We now know that a–1a a–1a = a–1a . Applying Lemma 2. we then have: a–1ae= .

(b) ea==== aa–1 aaa–1a ae a

Axiom 3 Axiom 1 part (a) Axiom 3

Axioms 2 and 3 stipulate the existence of an identity and of inverses in a group. Are they necessarily unique? Yes: THEOREM 2.8 (a) There is but one identity in a group G. (b) Every element in G has a unique inverse. 2.2 Elementary Properties of Groups 55

PROOF: (a) We assume that e and e are identities, and go on to show that ee= : Since e is an identity eeee== Since e is an identity

(b) We assume that a–1 and a–1 are inverses of aG , and show that a–1 = a–1 : Since aa–1 and aa–1 are both equal to the identity they must be equal to each other: aa–1 = aa–1 Multiply both sides by a–1: a–1aa–1 = a–1aa–1 Associativity: a–1a a–1 = a–1a a–1 ea–1 = ea–1 a–1 = a–1

CHECK YOUR UNDERSTANDING 2.6

Show that if abc are elements of a group such that abc= e , then Answer: See page A-9. bca= e. Both the left and right cancellation laws hold in groups: In any group G: Sum form: THEOREM 2.9 ab+ = cb+  a = c (a) If ab = cb , then ac= . ba+ ==bc+  a c (b) If ba = bc , then ac=

PROOF: Just in case you are ask- (a) ab= cb (b) ba= bc ing yourself: What if b is 0 and –1 –1 –1 –1 has no inverse? ab b = cb b b ba = b bc Tisk, every element in a group has an inverse. abb–1 = cbb–1 b–1b ab= –1b c ae= ce ea= ec ac= ac=

CHECK YOUR UNDERSTANDING 2.7

PROVE OR GIVE A COUNTEREXAMPLE: (a) In any group G, if ab= bc then ac= .

Answer: See page A-9. (b) In any abelian group G, if ab+ = bc+ then ac= . 56 Part 2 Groups

In the real number system, do linear equations ax= b have unique solutions for every ab   ? No: the equation 0x = 5 has no solution, while the equation 0x = 0 has infinitely many solutions. This observation assures us once more that the reals is not a group under multiplication, since:

THEOREM 2.10 Let G be a group. For any ab  G , the linear equations ax= b and ya= b have unique solutions in G.

PROOF: Existence: (a) ax= b (b) ya= b a–1ax = a–1b ya a–1 = ba–1 a–1a xa= –1b yaa–1 = ba–1 ex= a–1b ye= ba–1 xa= –1b yba= –1

Uniqueness: We assume (as usual) that there are two solutions, and then proceed to show that they are equal: Theorem 2.9(b)

(a) ax1 ==b and ax2 b  ax1 ==ax2  x1 x2 Theorem 2.9(a)

(b) y1ab== and y2ab y1a ==y2ay 1 y2

CHECK YOUR UNDERSTANDING 2.8 Since the set of real numbers under addition is a group, 2.9 applies. Show, directly, that any linear equation in  + has a unique solu- Answer: See page A-9. tion. The inverse of a product is the product of the inverses, but in reverse order:

This is another shoe-sock THEOREM 2.11 For every ab in a group G: theorem (see page 8). ab –1 = b–1a–1

PROOF: To show that b–1a–1 is the inverse of ab is to show that b–1a–1 ab = e . No problem:

b–1a–1 ab ====b–1a–1a bb–1eb b–1be 2.2 Elementary Properties of Groups 57

CHECK YOUR UNDERSTANDING 2.9 Give an example of a group G for which ab –1 = a–1b–1 does not Answer: See page A-9. hold for every ab  G . The axiom of a group G assures us that an expression such as abc , sans parentheses, is unambiguous [since ab c and abc yield the same result]. It is plausible to expect that this nicety extends to any

product a1a2an of elements of G. Plausible, to be sure; but more importantly, True:

THEOREM 2.12 Let a1a2an  G . The product expression

a1a2an is unambiguous in that its value is independent of the order in which adjacent factors are multiplied.

PROOF: [By induction (page 13)]: I. The claim holds for n = 3 (the axiom). II. Assume the claim holds for nk= , with k  3 . III. (Now for the fun part) We show the claim holds for nk= + 1 :

Let x denote the product a1a2ak + 1 under a certain pairing of its elements, and y the product under another pairing of its elements. We are to show that xy= . Let’s do it: Assume that one starts the two multiplication processes with the following pairing for x and y: A B C D xa==1a2ai ai + 1ak + 1 and ya1a2aj aj + 1ak + 1 Case 1. ij= : By the induction hypothesis (II), no matter how the products in A and C are performed, A will equal C. The same can be said concerning B and D. Consequently xABCDy== =. Case 2. Assume, without loss of generality, thatij . Break- ing the “longer” product B into two pieces M and D we have: A M D xa= 1a2ai ai + 1aj aj + 1ak + 1 By the induction hypothesis, A, M, and D are well defined (independent of the pairing of its elements in their products). Bringing us to: I: Claim holds for n = 3 xABAMD== =AM DCDy = = 58 Part 2 Groups

CHECK YOUR UNDERSTANDING 2.10 Use the Principle of Mathematical Induction, to show that for any a1a2an  G : Answer: See page A-9. –1 –1 –1 –1 ana2a1 = a1 a2 an

THEOREM 2.13 For any given element a of a finite group G: am = e for some mZ + .

PROOF: Let G be of order n. Surely not all of the n + 1 elements aa2 a3  an + 1 can be distinct. Choose 1  stn  + 1 such that at = as . Since ata–s ==ats– e : am = e , for mts= – .

DEFINITION 2.5 Let G be a group, and let aG be such that In the additive notation, am = e translates to am = e for mZ + . The smallest such m is ORDER OF AN na = 0 ; which is to say: ELEMENT OF G called the order of a and is denoted by aa++ +a = 0 oa. If no such m exists, then a is said to sum of n as have infinite order.

EXAMPLE 2.3 (a) Determine the order of the element 4 in the group Z6 +6 .

(b) Determine the order of the element 1 2 3 4 5  =  3 2 4 1 5

in the symmetric group S5 . SOLUTION: (a) Since: 14= 4

24==4+642

34===4+64+642+640

The element 4 has order 3 in Z6 . (b) Since: 1 2 3 4 5   3 2 4 1 5 2 e 4 2 1 3 5 3   1 2 3 4 5

1 2 3 4 5 The element  =  has order 3 in S . 3 2 4 1 5 5 2.2 Elementary Properties of Groups 59

CHECK YOUR UNDERSTANDING 2.11 1 2 3 4 (a) Determine the order of the element  =  in S . 2 3 4 1 4

(b) Determine the order of the element 4 in Z24 .

n (c) Let aZ n . Prove that om= ------Answer: (a) 4 (b) 6 gcd a n (c) See page A-10. See Definition 1.8, page 22

Note: There is no “subtraction” in a group G + . For convenience, however, for given ab  G , we define the symbol ab– as follows: ab– = ab+ – (add the additive inverse of b to a) There is no “division” in a group G . . In this setting, a however, one does not ever substitute the symbol --- for b ab–1 . Why not? Convention. 60 Part 2 Groups

EXERCISES

1. Let G be a group and abc  G . Solve for x, if: (a) axa–1 = e (b) axa–1 = a (c) axb= c (d) ba–1xab–1 = ba

2. Let G be a group. Prove that a–1 –1 = a for every aG .

3. Prove that for any element a in a group G the functions fa: GG given by fab = ab and

the function ga: GG given by gab = ba are bijections.

4. Let a be an element of a group G. Show that GabbG= 

5. Let G be a group and let aG . Show that if there exists one element xG for which ax= x , then ae= .

6. Let a be an element of a group G for which there exists bG such that ab= b . Prove that ae= .

7. Prove that a group G is abelian if and only if ab –1 = a–1b–1 for every ab  G .

8. Let G be group for which a–1 = a for every aG . Prove that G is abelian.

9. Let G be group for which ab 2 = a2b2 for every ab  G . Prove that G is abelian.

10. Let G be a finite group consisting of an even number of elements. Show that there exists aG , ae , such that a2 = e .

11. Let G be a group. Show that if, for any ab  G , there exist three consecutive integers i such that ab i = aibi then G is abelian.

12. Let * be an associative operator on a set S. Assume that for any ab  S there exists cS such that a*cb= , and an element dS such that d*ab= . Show that S * is a group.

–1 13. Let G be a group and aG . Define a new operation * on G by b*cba= c for all bc  G . show that G * is a group.

14. Let G be a group and ab  G . Use the Principle of Mathematical Induction to show that for any positive integer n: a–1ba n = a–1bna . 15. Let a and b be elements of a group G. Show that if ab has finite order n, then ba also has order n. 2.2 Elementary Properties of Groups 61

16. Let G be a cyclic group of order n. Show that if m is a positive integer, then G has an element of order m if and only if m divides n.

17. Let G be a group. Show that for every element aG and for any nZ : a–n = a–1 n .

18. Let G be a finite group, and ab  G . Prove that the elements aa –1 and bab–1 have the same order.

19. List the order of each element in the Symmetric group S3 of Figure 2.6, page 47.

20. Let aG be of order n. Prove that as = at if and only if n divides st– .

21. Prove that if a2 = e for every element a in a group G, then G is abelian.

22. Let * be an associative operator on a finite set S. Show that if both the left and right cancellation laws of Theorem 2.8 hold under *, then S * is a group.

PROVE OR GIVE A COUNTEREXAMPLE

23. If abc are elements of a group such that abc= e , then cba= e .

24. In any group G there exists exactly one element a such that a2 = a .

25. In any group G, ab –2 = b–2a–2 .

26. Let G be a group. If abc= bac then ab= ba .

27. Let G be a group. If abcd= bacd then ab= ba .

28. Let G be a group. If abc –1 = a–1b–1c–1 then ac= . 62 Part 2 Groups

§3. SUBGROUPS

DEFINITION 2.6 A subgroup of a group G is a nonempty SUBGROUP subset H of G which is itself a group under the imposed binary operation of G. As it turns out, apart from closure, to determine whether or not a non- GROUP AXIOMS empty subset of a group is a subgroup you need but challenge Axiom3: Closure: ab G Axiom 1. abc= ab c THEOREM 2.14 A nonempty subset S of a group G is a subgroup Axiom 2. ae= a of G if and only if: Axiom 3. aa–1 = e (i) S is closed with respect to the operation in G. (ii) sS implies that s–1  S .

PROOF: If S is a subgroup, then (i) and (ii) must certainly be satisfied. Conversely, if (i) and (ii) hold in S, then Axioms 1 and 2 also hold: Axiom 1: Since abc= ab c holds for every abc  G , that associative property must surely hold for every abc  S. Axiom 2: Since ae= a for every aG , then surely se= s for every sS . It remains to be shown that eS . Lets do it: Choose any sS . By (ii): s–1  S . By (i): ss–1 = eS .

When challenging if SG is a subgroup, we suggest that you first determine if it contains the identity element. For if not, then S is not a subgroup, period. If it does, then S   and you can then proceed to challenge (i) and (ii) of Theorem 2.14.

EXAMPLE 2.4 Show that for any fixed nZ the subset For example: 5Z = –510 –0510 nZ=  nm m Z is a subgroup of Z + .

SOLUTION: Since 0 = n0  nZ, nZ   . (i) nZ is closed under addition:

nm1 + nm2 = nm1 + m2  nZ

We remind you that, under (ii) For any nm nZ : addition, –a rather than –nm = nm–  nZ a–1 is used to denote the inverse of a. Conclusion: nZ is a subgroup of Z (Theorem 2.14). 2.3 Subgroups 63

CHECK YOUR UNDERSTANDING 2.12 The previous example assures us that 3Z is a subgroup of Z + . As Answer: See page A-10. such, it is itself a group. Show that 6Z is a subgroup 3Z . You are invited to show in the exercises that the following result holds for any collection of subgroups of a given group: THEOREM 2.15 If H and K are subgroups of a group G, then HK is also a subgroup of G.

PROOF: Since H and K are subgroups, each contains the identity element. It follows that eHK  and that therefore HK . We now verify that conditions (i) and (ii) of Theorem 2.13 are satisfied: (i) (Closure) If ab  H K , then ab  H and ab  K . Since H and K are subgroups, ab H and ab K . It follows that ab H K. (ii) (Inverses) If aHK  , then aH and aK . Since H and K are subgroups, a–1  H and a–1  K . consequently, a–1  HK .

CHECK YOUR UNDERSTANDING 2.13 Let H and K be subgroups of a G for which HK = e . Prove:

Answer: See page A-10. h1k1 ===h2k2  h1 h2 and k1 k2 , We recall the definition of a cyclic group appearing on page 48: A group G is cyclic if there exists aG such that Ga= n nZ .

DEFINITION 2.7 Let G be a group, and aG . The cyclic group a = an nZ is called the cyclic subgroup of G generated by a. (In sum form: a = na n Z )

CHECK YOUR UNDERSTANDING 2.14 Answer: 3 = Z8 4 = 04 For GZ= 8 , determine 3 and  4 . (Use sum notation.) 64 Part 2 Groups

THEOREM 2.16 Every subgroup of a cyclic group is cyclic.

PROOF: Let H be a subgroup of Ga=  . If ae= , then He=  , which i s cyclic. If ae then let m be the smallest positive integer such that am  H . We show Ha= m by showing that every an  H is a power of am : Employing the Division Algorithm of page 21, we chose integers q and r, with 0  rm , such that: nmqr= + . And so we have: an ==amq+ r am qar (*) or: ar = am –qan (**) Since an and am are both in H, and since H is a group: am –qan  H . Consequently, from (**): ar  H . Since 0  rm and since m is the smallest positive integer such that am  H : r = 0 . Consequently, from (*): an ==am qa0 am q — a power of am .

CHECK YOUR UNDERSTANDING 2.15 Let Ga=  with Gn= . Let bG with ba= s . Prove that: n ob= ------Answer: See page A-11. gcd n s

SUBGROUPS GENERATED BY SUBSETS OF A GROUP We have seen that any element a in a group G can be used to generate a subgroup of G — namely the cyclic group generated by a: a = an nZ Generalizing the above concept, we start off with a nonempty subset A of G, and consider the set A of all elements of G consisting In the event that G is abelian, the elements of of finite products of elements of A , wherein repetitions of its ele- A can be expressed ments may occur. For example, if A=  abc , then: in a non-repetition form, a3 , c–2b3 , aa–1 = e , and ab2c–3a3c2a are all in A . as with: ab2c–3a3c2aa= 5b2c–1 Note that, by its very definition, A is a subgroup of G: A is certainly not empty and closed under multiplication. Moreover, the inverse of any element in A is again of the form which positions it in A . For example: ab2c–3a3c2a –1 = a–1c–2a–3c3b–2a–1 (see CYU 2.10, page 58) 2.3 Subgroups 65

Bringing us to: DEFINITION 2.8 Let A be a nonempty subset of a group G. GENERATED The subgroup of G generated by A, SUBGROUP denoted by A , consisting of all finite products of elements of A In particular, here is the subgroup of Z + generated by 212 : Note that commutativity enables us to gather all of 212 ==2n12m nm  Z 2n3m nm  Z the 2s and 3s together. since 12= 22  3

THEOREM 2.17 Let A be a nonempty subset of a group G. The following are equivalent: (i) SA=  (ii) S is the intersection of all subgroups of G containing A.

PROOF: i  ii : Since subgroups are closed under multiplication, any subgroup of G that contains A, including the subgroup A has to contain A . It follows that A is the intersection of all subgroups of G that contain A. ii  i Your turn:

CHECK YOUR UNDERSTANDING 2.16 Answer: See page A-11. Verify that ii  i . Here is a particularly important result: THEOREM 2.18 If G is a finite group and H is a subgroup of Joseph-Louis Lagrange G, then the order of H divides the order of G: (1736-1813). (Lagrange) H G (see Definition 2.2, page 43) We will eventually show that the converse of Lagrange’s Theorem holds for abelian To illustrate: If a group G contains 35 elements, it cannot groups. It does not, however, contain a subgroup of 8 elements, as 8 does not divide 35. hold in general (Exercise 28, page 102). A proof of Lagrange’s Theorem is offered at the end of the section. At this point, we turn to a few of its consequences, beginning with: THEOREM 2.19 Any group G of prime order is cyclic.

PROOF: Let Gp= , where p is prime. Since p  2 , we can choose an element aG distinct from e . By Lagrange’s theorem, the order of the cyclic group a = an nZ must divide p. But only 1 and p 66 Part 2 Groups

divide p, and since a contains more than one element, it must contain p elements, and is therefore all of G.

The symmetric group S3 THEOREM 2.20 Every group of order less than 6 is abelian. is an example of a non- abelian group of order 6. PROOF: We know that Z4 and the Klein group are the only groups of order 4, and that each is abelian. The trivial group e of order 1 is clearly abelian. Any group or order 2 or 3, being of prime order, must be cyclic (Theorem 2.19), and therefore abelian (Theorem 2.6, page 49).

We remind you that oa THEOREM 2.21 For any element a in a finite group G: denotes the order of a (Definition 2.5, page 58). oa G

PROOF: If oa= m , then a = aa2  am – 1 am = e is a subgroup of G consisting of m elements. Consequently: oa G .

THEOREM 2.22 If G is a finite group of order n, then an = e for every aG . (Sum notation: na = 0 for every aG )

PROOF: Let aG , with oa= m . Since m divides n (Lagrange’s Theorem), ntm= for some tZ + . Thus: an ==atm am t ==et e

CHECK YOUR UNDERSTANDING 2.17

Determine the subgroup of the symmetric group S3 : 1 2 3 1 2 3 1 2 3 e = , = ,  =  1 2 3 1 2 3 1 2 3 1 2

1 2 3 1 2 3 1 2 3  = ,  = ,  =  3 1 3 2 4 3 2 1 5 2 1 3

Answer: S3 generated by the set 2 3 .

PROOF OF LAGRANGE’S THEOREM We begin by recalling some material from Chapter 1: See Definition 1.11 page 29. An equivalence relation ~ on a set X is a relation which is Reflexive: x~x for every xX , Symmetric: If x~y , then y~x , Transitive: If x~y and y~z , then x~z . See Definition 1.13 page 31. For x0  X the equivalence class of x0 is the set:

x0 = xXx ~x0 . 2.3 Subgroups 67

LEMMA 2.2 Let H be a subgroup of a group G. The relation a~b if ab–1  H is an equivalence relation on G. Moreover, the equivalence class containing aG is the set: a = ha h H

PROOF: ~ is reflexive: x~x since xx–1 = eH . ~ is symmetric:. a~b  ab–1 = h for some hH H is a group  ab–1 –1 = h–1 Theorem 2.11, page 56:  b–1 –1a–1 = h–1 Exercise 2, page 60:  ba–1 = h–1  b~a since h–1  H ~ is transitive: If a~b and b~c , then: ab–1  H and bc–1  H  ab–1 bc–1  H  ab –1b c–1  H

 aec–1  H  ac–1  H  a~c Having established the equivalence part of the theorem, we now verify that a = ha h H : ba  b~a  ba–1 = h for some hH  b = ha for some hH

NOTE: The above set ha h H will be denoted by Ha , and is said to be a right coset of H: Ha=  ha h H

We are now in a position to offer a proof of Lagrange’s Theorem: If H is a subgroup of a finite group G, then H G .

PROOF: Theorem 1.13(a), page 32, and Lemma 2.2, tell us that the right cosets of H, Ha a G , partition G. Since G is finite, we can k choose a1a2  ak such that GHa=  i with Hai Haj = i = 1 if ij . 68 Part 2 Groups

We now show that each Hai has the same number of elements as H,

by verifying that the function fi: HHa i given by fih = hai is a bijection:

fi is one-to-one: fih1 = fih2  h1a = h2a –1 –1  h1a a = h2a a  h aa–1 ==h aa–1  h h 1 2 1 2 fi is onto: For any given hai  Hai , fih = hai .

Since G is the disjoint union of the k sets Ha1Ha2  Hak , and since each of those sets contains H elements: GkH= , and therefore: H G . 2.3 Subgroups 69

EXERCISES

Exercise 1-5. Determine if the given subset S is a subgroup of Z + . 1.Snn=  is even 2.Snn=  1 3. Snn=  is odd 4.Snn=  is divisible by 2 and 3 5. Snn=  is divisible by 2 or 3

Exercise 6-8. Determine if the given subset S is a subgroup of Z8 +n (see Theorem 2.1, page 42). 6.S = 0246 7.S = 036 8. S = 0234

Exercise 9-12. Determine if the given subset S is a subgroup of  + . 9.Sxx= = 7y for y   10. Sxx= = 7y for y  0 11.Sxx= = 7 + y for y   12. Sxx= = 7 + y for y  0

Exercise 13-18. Determine if the given subset S is a subgroup of S3  where: 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3  = ,  = ,  =   = ,  = ,  =  0 1 2 3 1 2 3 1 2 3 1 2 3 1 3 2 4 3 2 1 5 2 1 3

13.S = 0 1 14.S = 0 2 15. S = 0 3

16.S = 01 2 17.S = 03 5 18. S = 12 3 4 5

Exercise 19-21. Determine if the given subset S is a subgroup of R3 + . 19.Sab= 0 ab   20. Sab= 1 ab   21.S=  abc cab= + 22. S=  abc cab=

Exercise 23-26. Determine if the given subset S is a subgroup of M22  + .

a b a b 23.S = ab   24. S = ab   ab+ 0 ab+ 1

a b a b 25.S = ab   26. S = abc  ab+ ab c 2ac+

Exercise 27-30. Determine if the given subset S is a subgroup of F  + (see Exercise 49, page 52). 27.Sff=  is continuous 28. Sff=  is differentiable 29.Sff= 1 = 1 30. Sff= 1 = 0

Exercise 31-34. Determine if the given subset S is a subgroup of S  (see Theorem 2.4, page 46). 31.Sff=  is continuous 32. Sff=  is differentiable 33.Sff= 1 = 1 34. Sff= 1 = 0 70 Part 2 Groups

35. Prove that all subgroups of Z are of the form nZ .

36. Find all subgroups of Z6 +n . 37. Prove that if e and G are the only subgroups of a group G, then G is cyclic of order p, for p prime. 38. Show that a nonempty subset S of a group G is a subgroup of G if and only if ab  S ab–1  S 39. Show that for any ab  Z+ , Snamb= + nm  Z is a subgroup of Z. 40. Show that for any group G the set ZG= aGag = ga gG is a subgroup of G. 41. Let G be an abelian group. Show that for any integer n, aGa n = e is a subgroup of G. 42. Prove that the subset of elements of finite order in an abelian group G is a subgroup of G (called the torsion subgroup of G). 43. Let G be a cyclic group of order n. Show that if m is a positive integer, then G has an element of order m if and only if m divides n. 44. Let a be an element of a group G. The set of all elements of G which commute with a: Ca= bGab = ba is called the centralizer of a in G. Prove that Ca is a subgroup of G. 45. Let H be a subgroup of a group G. The centralizer CH of H is the set of all elements of G that commute with every element of H: CH= aGah = ha for all hH . Prove that CH is a subgroup of G. 46. The center ZG of a group G is the set of all elements in G that commute with ever element of G: ZG= aGab = ba for all bG . (a) Prove that ZG is a subgroup of G. (b) Prove that aZG  if and only if Ca= G (see Exercise 43.) (c) Prove that ZG=  Ca . aG 47. Show that Table C in Figure 2.4, page 45, can be derived from Table B by appropriately relabeling the letters e, a, b, c in B. 48. Let H and K be subgroups of an abelian group G. Verify that HK=  hk h H and kK is a subgroup of G. 49. Let H and K be subgroups of a group G such that k–1Hk H for every kK . Show that HK=  hk h H and kK is a subgroup of G.

50. Prove that H is a subgroup of a group G if and only if HH–1 = ab–1 ab  H  H . 51. Let H and K be subgroups of an abelian group G of orders n and m respectively. Show that if HK = e , then HK=  hk h H and kK is a subgroup of G of order nm. 2.3 Subgroups 71

52. (a) Prove that the group Z + contains an infinite number of subgroups. (b) Prove that any infinite group contains an infinite number of subgroups. 53. Let S be a finite subset of a group G. Prove that S is a subgroup of G if and only if ab S for every ab  S . n 54. (a) H n be subgroups of a group G. Show that H is also a subgroup of G. i i = 1  i i = 1  (b) Let H  be a collection of subgroups of a group G. Show that H is also a i i = 1  i subgroup of G. i = 1 (c) Let H be a collection of subgroups of a group G. Show that H is also a    A     A subgroup of G.

PROVE OR GIVE A COUNTEREXAMPLE

55. If H and K are subgroups of a group G, then HK is also a subgroup of G. 56. It is possible for a group G to be the union of two disjoint subgroups of G.

57. In any group G, aGa n = e for some nZ is a subgroup of G.

58. In any abelian group G, aGa n = e for some nZ + is a subgroup of G.

59. Let G be a group with ab  G . If oa= n and ob= m , then ab nm = e . 60. If a group G has only a finite number of subgroups, the G must be finite.

61. If H and K are subgroups of a group G, then HK=  hk h H and kK is also a subgroup of G.

62. In any group G, aGa 3 = e is a subgroup of G.

63. No nontrivial group can be expressed as the union of two disjoint subgroups. 72 Part 2 Groups

§4. HOMOMORPHISMS AND ISOMORPHISMS Up until now we have focused our attention exclusively on the internal nature of a group G. The time has come to consider links between them: The word homomorphism comes from the Greek DEFINITION 2.9 A function : GG  from a group G to a homo meaning “same” and HOMOMORPHISM group G is said to be a homomorphism if morph meaning “shape.” ab = a b for every ab  G . Let’s focus a bit on the equation: ab = a b (*) The operation, ab, on the left side of equation (*) is taking place in the group G while that on the right, a b , occurs in the group G . What (*) is saying is that you can perform the product in G and then carry the result over to G (via  ), or you can first carry a and b over to G and then perform the product in that group. Those groups and products, however, need not resemble each other. Consider the following examples:

You can easily verify that EXAMPLE 2.5 Let GZ=  + , and let G = –11 under G = –11 , under stan- standard integer multiplication (see margin). dard multiplication Show that f: GG  given by: *1– 1  1if n is even 11– 1 n =  –11 –1  –if1 n is odd is a group. is a homomorphism. SOLUTION: We consider three cases: Case 1. (Both integers are even). If a = 2n and b = 2m , then: ab+ ==2n + 2m 1 (since 2n + 2m is even) And also: a b === 2n 2m 11 1 . Case 2. (Both are odd). If a = 2n + 1 and b = 2m + 1 , then: ab+ ===2n + 1 + 2m + 1 2n ++2m 2 1 And also: a b === 2n + 1 2m + 1 –1 –1 1 .

Since Z + is abelian, we Case 3. (Even and odd). If a = 2n and b = 2m + 1 , then: need not consider a = 2n + 1 ab+ ===2n + 2m + 1 2nm+ + 1 –1 and b = 2m , And also: a b === 2n 2m + 1 1 –1 –1 .

See page 42 for a discussion EXAMPLE 2.6 Show that the function : Z +  Zn +n of the group Z  + . n n given by m = r where mnqr= + with 0  rn is a homomorphism. 2.4 Homomorphisms and Isomorphisms 73

SOLUTION: Let anq= 1 + r1 , bnq= 2 + r2 with 0  r1  n and 0  r2  n , and let r1 + r2 = nq3 + r3 with 0  r3  n , then:

ab+ = nq1 + r1 + nq2 + r2 = nq1 + q2 + r1 + r2

= nq1 + q2 + nq3 + r3 with 0  r3  n

==nq1 ++q2 q3 + r3 r3 (since 0  r3  n) same

And: a +n b = nq1 + r1 +n nq2 + r2 ==r1+n r2 r3 (since r1 + r2 =nq3 + r3 with 0  r3  n)

EXAMPLE 2.7 For any fixed element a in a group G, let fa: G  G be given by fag = ag . Show See page 46 for a discussion on that the function : GS G  given by the symmetric group SG  . a = fa is a one-to-one homomorphism. SOLUTION:  is one-to-one:

a = b  fa = fb  fae = fbe  ae ==be  a b in particular To show that  is a homomorphism we need to show that

ab = a b , which is to say, that the function fab: G  G is equal to the function fafb: G  G . Let’s do it: For any xG : fabx = ab x

and fafb x ===fafbx fabx abx By associativity, ab xabx=  , and we are done.

CHECK YOUR UNDERSTANDING 2.18 Show that for any two groups G and G the function : G  G given by a = e for every aG is a homomorphism (called the Answer: See page A-11. trivial homomorphism from G to G ). Homomorphisms preserve identities, inverses, and subgroups: THEOREM 2.23 Let : GG  be a homomorphism. Then: (a) e = e (b) a–1 =  a –1 (c) If H is a subgroup of G, then: H =  h hH is a subgroup of G . (d) If H is a subgroup of G , then: –1H = gG g  H is a subgroup of G . 74 Part 2 Groups

PROOF: (a) Since  is a homomorphism: e == ee e e . Multiplying both sides by e –1 yields the desired result:

e –1e =  e –1e e e= e –1e e e ==ee e

(b) Since a–1 a === a–1a e e : (a) a–1 =  a –1 . (c) We use Theorem 2.14, page 62, to show that the nonempty set H is a subspace of G : Since a b = ab :H is closed with respect to the operation in G . Since, for any aG , a–1 =  a –1 : a –1  H for every a  H . (d) We use Theorem 2.14 to show that the nonempty set –1H is a subspace of G : Let ab  –1H . To say that ab  –1H is to say that ab  H , and it is: Since ab = a b , and since H , being a subgroup of G , is closed with respect to the operations in G : ab  H . Let a  –1H . To say that a–1  –1H is to say that a–1  H , and it is: Since a–1 =  a –1 , and since H contains the inverse of each of its elements: a–1  H .

CHECK YOUR UNDERSTANDING 2.19 Let : GG  and : G  G be homomorphisms. Prove that the Answer: See page A-11. composite function : GG  is also a homomorphism. 2.4 Homomorphisms and Isomorphisms 75

IMAGE AND KERNEL For any given homomorphism : GG  , we define the kernel of  to be the set of elements in G which map to the identity e  G [see Figure 2.7(a)]. We define the set of all elements in G which are “hit” by some a to be the image of  [see Figure 2.7(b)].

G G G G  .  e Kernel of  Image of  (a) (b) Figure 2.7 More formally:

DEFINITION 2.10 Let : GG  be a homomorphism. KERNEL The kernel of  , denoted by Ker , is Utilizing the notation of Definition 1.3, page 2: given by: Ker = –1e Ker = aG a = e Im = G IMAGE The image of  , denoted by Im , is given by: Im =  a aG

Both the kernel and image of a homomorphism turn out to be subgroups of their respective groups:

THEOREM 2.24 Let : GG  be a homomorphism. Then: (a)Ker is a subgroup of G. (b)Im is a subgroup of G .

PROOF: (a) A consequence of Theorem 2.23(d) and the fact that e is a subgroup of G (b) A consequence of Theorem 2.3(c).

CHECK YOUR UNDERSTANDING 2.20 Show that the function : 2Z  4Z given by 2n = 8n is a Answer: See page A-11. homomorphism. Determine the kernel and image of  . 76 Part 2 Groups

Definition 2.10 tells us that a homomorphism : GG  is onto if A homomorphism : GG  and only if Im = G . The following result is a bit more interesting, must map e to e . What this theorem is saying is that if e is in that it asserts that in order for a homomorphism to be one-to-one, it the only element that goes to need only behave “one-to-one-ish” at e (see margin): e , then no element of G is going to be hit by more that one THEOREM 2.25 A homomorphism : GG  is one-to-one element of G. This is certainly not true for arbitrary functions: if and only if Ker = e . y PROOF: Suppose  is one-to-one. If a  Ker , then both a = e and e = e [Theorem 2.22(a)]. Consequently ae= (since  is assumed to be one-to-one). Hence: Ker = e . x Conversely, assume that Ker = e . We need to show that if fx= x a = b , then ab= . Let’s do it: a = b a b –1 = e Theorem 2.22(b): a b–1 = e –1  is a homomorphism: ab = e Ker = e : ab–1 = e ab–1 beb= ab=

CHECK YOUR UNDERSTANDING 2.21

Let : GG  be a homomorphism. Show that if there exists an element c  G (not necessarily the identity e) such that if c = a then c = a , then  is one-to-one. In other words: for a homomorphism : GG  to be one-to-one, it Answer: See page A-11. need only behave “one-to-one-ish” at any one-point in G.”

The word isomorphism ISOMORPHISMS comes from the Greek iso meaning “equal” and As previously noted, a homomorphism : GG  preserves the morph meaning “shape.” algebraic structure in that ab = a b . An isomorphism also preserves set structures, in that it pairs of the elements of the set G with those of the set G . More formally:

DEFINITION 2.11 A homomorphism : GG  which is also a bijection is said to be an isomor- ISOMORPHISM phism from the group G to the group G . An isomorphism : GG is said to be AUTOMORPHISM an automorphism on G. ISOMORPHIC Two groups G and G are isomorphic, written GG  , if there exists an isomorphism from one of the groups to the other. 2.4 Homomorphisms and Isomorphisms 77

EXAMPLE 2.8 Show that the group  + of real numbers under addition is isomorphic to the group + . of positive real numbers under multiplication. SOLUTION: We show that the function :  +  + . given by In this discussion we are not a = ea is an isomorphism: using e to denote the identity element in + . (which is Homomorphism: 1). Here, e is the transcen- ab+ ===eab+ eaeb a b dental number e  2.718 . One-to-one: (See Theorem 2.24)

The identity in + . The identity in  + a ===1  ea 1  a 0 Onto: For a  + . , we have: lna ==elna a .

CHECK YOUR UNDERSTANDING 2.22 (a) Prove that  is an equivalence relation on any set of groups (see Definition 1.12, page 29). (b) Prove that nZ mZ for any nm  Z+ .

(c) Let gG . Prove that the map ig: GG given by –1 igx = gxg xG Answer: See page A-12. is an automorphism (called an inner automorphism.) Algebraically speaking, there is but one cyclic group of order n, and but one infinite cyclic group: THEOREM 2.26 (a) If the cyclic group Ga=  is of order n, the GZ n +n . (b) If Ga=  is infinite, the GZ  + .

PROOF: (a) We show that the function: : 012 n – 1  a0a1 a2  an – 1 e i given by i = a is an isomorphism from Zn +n to Ga=  : One-to-one. For 0 ijn : i = j  ai = aj  aij– ===a0  ij– 0  i j Onto. For ai  a0a1 a2  an – 1 , i = ai Homomorphism: ij+ ===aij+ aiaj i j (b) Your turn: CHECK YOUR UNDERSTANDING 2.23 Answer: See page A-12. Show that every infinite cyclic group is isomorphic to Z + .

78 Part 2 Groups

A ROSE BY ANY OTHER NAME

Let : GG  be an isomorphism. Being a bijection it links every element in G with a unique element in G (every element in G has its

a –1 a own G counterpart, and vice versa). Moreover, if you know how to –1 b b function algebraically in G, then you can also figure out how to func- ab tion algebraically in G (and vice versa). Suppose, for example, that ab you forgot how to multiply in the group G , but remember how to mul-  tiply in G. To figure out ab in G you can take the “–1 -bridge” G G back to G to find the elements a and b for which a = a and b = b , perform the product ab in G, and then take the “ -bridge” back toG to determine the product ab : ab . Basically, if a group G is isomorphic to G , then the two groups can only differ in appearance, but not algebraically. Consider, for example, the two groups which previously appeared in Figure 2.1, page 43:

+4 0123 * eabc Z4: K: 00123 eeabc 11230 aaecb 22301 bbcea 33012 ccbae (a) (b) Both contain four elements ({0,1,2,3} and {e,a,b,c}); so, as far as sets are concerned, they “are one and the same” (different element-names, that’s all). But as far as groups go, they are not the same (not isomorphic). Here are two algebraic differences (either one of which would serve to prove that the two groups are not isomorphic):

1.Z4 is cyclic while the Klein 4-group, K, is not. 2. There exist three elements in K of order 2 (see Definition

2.5, page 58), while Z4 contains but one (the element 2). To better substantiate the above claims: THEOREM 2.27 If GG  , then: (a) G is cyclic if and only if G is cyclic. (b)For any given integer n, there exists an element aG such that an = e if and only if there exists an element a  G such that a n = e . 2.4 Homomorphisms and Isomorphisms 79

PROOF: Let : GG  be an isomorphism. (a) Suppose G is cyclic, with Ga=  . We show G= a by showing that for any b  G , there exists nZ such that b= a n : Let bG be such that b = b . Since Ga=  , there exists nZ such that ba= n . Then: b==b an = a n Exercise 16 The “only-if” part follows from the fact that if G is isomorphic to G , then G is isomorphic to G [see CYU 2.23(a)]. (b)Let aG be such that an = e . Then: a n ===an e e The “only-if” part follows from CYU 2.22(a).

CHECK YOUR UNDERSTANDING 2.24 Answer: See page A-12. Prove that if GG  , then G is abelian if and only if G is abelian.

A property of a group G that is shared by all groups isomorphic to G is said to be a group invariant property. For example, abelian and cyclic are group invariant properties. Other group invariant properties are cited in the exercises. In general, one can show that two groups are not isomorphic by exhibiting a group invariant property that holds in one of the groups but not in the other. For example, the permutation

group S3 is not isomorphic to Z6 +6 as one is abelian while the other is not.

The following results underlines the importance of symmetric groups (see discussion on page 46).

Arthur Cayley (1821-1885) THEOREM 2.28 Every group is isomorphic to a subgroup of (Cayley) a symmetric group.

PROOF: The function : GS G given by

g ==fg: GG where fgx gx (xG  was shown to be a one-too-one homomorphism in Example 2.7.

Since  it is onto the subspace G of SG : : G  G is an isomorphism. 80 Part 2 Groups

EXERCISES

Exercise 1-10. Show that the given function : GG  from the group G to the group G is a homomorphism. 1.GG== Z + and n = 2n .

2.GG== + and rx = rx for r   . 3.GZ== + G  + and n = n .

4.GZ== +  G Z3 and n = r where n = 3mr+ with 0  r  3 .

5.GZ== +  G –11  . and n = 1 if n is even and fn= –1 if n is odd.

6.GZ= 6 , G = Z2 and n = r where n = 2dr+ with 0  r  2 .

ab ab+ d 7.GG== M22  + and = . cd –c 0

 i if i  4 8. GS= 3 , G = S4 and i =  ,  4ifi = 4

9.GG=  with G abelian, and a = a–1 for aG .

10.GG=  with G abelian, nZ + , and a = an for aG . Exercise 11-15. Show that the given function : GG  from the group G to the group G is not a homomorphism. 11.GG== Z + and n = n + 1 .

12.GZ= 5 , G = Z2 and n = r where n = 2dr+ with 0  r  2 .

ab ab+ d 13.GG== M22  + and = . cd –c 1

ab 14.GM==22  + G and = ad– bc . cd

–1 15.GG== S3 and a = a for aG .

Exercise 16-27. Find the kernel and image of the homomorphism of: 16. Exercise 1. 17. Exercise 2. 18. Exercise 3. 19. Exercise 4. 20. Exercise 5. 21. Exercise 6. 2.4 Homomorphisms and Isomorphisms 81

22. Exercise 7. 23. Exercise 8. 24. Exercise 9. 25. Exercise10. 26. Exercise 11. 27. Exercise 12.

28. Let  + denote the group of all real numbers under addition, and + . the group of all positive real numbers under multiplication. Show that the map : +   given by x = lnx is an isomorphism. 29. Let : GG  be a homomorphism and let aG . Prove that an =  a n for every nZ . 30. Let : GG  be a homomorphism and let aG . Show that the map : ZG given by n = an is a homomorphism. 31. Let : GG  be a homomorphism with G finite. Show that G is a divisor of G . 32. Let : GG  be a homomorphism. Prove that for all ab  G : ab–1 ==a b –1 and a–1b a –1b 33. Let : GG  be a homomorphism, Show that: (a) If  is onto and if G is abelian, then G is abelian. (a) If  is one-to-one and if G is abelian, then G is abelian. 34. Prove that a group G is abelian if and only if the function f: GG given by fg= g–1 is a homomorphism. 35. Let Ga=  be cyclic and let G be any group. Let : GG  be a homomorphism. Prove that Im is cyclic. 36. Let : GG  be a homomorphism. Show that if k  Ker , then gkg–1  Ker for every gG . 37. Let GG  and G be groups. Show that if : GG  and : G  G are homomorphisms, then so is : GG  . 38. Let : GG  be a homomorphism. Show that G is abelian if and only if for all ab  G: aba–1b–1  Ker . 39. Let : GG  be a homomorphism. Prove that, for any given xG : gGg = x = xk k  Ker 40. Let Ga=  be cyclic and let H be any group. Prove that for any chosen hH there exists a unique homomorphism : GH such that a = h . So, a homomorphism on a cyclic group Ga=  is completely determined by its action on a. 41. Let : GG  be a homomorphism. Prove that, for any given xG : gGg = x = xk k  Ker 42. Let A, B, C, and D be groups. Show that if AB and CD , then ACBD   (see Exercise 52, page 52). 82 Part 2 Groups

43. Let G and G be groups. Show that GG   G  G (see Exercise 52, page 52). 44. (a) Show that the set ZZ = ab ab  Z , with ab *cd = ac+  bd+ is a group.

(b)Verify that the functions 1: ZZ  Z and 2: ZZ  Z given by 1ab = a and

2ab = b , respectively, are homomorphisms.

(d)Show that the function : ZZ  ZZ given by ab =  2ab  1ab is an isomorphism.

45. For mZ , m  0 , let m: ZZ be given by mn = mn .

(a) Show that m is a one-to-one homomorphism.

(b) Show that m is an isomorphism if and only if m = 1 . 46. Let F denote the additive group of real valued function (see Exercise 49, page 52), and let  denote the additive group of real numbers. Prove that for any c   the function c: F   given by cf = fc for fF  is a homomorphism (called an evalu- ation homomorphism.) 47. Let D denote the set of differentiable functions from  to  . (a) Show that D  + is a group.

(b) Show that for any c   the function c: D   given by cf = fc is a homomorphism.

(d) Is c onto for any c? 48. Let C denote the set of continuous real valued functions. (a) Show that C  + is a group. (b) Show that for any closed interval ab in  the function : C   given by b f = fxdx is a homomorphism. a 1 3 (c) Show that the function : C   given by f = fxdx + 2 fxdx is a 0 2 homomorphism.

49. Show that for any   S3 , the function : S3  S3 given by =  is a homomorphism. Is it necessarily an isomorphism? 50. Let G be a group. Prove that AutG =  : GG is an automorphism   is a group. 2.4 Homomorphisms and Isomorphisms 83

Exercise 34-40. Show that the give property on a G is an invariant.

51.G — the order of a finite group G. 52. G contains a nontrivial cyclic subgroup.

53. G contains an element of order n for given n  1 .

54. G contains m elements of order n for given n  1 .

55. G contains a subgroup of order of order n for given n  1 .

56. The number of elements in Tn = gGog = n (see Definition 2.5, page 58).

57. The number of elements in ZG — the center of a finite group G. (See Exercise 45, page 70.)

PROVE OR GIVE A COUNTEREXAMPLE

58. The additive group  is isomorphic to the additive group Q of rational numbers) 59. The additive group Z is isomorphic to the additive group Q of rational numbers)

60. If  is a homomorphism from a group G to a cyclic group G = a , then Ker is a cyclic subgroup of G.

61. If  is an isomorphism from a group G to a cyclic group G = a , then Ker is a cyclic subgroup of G.

62. For C the group of continuous real valued functions under addition the function 1 3 : C   given by f = fxdx fxdx is a homomorphism. 0 2

63. If nm , Sn and Sm are not isomorphic. 84 Part 2 Groups

§5. SYMMETRIC GROUPS Cayle’s Theorem asserts that every group is isomorphic to a subgroup of a symmetric group. It follows that if one knew everything about symmetric groups, then one would know everything about groups in general. Alas, however, symmetric groups SX  are not “easy to own,” especially if X is an infinite set. In this section we focus our attention on finite symmetric groups, specifically on the groups Sn of section 2.1 (see page 46). CYCLE DECOMPOSITION Consider the permutation: 1234567  =  3547261 To get a sense of its action, let’s use the symbol 13 to indicate that  maps 1 to 3. We then have: 7. .1 13471 ; or, better yet: 4. . 3 Adhering to convention we let the symbol (1,3,4,7) represent the permutation in S7 that acts like  on the integers 1, 3, 4, and 7, and leaves 2, 5, and 6 fixed:

1 2 34567 1347 =  3247561 In general, a cycle of the form (said to be a 4-cycle of the permutation  ) n1n2  nk is said to be a k-cycle, Proceeding as above, but starting with 2 (or 5) we arrive at the 2- cycle: 1 2 345 67 25 =  1534267 All that remains is 6. But 6 is stationary under  , so; 123456 7 6 =  the identity permutation 1234567 In writing a permutation

  Sn as a product of At this point we can express  as a product of cycles; specifically: cycles, we generally don’t include cycles of length 1,  = 1347 25 i.e: as any such cycle is the identity in S . In particular, 1234567 1234567 1234567 n =  it is understood that the permutation 3547261 3247561 1534267  = 1347 25  S7 Since 1347 does not move 2 or 5, and since (2,5) does not effect leaves 6 fixed. 1, 3, 4, or 7, the two cycles are said to be disjoint and must commute (see Exercise 21): 1347 25 = 25 1347 2.5 Symmetric Groups 85

In general: Just as any integer can be expressed as a product of THEOREM 2.29 Every permutation in Sn can be expressed as primes, so then can permutations be expressed as prod- a product of disjoint cycles. ucts of cycles. PROOF: [By (APM) induction on n—see page 16]

Let Pn be the proposition that every permutation in Sn can be expressed as a product of disjoint cycles. I.P1 is true: S ==1 1 . 1 1 II. Assume Pm is true for 1 mk III. We show that Pk+ 1 is true, thereby completing the proof: Let   Sk + 1 . Since Sk + 1 is a finite group,  has finite order, say o = i . Note that m1 cycles 2 i – 1 back to 1. If ik= + 1 , then  = 11 1  1 — a cycle. If ik + 1 , then consider the set 2 i – 1 O1 = 11 1  1 (called the orbit of 1 under  ) Pulling the above orbit out from 12 k + 1 :

12 k + 1 – O1

we arrive at a permutation s on a set of sk= + 1 – i elements, with 1 sk. By II, s can be written as a product of disjoint cycles c1c2  ct . It follows that:   = c1  c2 ct  O1

(note that the orbit O1 is disjoint from all of the cis ) In the next example we again focus on the cycle-decomposition-procedure, but in reverse.

EXAMPLE 2.9 Construct a permutation   S10 that can be expressed as a product of a 2-cycle, a 3-cycle, and a 4-cycle. SOLUTION: Any such permutation must leave 10–1 2++ 3 4 = element fixed. We decide to go with the element 3 [see Figure 2.8(a)].

123 45678910 1234567 8 9 10   3 3 97 (a) (b)

123 4 5678910 1234 5 678910  tu 243 1 97 24311059678 (c) (d) Figure 2.8 86 Part 2 Groups

We then choose 7, along with 9, to generate the 2-cycle 79 [see Figure 2.8(b)]. Of the remaining 7 elements we decide to go with 1, 2, and 4 to create the 3-cycle 124 [Figure 2.8(c)]. All that’s left are the elements 5, 6, 8, 10, and decide to mold them into the cycle

86510 — bringing us to the completed permutation   S10 in Figure 2.8(d) with cycle decomposition:  = 79 124 86510

CHECK YOUR UNDERSTANDING 2.25

12345678910 (a) Express  =  as a product of cycles. 39415627810 Answer: (a) 134 2987 (b) Construct a permutation   S10 that can be expressed as a (b) See page A-13. product of two 2-cycles and a 5-cycle.

ORDER OF PERMUTATIONS

We already know that the symmetric group Sn has order n! . We now turn our attention to the task of determining the order of permutations in Sn . Let’s start off by considering the 4-cycle:  = 2635  Sn for n  6 Focusing on the element 2 we have: 2 = 6 22 = 3 32 = 5 42 =2 So, the smallest power s of  such that s2 = 2 is s = 4 , and the same can be said for the elements 6, 3, and 5. It follows, since all of the remaining elements in 12 n are held fixed by  , that o = 4 . Indeed, as you are invited to establish in the exercises:

THEOREM 2.30 Every k-cycle in Sn has order k. Moving things along we reconsider the permutation 1234 5 678910  ==79 124 86510 24311059678 that surfaced in Example 2.9. We know, from Theorem 2.30, that: o79 ==2 o124 3, and o86510 =4 It follows that s = e for any s that is divisible by 2, 3, and 4. In particular, s ==lcm234 12 will work. Moreover, since the three cycles are disjoint, no positive integer smaller that lcm234 will do the trick, bringing us to: 2.5 Symmetric Groups 87

THEOREM 2.31 If   Sn has a cycle decomposition of disjoint cycles of order (length) k1k2  ks , Note the “disjoint cycles” then: condition in the theorem. o = lcmk k  k A case in point: 1 2 s The 2-cycles 12  13 are not disjoint, and their product is not of order 2: CHECK YOUR UNDERSTANDING 2.26 o12 13 ==o132 3 Determine the order of the permutation: 12345678  =  Answer: 6 38674152

A cycle in Sn is, in a sense, a primitive object in that it cannot be decomposed into a product of smaller disjoint cycles. It can, however, always be decomposed into a product of 2-cycles, called transpositions. Consider, for example the cycle  = 3251 :

While the above 4-cycle could reside in any Sn with n  5 , all elements other that 1, 2, 3, and 5 are immune to its action. That being the case, we might as well embed it in S5 . We then have: Note that the decomposition 12345 of a cycle as a product of 32  1 3245 transposition is not unique. 35   = 31 35 32 A case in point: 1 5 243 3251 = 31 35 32 31  Note the pattern: 3 5241 first switch 3 with 2 = 31 35 32 12 12 then switch 3 with 5 finally switch 3 with 1 Generalizing the above pattern, we have (Exercise 29): THEOREM 2.32 Any cycle can be expressed as a product of transpositions.

Merging the above result with Theorem 2.29 we come to THEOREM 2.33 Every permutation can be expressed as a product of transpositions. Specifically:  a1a2 am = a1am a1am – 1 a1a2

CHECK YOUR UNDERSTANDING 2.27 (a) Express th cycle 31647 as a product of transpositions. Answers: 1234 5 678910 (a) 37 34 36 31 (b) Express the permutation  as a prod- (b) 14 12 56 58 510 79 24311059678 (c) Se page A-13. uct of transpositions. (c) Show that for any transpositions  : –1 =  . 88 Part 2 Groups

As previously noted, the decomposition of a permutation as a product of transposition is not unique. However: THEOREM 2.34 No permutation can be expressed as both a product of an even number of transpositions and as a product of an odd number of transpositions.

PROOF: Chances are that you are familiar with the matrix space Mnn , along with the determinant function det: Mnn   . You may also recall that:

If two rows of AM nn are interchanged, then, then (*)  Here is the identity matrix the determinant of the resulting matrix is – detA . in M4 : (A brief development of the above result appears in Appendix B.) I 1000 1 At this point, rather then focusing on a permutation i on the set I I = 0100 2 12 n of integer, we turn our attention to a permutation I on 0010 I3 the n rows I I  I of the identity matrix IM (see mar- 0001 I4 1 2 n nn And here is the transposi- gin). In this new environment, a transposition is the switching of two tion I1 I3 : rows. Let AM nn be achieve by permuting the rows of I. Can that 0010 switched be done by both an even number and an odd number of transposi- 0100 first and third row tions? No, for by (*), detA would have to equal both 1 and –1 : 1000 of I. detA ==–1 2kdetI 1 while detA =1–1 2k + 1detI =– 0001 (Note: detI = 1 ) DEFINITION 2.12 A permutation is even, or odd, if it can be Even and Odd expressed as the product of an even, or odd, Permutations number of transpositions, respectively

CHECK YOUR UNDERSTANDING 2.28

(a) Show that the identity permutation eS n is even. 1234 5 678910 Answers: (a) See page A-13. (b) Is the permutation  even or odd? (b) Even. 24311059678

At this point we know that any symmetric group Sn can be partitioned into the set of even permutations and the set of odd permutations. Since the set of odd permutations does not contain the identity element [CYU

2.28(a)], it cannot be a subgroup of Sn . On the other hand: DEFINITION 2.13 The alternating group of degree n is the Alternating Group subgroup An of even permutations of the

symmetric group Sn . Lest there be any doubt: 2.5 Symmetric Groups 89

THEOREM 2.35 For n  2 , the set An of even permutations is n! a subgroup of S of order ----- . n 2

PROOF: Since the identity permutation is even, An   . Closure: If  and  are even permutations, then each can be expressed as a product of an even number of transpositions, say:  = 12 2k and = 122h It follows that  can be expressed as a product of 2kh+ transpo-

sitions; namely: = 122k122h –1 Inverse: If = 122k , then  can also be expressed as a product of 2k transpositions; namely: –1 –1 –1 –1  ==2k 2 1 2k 21 CYU 2.10, page 58 C YU 2.27(c).

Conclusion: An is a subgroup of Sn (Theorem 2.14, page 62).

Verifying that A = ----n-! : n 2 Let Bn denote the set of odd permutations in Sn . We show that the

function f: An  Bn given by f = 12 is a bijection. One-to-one: f ==f  12 12

 12 12 ==12 12 

Onto: For   Bn , 12  An and: f12 ==12 12

We now know that An and Bn have the same number of elements.

The fact that An Bn = and that An  Bn = Sn with Sn = n! assures us that A = ----n-! . n 2

CHECK YOUR UNDERSTANDING 2.29

Determine A3 utilizing the notation: 1 2 3 1 2 3 1 2 3 e = ,  = ,  =  1 2 3 1 2 3 1 2 3 1 2 S3 =  Answer: 1 2 3 1 2 3 1 2 3   =  3 = 1 3 2 , 4 = 3 2 1 , 5 2 1 3 A3 = e1 2    90 Part 2 Groups

EXERCISES

Exercise 1-9. Express the given permutation as a product of disjoint cycles and also as a product of transpositions.

2 12345 12345 12345 1. 2. 3.  35412 52413 25431

2 123456 123456 123456 4. 5. 6.  654123 153264 543126

2 1234567 1234567 1234567 7. 8. 9.  6541237 5741236 3421576

Exercise 10-16. Find the order of the permutation in Exercise: 10. 1 11. 3 12. 4 13. 5 14. 7 15. 8 16. 9

Exercise 17-20. Solve for  in the symmetric space S5 . 17.135 = 241 18. 135 = 241 45

19. 241 45 = 135 20. 135 2 = 241 3

21. Let a be an element of a group G. Show that the map a: GG given by agag= is a permutation on the set G.

22. Referring to Exercise 21, show that H = a aG is a subgroup of SG (the group of all permutations on G).

23. Prove that if  are disjoint cycles in Sn , then =  .

24. Prove that there is no permutation  such that 12 –1 = 123 .

25. Prove that for any permutation  and any transposition  : –1 is a transposition.

26. Prove that if  is a k-cycle, then –1 is also a k-cycle for any permutation  .

27. Prove that there is a permutation  such that 123 –1 = 456 .

28. Prove that every k-cycle in Sn has order k.

29. Use induction to show that any cycle a1a2  as in Sn can be expressed as a product of transpositions as follows: a1a2  as = a1 as a1 as – 1 a1a2 2.5 Symmetric Groups 91

30. Show that if  is a cycle of odd length, then 2 is a cycle.

31. List all the elements in the alternating group of degree 4: A4 .

32. Let H be a subgroup of Sn . Prove that either all of the elements of H are even, or that exactly one-half the elements in H are even.

33. Express the k-cycle a1a2  ak as a product of k + 1 transpositions.

34. Let 1 2 be transpositions with 1  2 . Show that:

(a) If 1 and 2 are disjoint, then 21 can be expressed as the product of two 3-cycles.

(b) If 1 and 2 are not disjoint, then 21 can be expressed as a product of 3-cycles.

35. Show that every even permutation   An , with n  3 , is a product of 3-cycles. Suggestion: consider Exercise 34.

–1 36. Let  be a k-cycle. Show that   An if and only if   An for every transposition  .

PROVE OR GIVE A COUNTEREXAMPLE

37. The permutation equation 123 –1 = 124 567 has a solution.

38. The transposition 12 in S3 can be expressed as a product of 3-cycles.

39. The identity in Sn cannot be expressed as a product of three transpositions. 92 Part 2 Groups

§6. NORMAL SUBGROUPS AND FACTOR GROUPS An important recollection from page 67: If H is a subgroup of a group G, then a~b if ab–1  H is an equivalence relation on G. Moreover, the equivalence class a is the set: Ha=  ha h H — a right coset of H. Let’s switch from right to left, and replay the above development: THEOREM 2.36 If H is a subgroup of a group G, then a~b if a–1bH is an equivalence relation on G. Moreover, the equivalence class a is the set: aH=  ah h H — a left coset of H. If G is finite: The number of elements in each aH equals H .

ROOF –1 This proof mimics that of P : ~ is reflexive: x~x since x xeH=  . Lemma 2.2, page 67. ~ is symmetric: a~b  a–1b = h for some hH  a–1b –1 = h–1  b–1a = h–1  b~a since h–1  H ~ is transitive: If a~b and b~c , then: a–1bH and b–1cH   ab–1 bc–1  H –1 –1  aec  Hac  H  a~c Having established the equivalence part of the theorem, we now verify that a = ah h H : ba  b~a  ab  a–1b = h for some hH  b = ah b aH As for the rest of the proof: CHECK YOUR UNDERSTANDING 2.30 Let H be a subgroup of a finite group G. Show that each left coset aH contains H elements. Answer: See page B-14. Suggestion: Consider the function f: HaH . If H is a subgroup of an abelian group G then for any aG : aH=== ah h H ha h H Ha (every left coset is also a right coset) This need not be so if G is not abelian. A case in point: 2.6 Normal Subgroups and Factor Groups 93

1 2 3 1 2 3 e ==1 2 3   2 3 1 1 EXAMPLE 2.10 Find the partition of S3 (margin) into both the 1 2 3 1 2 3 left and right cosets of the subgroup 2 ==3 1 2  3 1 3 2 He=  3 . 1 2 3 1 2 3 4 ==3 2 1  5 2 1 3 SOLUTION: Since S3 = 6 and H = 2 , each partition is composed of 3 subsets, one of which is the subgroup H itself (eH== He H ). As for the rest of the story: Left Cosets: Right Cosets: 123 eH== H e   123 3 3 He== H e 3 1 132 = 5 231 =  1  4 213 3 321  H ==  a    1 1 1 3 1 5 H1 ==1 31 1 4  H ==      2 2 2 3 2 4 H2 ==2 32 2 5 Left-Cosets Partition Right-Cosets Partition

 H =  H   1 5 1 5 1 4 H1 = H4 e  eH = 3H 3 e 3 He= H3   2H = 4H 2 4 2 5 H2 = H5 S 3 S3 The above example illustrates the fact that a left cosets aH of a subgroup H of G need not equal the right coset Ha . Of particular importance are those subgroups for which left and right cosets are one and the same: DEFINITION 2.14 A subgroup N of a group G is said to be Clearly both G and e NORMAL SUBGROUP normal in G if for every aG : are normal subgroups of any group G. aN= Na (The symbol N  G is read: N is normal in G) If G is finite, then the number of cosets of Index of N in G N in G, namely GH , is called the index of H in G.

THEOREM 2.37 Let H be a subgroup of a group G. The following are equivalent: (i)aH= Ha for every aG . (i.e: H is normal in G) (ii)aH Ha for every aG . (iii)aha–1  H for every hH and aG . 94 Part 2 Groups

PROOF: i  ii : Clear. ii  iii : Let hH and aG be given. Since aH Ha , ah= ha for some hH . Consequently: aha–1 = hH iii  i : We show that aH Ha . A similar argument can be used to show that Ha aH : gaHg  = ah for hH  ga–1 = aha–1 by (iii):  ga–1 = h for hH

 ghagHa=  

Note: If N  G then every element of nN almost commutes with every element of gG in that: ng= gn for nN (and not just ng= gn ) Note: One way or showing that a subgroup H of G is normal in G: grab any element from H G If aha–1  H –1 H.. . aha then H is normal and any element of G

We showed, in Theorem 2.23, page 71, that homeomorphisms preserves subgroups. They fair nearly as well when it comes to normal subgroups. Specifically:

THEOREM 2.38 Let : GG be a homomorphism. (a) If N is normal in G, and if  is onto, then N is normal in G . (b) If N is normal in G , then–1N is normal in G.

PROOF: (a) Assume that  is onto and that N is normal in G. We are to show that for any n  N and any aG , an a–1  N . Let’s do it: Choose aG such that a = a . Then: an a–1 = a n a –1 Theorem 2.23(b), page 73: = a n a–1  is a homomorphism: = ana–1  N N is normal in G 2.6 Normal Subgroups and Factor Groups 95

(b) Let N be normal in G . We show that the subgroup –1N = nG n  N is normal in G by showing that for aG and n  –1N , ana–1  –1N : ana–1 = a n a–1  N N is normal in G CHECK YOUR UNDERSTANDING 2.31

Give an example illustrating that a homomorphism : GG that is not onto need not carry normal subgroups of G to normal subgroups Answer: See page A-14. of G . (Suggestion: Consider Example 2.10.) Let’s reconsider the left-coset partition of the subgroup He=   of Example 2.10. That partition, appearing 3 1 5 on the right, broke the group S into three disjoint pieces; 3 e 3 each of which has “two names:” 2 4 1H = 5H , eH = 3H , and 2H = 4H Can we impose a group structure on that partition? Here is a noble attempt: iH jH =  ij H From Example 2-10: 123 Yes, the above product certainly yields another left coset, but there is 2 a fatal flaw — the “product” is simply not defined: 12: 312 = e 1 123  12 HeHH== while: 1H = 5H   BUT: while (see margin) 123  H =  H  4 2 4 54: 321 54 H = 2H  = 2  5 312 The above fatal flaw is averted whenever the coset-partition of a group G stems from a normal subgroup of G.

THEOREM 2.39 If N  G and GN = aN a G then GN is a group under the operation aN bN = ab N GN is said to be the factor group of G by N, read: G modulo H or G mod H Factor groups are also said to be quotient groups.

PROOF: We first show that the operation aN bN = ab N is well defined: For aN= aN and bN= bN we need to establish the set equality ab Nab= N . We show that ab Nab N and leave it for you to verify that ab Nab N : 96 Part 2 Groups

gab Ng = abn for some nN

Since aaN and bbN :  g = an1bn2n for some n1 n2  N

 g ==an1b n3 where n3 n2n

Since N is normal in G:  g = abn4 n3 for some n4  N

 g = ab n4n3  ab N Having legitimatized the operation aN bN = ab N , we now verify that, under that operation, the nonempty set GN is a group (see Definition 2.1, page 41): Closure: For every aN bN G N , aN bN = ab NGN  . Associative: aNbN cN = ab NcN ==ab c NabcN ==aN bc NaNbNcN Identity: For every aN G N , aNeN= aN . Inverses: For every aN G N , aNa–1Naa==–1NeN (=N) .

When confronted with the factor group GN it is important that you keep in mind that you are dealing with a set of sets! In particular, the identity element in GN is the set N itself, which may have many names: NeNaN== for any aN Similarly, the inverse of the element (set) aN is the element a–1N , which may also have many names: a–1Na= –1b N for any bN (after all, the set bN equals the set N for any bN )

THEOREM 2.40 Let N be normal in G. The natural projection map : GGN  given by g = gN is a homomorphism, and Ker = N .

PROOF: For gg  G : gg ==ggN gNgN =g g . Moreover: g  Ker  g ==gN N g N

Recall that N is the identity in GN

CHECK YOUR UNDERSTANDING 2.32

(a) Show that if G is a finite group and if N  G , then: G GN = ------N (b) Let N be a normal subgroup of a cyclic group G. Prove that GN Answer: See page A-14. is also cyclic. 2.6 Normal Subgroups and Factor Groups 97

THE CENTER AND COMMUTATOR SUBGROUPS Every group G contains two particularly important normal subgroups, the center of G and the commutator subgroup of G, where: DEFINITION 2.15 The center of G, denoted by ZG , is the set of elements of G that commute with every element of G: CENTER OF G ZG= aGag = ga gG COMMUTATOR The commutator subgroup of G, which we SUBGROUP OF G denote by CG , is the generated group: CG= aba–1b–1 ab  G As advertised: THEOREM 2.41 Both ZG and CG are normal subgroups of the group G.

PROOF: Turning to the center of G. Since eZG  , ZG  . Closure: For ab  ZG , and any gG : ab gabg===== agb ag bgabgab bZG  bZG 

Inverses: For gZG  and for any aG (same as “for any a–1  G )”: ag= ga  ag –1 ==ga –1  g–1a–1 a–1g–1 Replacing a with a–1 in the above argument we conclude that g–1 commutes with every element of G. Normal: Employing Theorem 2.37(ii) we show that for every aG aZ G  ZGa : Using Theorem 2.37(iii): For zZG  and aG : xaZG  x = ag with gZG  aza–1 ==aa–1zzZG    x = ga x Z G a

Now for CG . We already know that CG is a subgroup of G (see Definition 2.8, page 65). As for the rest of the story: Normal: For xaba= –1b–1 and any cG , let xcab= and ya= –1b–1c–1 . We then have: xyx–1y–1 = cab a–1b–1c–1 b–1a–1c–1 cba ==cab a–1b–1c–1 caba–1b–1 c–1 So: c aba–1b–1 c–1 = xyx–1y–1  CG Turning to Theorem 2.37(iii), and the Principle of Mathematical Induction, we now verify that CG is normal in G. 98 Part 2 Groups

–1 –1 –1 I. If xa= 1b1a1 b1 , then, for any cG : cxc  CG . II. Assume that for nk= and any cG : –1 –1 –1 –1  –1 –1 –1 xa= 1b1a1 b1 a2b2a2 b2 akbkak bk  cxc  CG III.Then:

–1 –1 –1 –1  –1 –1 –1 –1 –1 ca1b1a1 b1 a2b2a2 b2 akbkak bk ak + 1bk + 1ak + 1 bk + 1 c –1 –1 –1 –1  –1 –1 –1 –1 –1 –1 = ca1b1a1 b1 a2b2a2 b2 akbkak bk c cak + 1bk + 1ak + 1 bk + 1 c –1 –1 –1 –1  –1 –1 –1 –1 –1 –1 = ca1b1a1 b1 a2b2a2 b2 akbkak bk c cak + 1bk + 1ak + 1 bk + 1 c

(*) (**) By II: (*)  CG . By I: (**)  CG Consequently: –1 –1  –1 –1 –1 xa= 1b1a1 b1 ak + 1bk + 1ak + 1 bk + 1  cxc  CG

THEOREM 2.42 For N normal in G, the factor group GN is abelian if and only if CG N .

PROOF: GN is abelian if and only if for any ab  G : aNbN== bNaN abN baN  ba –1ab  N –1 –1  a b ab N C G  N

CHECK YOUR UNDERSTANDING 2.33 Let G be an abelian group. Show that ZG= G and that Answer: See page A-14. CG= e .

ISOMORPHISM THEOREMS

THEOREM 2.43 If : GG  is a homomorphism, then FIRST K = Ker is normal in G and: ISOMORPHISM THEOREM GK  G

PROOF: We utilize Theorem 2.37(iii) to establish the normality of K. For kK and gG we show that gkg–1  K ; which is to say, that gkg–1 = e:

the identity in G gkg–1 = g k g–1

Theorem 2.23(b), page 73: ===g eg –1 g g –1 e 2.6 Normal Subgroups and Factor Groups 99

We complete the proof by showing that the function : GK G given by gK = g is an isomorphism. To begin with, we need to verify  is well defined: aK== bK ab–1  K  ab–1 e  a b–1 = e a = b  aK = bK  is One-to-one: We are to show that: aK == bK  aK bK Which is to say: aK = bK  ab–1  K . Let’s do it: aK = bK a ==b  a b –1 e  a b–1 = e  ab–1 = e  ab–1  K

 is Onto: For given g  G , gK = g .

 is a homomorphism: aKbK = ab K ==ab a b = aK bK EXAMPLE 2.11 Show that: Zn +n  ZnZ 

SOLUTION: In Example 2.6, page 72, we showed that the function : Z +  Zn +n given by m = r where mnqr= + with 0  rn is a homomorphism. While  is not necessarily one-to-one it is certainly onto, as, for any sZ n s = s . Applying Theorem 2.43, we then have:

Zn  ZK where K = Ker . Noting that: Ker ===m m = 0 kn k Z nZ Example 2.4, page 62

we conclude that: Zn  ZnZ 

CHECK YOUR UNDERSTANDING 2.34 Represent the group G = –11 (under standard integer multipli- Answer: GS  A n n cation) as a factor group of the symmetric group Sn . 100 Part 2 Groups

Here are a couple more isomorphism theorems for your consideration: THEOREM 2.44 Let H be a subgroup of a group G, and N a normal SECOND subgroup of G. Then: ISOMORPHISM HN=  hn h H n N THEOREM is a subgroup of G, HN is normal in G, and: HH  N HN  N

PROOF: See Exercise 29.

THEOREM 2.45 Let : GG  be an onto homomorphism with THIRD kernel K. If N is normal in G , then: ISOMORPHISM THEOREM N ==–1N aG a  N is normal in G and: GN  G  N

PROOF: See Exercise 30.

CHECK YOUR UNDERSTANDING 2.35 Use Theorem 2.42 to verify that the isomorphism GN  G  N in Theorem 2.44 can also be expressed in the form: GN  GK  NK Answer: See page A14. (“cancel” the K in the numerator and denominator) 2.6 Normal Subgroups and Factor Groups 101

EXERCISES

1-4. Determine if the give subgroup H is normal in the symmetric group S3 .

1.H = 12 2.H = 123 3.H = 132 4. HA= 3

5. (a) Show that ZZ  11 is an infinite cyclic group. (b) Show that ZZ  22 is not a cyclic group

6. Show that if N1 and N2 are normal subgroups of G, then N1  N2 is also normal in G. 7. Let N be a normal subgroup of G and let H be any subgroup of G. Show that NH=  nh n N and hH is a subgroup of G. 8. Let G be abelian and let H be a subgroup of G. Show that GH is abelian. 9. Let G be cyclic and let H be a subgroup of G. Show that GH is cyclic.

10. Let N be a collection of normal subgroup of G. Prove that N is normal in G.    A     A 11. Show that if there are exactly 2 left (or right) cosets of a subgroup H of a group G, then H  G.

12. Show that if a finite group G has exactly one subgroup H of a given order, then H  G . 13. Show that if H is a finite subgroup of G and if H is the only subgroup of G with order H , then H  G. 14. Let n be the index of the normal subgroup N in G. Show that an  N for every aG . 15. Let G be a group containing at least one subgroup of order n. Show that the intersection of all subgroups of order n in G is normal in G. Hint: first show that if a group H if of order n, then show that gHg–1 is also a subgroup of order n for all gG . 16. Show that the set of inner automorphisms of a group G is a normal subgroup of the group of all automorphisms of G. [see CYU 2.22(c), page 77]

17. Let G be a group. Show that the set SgGgxg=  –1 = x xG is a normal subgroup of G.

18. Let N be a normal subgroup of G, and let abcd  G be such that aN= cN and bN= dN. Show that abN= cdN .

19. Let G be a finite group of even order with n elements, and let H be a subgroup with n  2 ele- · ments. Prove that H must be normal. Suggestion: Consider the map : G  –11  . . 20. Let H and K be normal subgroups of G with HK = e . Show that hk= kh for all hH and kK . 102 Part 2 Groups

21. Let N be a normal subgroup of G such that GN is cyclic. Show that G is cyclic. 22. Let G be a group. Show that any subgroup of ZG is a normal subgroup of G. 23. Let G be a group. show that Ca= gGag = ga is a subgroup of G (called the centralizer of a). 24. Prove that the center of a group G is the intersection of all the centralizers in G; that is: ZG=  Ca (See Exercise 22). aG 25. Show that aZG  if and only if Ca= G . (See Exercise 22).

26. Find both the center and the commutator subgroup of S3 . 27. Let : GG  be an onto homomorphism with kernel K. Prove and if H is a subgroup of G , and if H = –1H , then HK  H .

28. Verify that there is no subgroup of order 6 in the alternating group A4 . (Note that A4 = 12 ). 29. Sow that if N is not a normal subgroup of G, then the coset operation aN bN = ab N is not well defined. 30. Prove Theorem 2.44. 31. Prove Theorem 2.45.

PROVE OR GIVE A COUNTEREXAMPLE 

32. If N G and if H is a subgroup of G, then HN  G .

33. If H  G and K  H, then K  G.

34. If HN  G then either H or N must be normal in G.  35. Le : GG  be a homomorphism. If N G , then N  G .

 –1 36. Le : GG  be a homomorphism. If N G , then  N  G .

 –1 37. Le : GG  be an onto homomorphism. If N G , then  N  G . 2.7 Direct Products 103

§7. DIRECT PRODUCTS We begin by extending the Cartesian product definition of page 2: DEFINITION 2.16 The Cartesian Product of n nonempty sets:

CARTESIAN X1X2  Xn PRODUCt is denoted by: n X  X    X or: X 1 2 n  i i = 1 and consists of all n-tuples:

x1x2  xn where xi  Xi for 1 in In particular:  is the familiar Cartesian plane while   is the Euclidean three-dimensional space. Imposing a group structures we arrive at: DEFINITION 2.17 The (external) Direct Product of the n groups G G  G is denoted by: DIRECT PRODUCT 1 2 n (EXTERNAL) n  G1  G2   Gn or by  Gi i = 1 consists of all ordered n-tuples

a1a2  an where ai  Gi for 1 in and where their multiplication is defined component-wise; that is:

a1a2  an b1b2  bn = a1b1a2b2  anbn

CHECK YOUR UNDERSTANDING 2.36  (a) Verify that G1  G2   Gn is a group.  (b) Prove that the group G1  G2   Gn is abelian if and only if Answer: See page A-15. each Gi is abelian.

EXAMPLE 2.12 (a) Verify that Z2  Z3 is cyclic.

(b) Verify that Z2  Z3  Z4 is not cyclic. 104 Part 2 Groups

SOLUTION: (a) We know that Z2  Z3 ==23 6 . Using the sum notation in the abelian group, we simply observe that the element 11 has order 6: 211 = 02 311 ===11 +11211  +1002  411 ===11 +11311  +0110  511 ===11 +11411  +1201  611 ===11 +11511  +0012  Ah!

(b) We show that the group Z2  Z3  Z4 , which is of order 24, contains no element of order greater than 12: Let abc  Z2  Z3  Z4. Since:

12a ===012 b 012 c 0 in the groups Z2Z3 Z4 , respectively: 12abc = 000 . In the above argument, 12 is the smallest positive integer that is divisible by 2, 3, and 4. In general: DEFINITION 2.18 The least common multiple of nonzero inte- LEAST COMMON gers a1a2  an , written lcma1a2  an MULTIPLE is the smallest positive integer that is a multiple of each ai ; i.e. is divisible by each ai .

THEOREM 2.46 n Let a1a2  an   Gi . If the order of ai i = 1 in Gi is ri , then the order of a1a2  an in  G1  G2   Gn is lcmr1r2  rn .

PROOF: Let M = lcmr1r2  rn . Since each ri M : M a1a2  an = e1e2  en Moreover, for any positive integer 0 mM : m m m m a1a2  an = a1 a2  an  e1e2  en Why is that so? Because since M is the smallest positive integer divis- m ible by each ri , some ai  ei .

EXAMPLE 2.13 Find the order of 154 in Z2  Z6  Z30 .

SOLUTION: Let r1r2 r3 denote the order of 1 in Z2 , the order of 5 For mZ n : in Z6 , and the order of 4 in Z30 , respectively. n om= ------Employing CYU 2.11(c), page 59 (margin) we find that: gcd m n r1 ===2 r2 6 r3 15 2.7 Direct Products 105

All that remains is to calculate the least common multiple of the above orders: o15r ==lcm2615 30

2 2 3 3 5: need one 2, one 3, and one 5: 2 3 5

CHECK YOUR UNDERSTANDING 2.37 Answer: 4 Determine the order of 334 in Z6  Z4  Z16 .

THEOREM 2.47 The group Zn  Zm is cyclic and isomorphic to Znm if and only if n and m are relatively prime. PROOF: Assume that n and m are relatively prime. Theorem 2.42 tels us that o11 = nm ; which is to say: Zn  Zm = 11 Since Zn  Zm = nm: Zn  Zm  Znm +n [see Theorem 2.26(a), page 77] As for the converse, assume that gcd n m = d  1 . Noting that nm ------is divisible by both n and m, we find that, for any d nm ab  Z  Z : ------ab = 00 . Since no element of Z  Z n m d n m nm has order greater than ------, Z  Z is not cyclic. d n m

CHECK YOUR UNDERSTANDING 2.38 Prove: The group Z  Z    Z is cyclic and isomorphic to n1 n2 ns Z if and only all pair of the numbers n n  n are n1n2ns 1 2 s Answer: See page A-15. relatively prime.

INTERNAL DIRECT PRODUCT On the surface, the following definition appears to be far removed from Definition 2.17: DEFINITION 2.19 A group G is said to be the (internal) direct product of n normal subgroups DIRECT PRODUCT (INTERNAL) N1N2  Nn if every gG has a unique representation of the form ga= 1a2an

where each ai  Ni for 1 in . 106 Part 2 Groups

CHECK YOUR UNDERSTANDING 2.39 Show that if G is the internal direct product of two normal subgroups Answer: See page A-16. H and K, then HK = e . Appearances aside, the internal and external direct product concepts are “algebraically equivalent,” in that every internal product space is isomorphic to an external product space, and every external product space is isomorphic to an internal product space. Taking the easy way out, we will content ourselves by establishing the above claim in the special case when the group G is the internal direct product of just two normal subgroups: THEOREM 2.48 (a) If G is the internal direct product of the normal subgroups H and K, then: HK H K

(b) If GG= 1  G2 , then there exist normal subgroups N1 and N2 in G such that: G1  G2  N1N2

PROOF: (a) We first show that for all hH and kK hk= kh (*): Since h–1  H and H  G : kh–1k–1  H . So: hkh–1k–1  H . Since K  G: hkh–1  K . So: hkh–1 k–1  K . Since, by CYU 2.39, hkh–1k–1  HK = e : hk= kh . Turning to the external product HK of the two groups H and K, we now show that the function : HK  HK given by hk = hk is an isomorphism:

One to one: h1 k1 ==h2 k2  h1k1 h2k2

 h1 ==h2 and k1 k2

 h1 k1 = h2 k2 Onto: Clear.

Homomorphism: For h1 k1  h2 k2  HK :

h1 k1 h2 k2 ==h1h2 k1k2 h1h2k1k2 By (*): = h1k1h2k2 = h  k h  k 1 1 2 2

(b) Let GG= 1  G2 . It is easy to see that: N1 ==e1 g gG and N2 ge 2 gG

are normal subgroups of G with N1  N2 = e e1 e2 , and that GN= 1N2 . That being the case, the identity map itself is an isomorphism from the external direct product GG= 1  G2 to the internal direct product GN= 1N2 . 2.7 Direct Products 107

THE FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS And here it be, presented without proof: THEOREM 2.49 Every finitely generated abelian group is isomorphic to a direct product of cyclic groups of the form:  m Z r1  Z r2   Z rn  Z p1 p2 pn

where the pi are primes, not necessarily distinct, and where the ri and m are positive integers. Moreover, the direct product is unique, up to order. For example: While an abelian group gener- GZ= 2  Z8  Z9  ZZ (*) ated by an element of order 2, one of order 8, another of is a finitely generated abelian group, and here is a particularly nice order 9, and a couple of gen- choice for its generators: erators of infinite order need not consist of 5-tuples, it is 10000 01000 00100 0001 0 00001 nonetheless isomorphic to (*). could have chosen any 1 i 7 could have chosen any non-zrto integer could also have chosen 2, 4, 5, 7, or 8

EXAMPLE 2.14 Find all abelian groups of order 600 (up to isomorphism). SOLUTION: Any finite abelian group G is surely finitely generated (the elements of G itself generate G). Employing Theorem 2.45 to: FUNDAMENTAL COUNTING PRINCIPLE: 600= 23 352 If each of n choices is fol- lowed by m choices, then the we arrive at the following six possibilities (see margin): total number of choices is given by nm . G1 = Z2  Z2  Z2  Z3  Z5  Z5 There are three choices for the number of 2’s in G2 = Z2  Z2  Z2  Z3  Z25 the direct product, a choice of one for the G3 = Z4  Z2  Z3  Z5  Z5 number of 3’s, and a choice of two for the G4 = Z8  Z3  Z5  Z5 number of 5’s. Total number of G5 = Z4  Z2  Z3  Z25 choices: 312 = 6 G6 = Z8  Z3  Z25 It can be shown that none of the above groups is isomorphic to any of the rest. For example, since G3 contains an element of order 4 while G1 does not: G1  G3 (see Exercise 36, page 82).

CHECK YOUR UNDERSTANDING 2.40

Answer: See page A-15. Referring to the above example, show that G3  G6 . 108 Part 2 Groups

Lagrange’s Theorem assures us that the order of any subgroup H of a

The alternating group A4 , of finite group G must divide the order of G. In the event that G is abelian, order 12, has no subgroup of the converse also holds: order 6. Yes, but A4 is not an abelian group. THEOREM 2.50 If m divides the order of an abelian group G, then G has a subgroup of order m.

PROOF: Theorem 2.45 enables us to express G in the form:

r r  Z 1  Z 2   Z rn p1 p2 pn

Since m divides the order of G:

s1 s2 sn mp= 1 p2 pn , where 0 si ri . By CYU 2.15, page 64: pri ri – si i ri – ri – si si opi ===------pi p ri ri – si gcd pi  pi It follows that:

r1 – s1 r2 – s2 rn – sn p1  p2   pn is a subgroup of G of order m. 2.7 Direct Products 109

EXERCISES

Exercise 1-6. Find the order of the given element if the give group.

1. 23 in Z4  Z9 2. 23 in Z5  Z12

3.228 in Z4  Z3  Z12 4. 228 in Z4  Z6  Z11

123 123 5.2  in Z4  S3 6. 3  in Z4  S3 213 231

Exercise 7-10. Find the order of each element if the given group.

7.Z2  Z3 8.Z2  Z4 9.Z2  Z2  S2 10. Z3  S2

Exercise 11-14. Find all proper subgroups of the given group.

11.Z2  Z3 12.Z2  Z4 13.Z2  Z2  S2 14. Z3  S2 Exercise 15-18. Find all abelian groups G of the give order (up to isomorphism). 15.oG= 36 16.oG= 100 17.oG= 180 18. oG= 243

19. Determine the number of elements of order 6 in Z6  Z9 .

20. Determine the number of elements of order 7 in Z49  Z7 .

21. Show that the Klein 4-group V (Figure 2.1, page 43) is isomorphic to Z2  Z2 .

22. Show that ZZ  11  Z .

23. Show that ZZZ   111  ZZ .

24. Use the Principle of Mathematica Induction to show that for finite groups G1G2  Gn :  G1  G2   Gn = G1 G2  Gn

25. Let G1 and G2 be groups. Show that G1  G2  G2  G1 .

26. Let G1 and G2 be groups. Show that ZG1  G2  ZG1  ZG2 (see Definition 2.15, page 96).  27. Let G1 and G2 be groups. Show that e1  G2 G1  G2 and that:

G1  G2  e1  G2  G2    28. Let H G1 and K G2 . Show that HK G1  G2 and that:

G1  G2  HK  G1  H  G2  K 110 Part 2 Groups

29. Let G1 and G2 be groups. Show that the order of ab  G1  G2 is the leas common multiple of og and oh . 30. Prove that the order of an element in a direct product of a finite number of finite groups n Gi i = 1 is the least common multiple of the orders of its components:

og1g2  gn = lcm o g1 og2 ogn

31. Let G be a group and Kgg=  gG  GG . Prove that (a) KG (b) K  GG if and only if G is abelian.

 32. Let GG= 1  G2   Gn be a direct product of groups. Show that the projection map i: GG i given by ig1g2  gi  gn = gi is a homomorphism

PROVE OR GIVE A COUNTEREXAMPLE

33. The groups Z2  Z12 and Z4  Z6 are isomorphic.

34. The groups Z2  Z4  Z8 and Z8  Z8 are isomorphic.

35. The groups Z2  Z3  Z8 and Z3  Z4  Z4 are isomorphic.

36. Let G, H, K denote groups. If GKHH   , then GH . 3.1 Definitions and Examples 111

3 Part 3

From Rings To Fields §1. DEFINITIONS AND EXAMPLES The familiar set of integers can boast or two operators: addition and multiplication. Though The integers under addition turns out to be an abelian group, the multiplication operator does not fair as well: (5, for example, has no multiplicative inverse). Multiplication is, however, an associative operator: abc= ab c abc  Z and it plays well with addition: ab+ c = ab+ ac abc  Z Just as the integers under addition directed us, in part, to the definition of a group (“in part,” as a group need not be abelian), so then do the integers under addition and multiplication direct us, in part, to the definition of a ring (“in part,” as a ring need not have a multiplicative identity). Specifically:

DEFINITION 3.1 A ring R +, . (or simply R) is a set R RING together with two binary operators, called From an axiomatic point of view, multiplication takes addition and multiplication; for which: a back seat to addition. Its only obligation, apart from Group Axiom: 1.R + is an abelian group. closure and the associative axiom, is to cooperate with Associativity Axiom: 2. For all abc  R : abc  = ab  c addition via the left and (multiplicative) right distributive property Distributive Axioms: 3. For all abc  R : of Axiom 3. abc + = ab + ac ab+  c = ac + bc The set Z of integers under standard addition and multiplication is a ring. The same can be said for the set Q or rational numbers, and the set  of reals. CHECK YOUR UNDERSTANDING 3.1 (a) Does there exist an operator “*” on the permutation group S3 = S3  for which S3 * is a ring?

Answer: (a) No (b) Let G + be an abelian group. Show that there exists an opera- (b) See page A-16. tor “*” on G for which G+ * is a ring. 112 Part 3 From Rings To Fields

EXAMPLE 3.1 Let R1 and R2 be rings. Prove that the group Noe that addition and mul- . tiplication in R1  R2 are R1  R2+ where both being defined in terms of their correspond- ab + cd = ac+  bd+ ing established operations ab cd = ac bd in R and R . 1 2 is a ring.

SOLUTION: Appealing directly to Definition 2.1, page 41, we first show that R1  R2 + is an abelian group: Associative:

For any a1 b1 a2 b2 a3 b3  R1  R2 :

a1 b1 + a2 b2 + a3 b3 = a1 b1 + a2 + a3 b2 + b3

= a1 + a2 + a3  b1 + b2 + b3

==a 1 + a2  + a3 b 1 + b2  + b3 a1 b1 + a2 b2 + a3 b3

Identity: For any given ab  R1  R2 : ab + 00 ==a + 0 b + 0 ab

Inverses: For any given ab  R1  R2 : ab + –a –b ==aabb–  – 00

Noting that for any a1 b1  a2 b2  R1  R2 :

a1 b1 + a2 b2 ===a1 + a2 b1 + b2 a2 + a1 b2 + b1 a2 b2 + a1 b1

we conclude that R1  R2 + is an abelian group. Moving on to the multiplicative axioms of Definition 3.1:

Associative: For any a1 b1 a2 b2 a3 b3  R1  R2

a1 b1 a2 b2 a3 b3 = a1 b1 a2a3 b2b3

= a1a2a3  b1b2b3

==a1a2 a3 b1b2 b3 a1 b1 a2 b2 a3 b3

Distributive: For any a1 b1 a2 b2 a3 b3  R1  R2

a1 b1 a2 b2 + a3 b3 = a1 b1 a2 ++a3 b2 b3

= a1a2 + a3  b1b2 + b3

= a1a2 + a1a3 b1b2 + bb3

==a1a2 b1b2 + a1a3 b1b3 a1 b1 a2 b2 + a1 b1 a3 b3 In a similar fashion once can show that:

a1 b1 + a2 b2 a3 b3 = a1 b1 a3 b3 + a2 b2 a3 b3

CHECK YOUR UNDERSTANDING 3.2

Answer: See page A-16 Sow that nZ , under standard addition and multiplication, is a ring. 3.1 Definitions and Examples 113

THEOREM 3.1 Let a and b be elements of a ring R. Then: (a) a00==a 0 (b) ab– ==–a bab–

There is only one product tak- (c) –a –b = ab ing place in nab ; namely the (d) nab==na banb for any integer n. ab . The n is not involved in a (see margin) product— it represents a sum. For example: 3ab = ab++ ab ab PROOF: (a) a0 ==a00+ a0 + a0 a0 – a0 = a0 0 = a0 (b) Since ab– + ab ===ab– + b a00: ab– = –ab. Given your arithmetic evolu- tion, you may be thinking Since –a bab+ ===– a + a b 0b 0 : –a bab= – . along these lines: –a –b ==–1a –1b ab Since ab– = –ab and –a bab= – : ab– = –a b . Tisk. For one thing, the ring R need not even have a unity. (c)–a –b ==––a bab (see margin). That –a , for example, is the by (b) additive inverse of a. That being the case: ––a = a As for (d): CHECK YOUR UNDERSTANDING 3.3 Let a and b be elements of a ring R. Show that for every nZ : Answer: See page A-16 nab==na banb While Definition 3.1 stipulates that addition is a commutative operator tin a ring R +, . , no such attribute is imposed on the product operator. Moreover, while every ring contains the additive identity “0”, a ring An element aR distinct from 0 is a multiplicative need not contain a multiplication identity “1” (see margin). identity (or unity) if for Bringing us to: every bR : ab== ba b . DEFINITION 3.2 A ring R +, . is said to be commutative Commutative Ring if ab= ba for every ab  R . Ring with Unity A ring with a multiplicative identity (or unity) is said to be a ring with unity.

For any n  1 , the commutative ring nZ of CYU 3.2 is an example of a ring that does not have a unity. Here is an example of a ring that is not commutative: 114 Part 3 From Rings To Fields

EXAMPLE 3.2 Show that the set of two-by-two matrices: ab M22 = abcd   cd Chances are that you are already familiar with the with addition and multiplication given by: matrix space Mnn which possesses both an additive ab + ab = aa+ bb+ and multiplicative structure. If so, then you already know cd cd cc+ dd+ that, for any n  2 , Mnn is a non-commutative ring. and ab ab = aa+ bc ab+ bd cd cd ca+ dc cb+ dd is a non-commutative ring.

SOLUTION: In CYU 3.4 below you are invited to show that . M22 = M22  +, is a ring. Is it a commutative ring? No:

10 01 ==01 while 01 10 00 00 00 00 00 00 00

CHECK YOUR UNDERSTANDING 3.4

. (a) Verify that M22 = M22  +, is a ring with unity. Answer: See page A-17. (b) Prove that if a ring contains a unity, then that unity is unique.

You need to distinguish DEFINITION 3.3 Let R be a ring with unity 1. An element between “unity” and “unit.” aR is a unit if there exists bR such unity: Multiplicative identity. Unit unit: An element that has a that ab== ba 1 . multiplicative inverse. The element b is called the inverse of a and is denoted by a–1 .

EXAMPLE 3.3 (a) Show that the ring M22 of Example 3.2 has a unity.

(b) Show that –25 is a unit, and that 94– 23 is not a unit in M22 . –64 –

10 SOLUTION: (a) is the unity in M22 : 01

10 ab ==ab 10 ab 01 cd cd 01 cd 3.1 Definitions and Examples 115

(b) Does there exist a matrix ab for which: A linear algebra approach: cd Since det –25  0 , –25 ab ==ab –25 10? 94– 94– cd cd 94– 01 Let’s see: –25 is invertible. 94– –25 ab = 10  2a +23c b + 3d = 10 94– cd 01 –64a –4c –6b – d 01 2a +13c =   2b +03d =    –64a –0c =  –6ab –1d =   If you take the time to solve the above system of equations you will find that: a ===2 b 3 c –64 and d =– ; leading us to:

–25 23 = 10 94– –64 – 01

Also, as you can easily check: 23 –25 = 10 . –64 – 94– 01

As for 23 : –64 –

CHECK YOUR UNDERSTANDING 3.5

23 Verify that is not a unit in M22 . Answer: See page A-18. –64 –

THEOREM 3.2 Let R be a ring with unity 1. Then: (a) –1 –1 = 1 (b) For any aR : –1 aa==–1 –a

PROOF: (a) –1 –1 ==1 1 1

Theorem 3.1(c) (b) –1 a ==1–a –a and a–1 ==–a1 –a Theorem 3.1(b) Theorem 3.1(b) 116 Part 3 From Rings To Fields

You are invited to establish the following result in the exercises: THEOREM 3.3 . For any integer n  1 , Zn+n n , under addition and multiplication modulo n; which is to

say, for ab  Zn : For example, in a +n br== where: ab+ qn+ r 0  rn Z = 012345 : 6 and (see margin) 1+6 3== 4 and 2+6 51 while a .n br== where: ab qn+ r 0  rn . . 1 n 3== 3 and 2 n 54 is a ring.

CHECK YOUR UNDERSTANDING 3.6

(a) Determine the units in the ring Z6

Answer: (a) 1, 5 (b) Show that mZ n is a unit if and only if m and n are relatively (b) See page A-18. prime. As might be anticipated: DEFINITION 3.4 A subring of a ring R is a nonempty subset S of R which is itself a ring under the imposed binary operations of R. As was the case with groups, the above subring definition can be recast in a more compact form:.

THEOREM 3.4 Let R +, . be a ring. A subset S of R is a subring of R if and only if: (i) S, + is a subgroup of R, + . (ii) S is closed under multiplication, i.e: ss  S ss S

PROOF: If S is a subring of R then (i) and (ii) clearly hold. Con- versely, if (i) and (ii) hold then, since abc= ab c along with ab+ c = ab+ ac and ab+ cacbc= + hold for all elements abc  R, they must surely hold for all elements abc  S .

CHECK YOUR UNDERSTANDING 3.7 Let R +, . be a ring. A subset S of R is a subring of R if and only if for every ss  S : (i) ss–  S and (ii) ss S Answer: See page A-18 Suggestion: Consider Exercise 38, page 70 3.1 Definitions and Examples 117

EXAMPLE 3.4 (a) Show that, for any nZ + , the additive group nZ under standard multiplication is a subring of Z.

(b) Is U22 = AA is a unit in M22 a subring of M22 ? (a) For any ab  Z : na– nb = na– b  nZ, and na nb = nnab nZ.

(b) No. U22 is not closed under addition: . Incidentally U22  is a group (under multipli- 10 –01 10 –01 00   U22 but + =  U22 . cation) (Exercise 40) 01 01– 01 01– 00

CHECK YOUR UNDERSTANDING 3.8

00 (a) Show that H = ab   is a subring of M22 . ab (b) Let a be and element of a ring R. Show that

Sa = xRax = 0 Answer: See page A-18. is a subring of R. 118 Part 3 From Rings To Fields

EXERCISES

Exercise 1-12. Determine if the given set it a ring under the give addition and multiplication operations. If it is a ring, indicated whether or not it is commutative, and whether or not it has a unity. 1. The set nZ under standard addition and multiplication.

2. The set 2nn Z+ of positive even integers under standard addition and multiplication.

3. The set 2nn 0 of nonnegative even integers under standard addition and multiplication. 4. The set ab+ 2 ab   under standard addition and multiplication.

5. The set ab+ 2 ab  Q under standard addition and multiplication. 6. The set 01 under standard addition and multiplication. 7. The set 2ZQ under component addition and multiplication. 8. The set 2Z  01 under component standard addition and multiplication.

ab 9. The set abc   under matrix addition and multiplication. (See Example 3.2.) 00

ab 10. The set abc   under matrix addition and multiplication. (See Example 3.2.) c 0

ab 11. The set abc   under matrix addition and multiplication. (See Example 3.2) 0 c

12. The set of polynomials, px , with real coefficients, of degree less than or equal to 5, under standard polynomial addition and multiplications.

Exercise 13-20. Determine if the given subset S of the giver ring R is a subring of R. 13.RQ= , and SQ= + . 14.RQ= and SZ= .

15.RZZ=  and Snn=  . 16.RQ= , and Sq= 2 qQ .

17.RZZ=  and Snn=  n  0 . 18.RZZ=  and Sn=  2n .

aa aab–  19.RM= 22 and S =  . 20.RM= 22 and  . 00 ab– b 3.1 Definitions and Examples 119

Exercise 21-27. Find the units in the give ring.

21.Z 22. 5Z 23.Z5 24. Z15 25.ZZ 26. ZQ 27. Z6  Z9

28. Show that any abelian group G + can be turned into a ring by defining ab = 0 for every ab  G.

29. Verify that for any ABC  M22 (see Example 3.2): (a) AB CABC=  (b) AB+ C = AB+ AC (c) AB+ CACBC= + 30. Let R and R be rings. Prove that R  R is a ring. 1 2 1 2 n 31. Let R n be a collection of rings. Prove that R is a ring. i i = 1  i i = 1 R 32. Let R   A be a collection of rings. Prove that   is a ring.   A 33. Let a and b be element in a ring R. Show that nm ab = na mb for any integer n and m. 34. Describe all of the subrings of the ring of integer. 35. Let the ring R be cyclic under addition. Prove that R is commutative. 36. Let F denote the set of all real-valued functions. For f and g in F , let fg+ be given by fg+ x = fx+ gx and fg x = fxgx . Show that under these operation F is a ring with unity. 37. The center of a ring R is the set xRax = xa aR . Sow that the center of R is a subring of R. 38. For a and element of a ring R, let Ca= xRxa = ax . Show that Ca is a subring of R containing a. 39. Show that the center of a ring R is equal to  Ca . (See Exercises 36 and 37.) aR ab 40. Prove that is a unit in M22 if and only if ad– bc  0 . cd 41. Prove that if aR is a unit, then it has a unique inverse.

42. Prove that the set U22 = AA is a unit in M22 is a group under multiplication.

43. Let R be a ring, and let aR . Show that the set Sa = axa x R is a subring of R. 44. Show that the multiplicative inveres of any unit in a ring with unity is unique. 45. Let R be a commutative ring with unity, and let UR denote the set of units in R. Prove that UR is a group under the multiplication of R. 46. Show that if there exists an integer n greater than 1 for which xn = x for every element x in a ring R, thenab ==0  ba 0 . 47. Let k be the least common multiple of the positive integers m and n. Show that mZ nZ = kZ. 120 Part 3 From Rings To Fields

48. Let R be a commutative ring. Prove that a2 – b2 = ab+ ab– . 49. An element a of a ring R is idempotent if a2 = a . Show that the set of all idempotent elements of a commutative ring is closed under multiplication. 50. An element a of a ring R is nilpotent if an = 0 for some nZ + . Show that if a and b are nilpotent elements of a commutative ring R, then ab+ is also nilpotent. 51. A ring R is said to be a Boolean ring if a2 = a for every aR . Prove that every Boolean ring is commutative. 52. Give an example of finite Boolean ring, and an example of an infinite Boolean ring (see Exer- cise 50). 53. Prove Theorem 3.3.

54. Prove that m is a unit in Zn if and only if gcd n m = 1 .

55. Let R1R2  Rn be rings. Show that:

(a)R1 R2   Rn with operations

a1a2  an + b1b2  bn = a1 + b1a2 + b2  an + bn

a1a2  an b1b2  bn = a1b1a2b2  anbn is a ring.

(b)R1 R2   Rn is commutative if and only if Ri is commutative for 1 in .

PROVE OR GIVE A COUNTEREXAMPLE

56. If R1 and R2 are rings, then R1  R2 is a ring. 57. In any ring R, ab ==0  ba 0 . 58. If x3 = x for all elements x in a ring R, then6x = 0 for all xR . 59. In any ring R: a2 – b2 = ab+ ab– .

60. If R1 and R2 are Boolean ring, then R1  R2 is a Boolean ring. (See Exercise 50).

61. If R1 and R2 are Boolean ring, then R1  R2 is a Boolean ring. (See Exercise 50). 62. The set of all idempotent elements in a ring R is a subring of R. (See Exercise 48). 3.2 Homomorphisms and Quotient Rings 121

§2. HOMOMORPHISMS AND QUOTIENT RINGS Moving the group-homomorphism concept of page 72 up a notch we So, a ring homomorphism come to:: preserves both the sum and product operations: DEFINITION 3.5 The function : R +, .   R   +, . is a You can perform sums and RING homomorphism if, for every ab  R : products in R and then HOMOMOrphism carry the results over to the (1) ab+ = a + b ring R (via  ), or you can first carry a and b over to and (2) ab = a b R and then perform the operations in that ring. A homomorphism : RR  which is also ISOMORPHISM a bijection is said to be an isomorphism from the ring R to the ring R . ISOMORPHIC Two rings R and R are isomorphic, written RR  , if there exists an isomorphism from one of the rings to the other. Condition (1) above assures us that a ring homomorphism is also a group homomorphism : R, +  R, + . That being the case, previously encountered group-homomorphic results remain in effect in the Why Ker = 0 rather current setting. In particular, Theorem 2.25, page 76, tells us that: than Ker = e ? Because we are dealing A ring homomorphism : R +, .  R +, . with abelian groups R, + and R, + is one-to-one if and only if Ker = 0 . that’s why. EXAMPLE 3.5 Let : ZZ n be given by a = ra , where aq= anr+ a , with 0  ra  n . Show that  is a ring homomorphism.

SOLUTION: A consequence of CYU 1.18, page 36, and the fact that: ab+  a + b mod n and ab  a b mod n

As it is with group homomorphisms, we have:

THEOREM 3.5 Let : RR  be a ring homomorphism. (a) If H is a subring of R, then H is a subring of R . (b) If H is a subring of R , then –1H is a subring of R .

PROOF: We establish (a) and invite you to verify (b) in CYU 3.9. Appealing to CYU 3.7, page 116, we show that the nonempty set H is closed under subtraction and multiplication:

h1 – h2 = h1 – h2  H

h1 h2 = h1h2  H 122 Part 3 From Rings To Fields

CHECK YOUR UNDERSTANDING 3.9

(a) Let : RR  be a homomorphism. Prove that if H is a subring of R , then –1H is a subring of R . (b) Show that the rings 3Z and 5Z are not isomorphic. Answer: See page A-19. (Compare with CYU 2.22(b), page 77) Every ring R +, . is, in part, an abelian group: R + . Choosing to use the sum notation in Theorem 2.39 (page 95) we have: For any subring H of the ring R +, . , the factor group GH = aH+ aG is a group under the coset operation: aH+ + bH+ = ab+ + H

Fine, but will that factor group GN evolve into a ring under the “natural” product operation aH+ bH+ = ab + H ? Not necessarily. Indeed, that coset “product” need not be defined. A case in point: EXAMPLE 3.6 00 Consider the subring H = ab   of ab CYU 3.8(a), page 117. Show that 10 01 10 01 + H + H = + H ab ab ab ab is not a well defined operation. SOLUTION: To be well define, the set products must yield the same result, independently of the chosen representative for the two given cosets. However, while:

10 – 10 = 00  H 10 10 01 01 00 01 01 + H = + H and + H = + H : 00 margin 01 00 01

10 01 + H is not equal to 10 01 + H . 00 00 01 01

Why not? Because: 10 01 = 01 , 01 01 = 02 00 00 00 00 01 00

and 01 – 02 = 01–  H 00 00 00

The above example illustrates the fact that for a given subring H of a Note that the factor group ring R, one can not expect that the factor group RH of Theorem 2.39, RH exists, as every sub- page 95, will become a ring under the (attempted) product group of an abelian group aH+ bH+ = ab+ H. Of particular importance are those subring is normal. for which that expectation will be realized: 3.2 Homomorphisms and Quotient Rings 123

DEFINITION 3.6 A subring I or a ring R is a (two-sided) ideal IDEAL if for any aI and every rR : ra I and ar I Justifying our expectation: THEOREM 3.6 If I is an ideal in R then the (additive) factor group RI turns into a ring under the imposed multiplication aI+ bI+ = ab + I .

RI is said to be the quotient ring of R by N. Quotient rings are also said to be factor rings. PROOF: The first order of business is to show that the above coset product operation is well defined; which is to say that: If aI+ = a + I and bI+ = b + I then: ab+ I = ab + I . Lets do it: aI+ = a + Iaa –   Iaa –  bI  aba– bI since I is an ideal bI+ = b + Ibb –   Ia bb–   Ia ba– b  I Since both ab– ab and aba– b are in I: ab– ab + aba– b = ab– ab  I Thus ab+ I = ab + I (as I, being a subgroup of an abelian group, is normal. As for the ring part of the proof, we need only establish the associative and distributive axioms of Definition 3.1 (page 111), as we already know that RI is an abelian group. Not a serious challenge, now that we know that the coset multiplications is well defined: Associative: aI+ bI+ cI+ = aI+ bc + I ==abc+ I ab cI+ = aI+ bI+ cI+ Distributive: aI+ bI+ + cI+ = aI+ bc+ I = ab+ c + I = ab+ ac + I = ab+ I + ac+ I = aI+ bI+ + aI+ cI+ In the same fashion, one can show that: bI+ + cI+ aI+ = bI+ aI+ + cI+ aI+ 124 Part 3 From Rings To Fields

CHECK YOUR UNDERSTANDING 3.10 (a) Show that I is an ideal in Z if and only if InZ= . (b) Let : RR  be an onto ring homomorphism. Show that if I is Answer: See page A-19. an ideal in R then I is an ideal in R Roughly speaking:

NORMAL SUBGROUPS ARE TO GROUPS AS IDEALS ARE TO RINGS A case in point (compare with Theorem 2.43, page 98): THEOREM 3.7 If : RR  is a ring homomorphism, then FIRST K = Ker is an ideal of R and: ISOMORPHISM THEOREM RK  R

PROOF: We already know that K ==Ker –10 is an additive subgroup of R. It is, in fact, an ideal since, for any aK and every rR , both ra and ar are in K: ra === r a r 0 0 and ar =0 The proof of Theorem 2.43 serves to show that the function : RK R given by rK+ = r is a group isomorphism. Indeed, it a ring isomorphism:

r1 + K r2 + K = r1r2 + K

==r1r2 r1 r2

= r1 + K r2 + K

CHECK YOUR UNDERSTANDING 3.11 Let : RR  be an onto ring homomorphism with kernel K. If I is an ideal of R , then I = –1I is an ideal in R containing K and: Answer: See page A-19. IK  I In CYU 3.10(a) you were invited to show that, under standard addition and multiplication, nZ is an ideal of Z. More can be said: EXAMPLE 3.7 Show that for any positive integer n: Zn  ZnZ  SOLUTION: In Example 2.6, page 72, we showed that the function : ZZ n given by m = r where mnqr= + with 0  rn is a group homomorphism. You are invited to show, in CYU 3.12, that the function also preservers products; in other words, that it is a ring homomorphism from the ring Z to the ring Zn . 3.2 Homomorphisms and Quotient Rings 125

While  is not one-to-one, it is certainly onto. As such we know, by Theorem 3.7, that:

Zn  ZK where K = Ker . Noting that: Ker ===m m = 0 kn k Z nZ Example 2.4, page 62 we conclude that: Zn  ZnZ  .

CHECK YOUR UNDERSTANDING 3.12

Let : ZZ n be given by m = r where mnqr= + with Answer: See page A-19. 0  rn . Show that for any ab  Z : ab = a b . You are invited to establish the next two isomorphism theorems in the exercises. THEOREM 3.8 Let H be a subring of a ring R, and I an ideal SECOND of R. Then: ISOMORPHISM THEOREM HI+ = hihHiI+    is a subring of R, I is an ideal of HI+ , and: HI+  I  HH  I (Compare with Theorem 2.44, page 99.)

THEOREM 3.9 Let : RR  be an onto homomorphism THIRD with kernel K. If I is an ideal of R , then: ISOMORPHISM THEOREM I ==–1I aR a  I is ideal of R and: RI  R  I (Compare with Theorem 2.45, page 99.) 126 Part 3 From Rings To Fields

EXERCISES

Exercise 1-6. Determine if the given map : RR  is a ring homomorphism.

1.RR== Z , and n = 3n . 2.RZ= , R = 3Z and n = 3n .

ab –d b ab ab 0 3.RR== M22 and  = . 4.RR== M22 and  = . cd dc cd 0 cd

ab ab 5.RM= 22 , R= and  = a . 6. RM= 22 , R =  and  = ad– bc . cd cd

Exercise 7-10. Determine if the given subset S of the ring R is an ideal of R.

a 0 a 1 7.RM= 22 , S = ad   . 8.RM= 22 , S = ad   . 0 d 0 d

9.RZZ=  , Saa=  – aZ . 10.RZZ=  , S = 2aa aZ .

11. Let : RR  be a homomorphism from R onto R . Show that: (a)an =  a n for all aR and n  0 . (b) If R possesses a unity then so does the ring R . (c) If a is a unit of R, then a is a unit of R . 12. Let : RR  and : R  R be ring homomorphisms. Prove that the composite function : RR  is also a homeomorphism. 13. Let S be a subset of a ring R. Show that S is an ideal of R if and only if the following two conditions hold: (i) S is an additive subgroup of R. (ii) For every sS and rR we have rs S and sr S . 14. Let F denote the ring of all real-valued functions of Exercise 36, page 119, and let

a   . Show that the map a: F   given by af = fa is a homomorphism. 15. Let I be an ideal is commutative ring R with unity 1. Show that RI is a commutative ring with unity. 16. Let R be a ring with unity. Show that if I is an idea of R that contains a unit, then IR= . 17. Let I be an ideal in a ring R. Show that there exist an onto ring homomorphism : RRI  with Ker = I . 18. Let I be an ideal in a ring R. Show that if K is an ideal in I, then KaIaK= +  is an ideal in RI . 3.2 Homomorphisms and Quotient Rings 127

19. Let I be an ideal in a ring R. Show that if K is an ideal in RI , then there exists an ideal K in R with IK such that KKI== aIaK+  . 20. Describe all ring homomorphisms from Z to Z. 21. Describe all ring homomorphisms from Z to ZZ . 22. Show that if nm , then the rings nZ and mZ are not isomorphic. 23. Prove that  (isomorphic) is an equivalence relation on any set of rings (see Definition 1.12, page 29).

24. Show that the function : ZZ n given by m = r where mnqr= + with 0  rn preservers products: st = s t st  Z . 25. Find a subring of ZZ that is not an ideal of ZZ . 26. Prove that I is an ideal of a ring R if and only if: (i) 0  I (ii) If ab  I. then ab–  I (iii) If aI and rR , then ra R 27. An element a of a ring R is nilpotent if an = 0 for some nZ + . Show the collection of nilpotent elements of commutative ring R is an ideal of R. 28. Let R be a commutative ring and aR . Show that xRxa = 0 is an ideal of R.

29. Prove that if I1 and I2 are ideals of R, then I1  I2 is an ideal of R. 30. Let A and B be ideals of R, and let I be the set of all elements of the form ab with aA and bB . Prove that I is an ideal of R. 31. Let H be a subring or R that is not an ideal of R. Verify that the operation aH+ bH+ = ab+ H is not well defined.. 32. Prove Theorem 3.8. 33. Prove Theorem 3.9.

PROVE OR GIVE A COUNTEREXAMPLE

34. If I1 and I2 are ideals of R, then I1  I2 is an ideal of R. 35. If I is an ideal of R and if H is a subring of R, then IH is an ideal of R.

36. For every element a of a ring R, the set xRxa = 0 is an ideal of R.

37. Let : RR  be a homomorphism from R onto R . Show that if R possesses a unity then so does R . 38. The collection of nilpotent elements n a ring R is an ideal of R. (See Exercise 27). 128 Part 3 From Rings To Fields

§3. INTEGRAL DOMAINS AND FIELDS The set Z of integers under addition led to the definition of a group on page 41. Tossing multiplication into the mix brought us to the definition of a ring on page 111. How about “division”? Can one perform (grade school) division in the ring Z, or Q, or  ? Absolutely not in Z, where you can only divide by 1 or –1 . Q and  fair much better in that one can divide by any number other then 0. As it turns out, Q and  are examples of fields: DEFINITION 3.7 A zero-divisor in a commutative ring R is a ZERO DIVISOR nonzero element a for which there exits a nonzero element bR with ab = 0 .

INTEGRAL DOMAIN An integral domain is a commutative ring R with unity that contains no zero-divisors. Note that Every field is an integral domain. FIELD A field is a commutative ring with unity in which every nonzero element is a unit. As is depicted below, fields are at the top of our algebraic pecking order, and groups are at the bottom: Fields: Q 

not a field: 2 has no multiplicative inverse

Integral Domains: QZ

not an integral domain: 26 = 0

Commutative Rings with unity: QZZ6 no unity not commutative

Rings: QZZ6 2ZM 22 not a ring, see CYU 3.1(a), page 111

Groups: QZZ6 2ZM 22 S3

CHECK YOUR UNDERSTANDING 3.13 Assign to the given group its highest algebraic rank (Field at the top, Answer: (a) Field. (b) Commutative ring and Group at the bottom). with unity. (a) Z5 (b) Z15 3.3 Integral Domains and Fields 129

The familiar high school cancellation law (margin) holds in any inte- ab b gral domain: ------= --- (if a  0) c ac THEOREM 3.10 Let D be an integral domain, and abc  D . If ab= ac and a  0 , then bc=

PROOF: ab=== ac ab– ac 0  ab– c 0 . Since D is an integral domain, and since a  0 : bc–0= .

CHECK YOUR UNDERSTANDING 3.14 (a) Prove that a commutative ring with unity is an integral domain if and only if the cancellation property of Theorem 3.10 holds. (b) Let D be an integral domain, and let a  0 be an element in D. Show that the function fa: DD given by fax = ax is one- Answer: See page A-20. to-one.

THEOREM 3.11 A commutative ring with unity R is a field if and only if 0 and R are the only ideals in R.

PROOF: Let R be a field and let I be an ideal in R with I  0 . Chose aI , a  0 . Since R is a field and I is an ideal, we then have: a–1a = 1  I . Since I is an ideal: r  1 rR = RI . Thus: IR= . To establish the converse, we show that if 0 and R are the only ideals in R, then every nonzero element in R is a unit: Let a  0 be an element of R. Consider the ideal Ia=  . Since I  0 , IR= . We then have 1  I = a . It follows that an = 1 for some nZ , and that an – 1 is the inverse if a.

THEOREM 3.12 Any finite Integral domain is a field.

PROOF: Let D be a finite integral domain, and let a  0 be an ele-

ment in D. CYU 3.14(b) assures us that the function fa: DD

given by fax = ax is one-to-one. It follows, since D is finite, that the function is also onto. In particular, there must exist some bD such that ab = 1 , and a is seen to be a unit. Since a was an arbitrary nonzero element in D, D is a field.

CHECK YOUR UNDERSTANDING 3.15

Answer: See page A-20. Prove that for every prime p, Zp is a field. 130 Part 3 From Rings To Fields

We focus briefly, and somewhat loosely, on the set PZx of polynomials with integer coefficients, as well as the sets P x of polynomi- Zn

als with coefficients taken from the rings Zn = 012 n – 1 . All turn out to be commutative rings (with unity) under the following standard sum and product operations: In P x In P x Z Z6

3x2 ++4x 5 +4x2 ++4x 1 = x2 ++8x 6 3x2 ++4x 5 +4x2 ++4x 1 = x2 + 2x

2x2 – 4x 5x – 2 = 2x25x – 2 – 4x5x – 2 2x2 – 4x 5x – 2 = 2x25x – 2 – 4x5x – 2 = 10x3 –204x2 –8x2 + x = 4x3 –24x2 –2x2 + x = 10x3 –824x2 + x = 4x3 + 2x

Note that while the coefficients of a polynomial px in P x are Zn elements of the ring Zn = 012 n – 1 , the degree of such a polynomial can be any nonnegative integer. In particular, while 3x9 – 5x2 + x – 1 might very well be a polynomial in P x (of Z6 degree 9), it can not be a polynomial in P x (5  Z ). Z5 5 Consider the polynomial x2 –6x – . Since the distributive property holds in both P x and in P x we can express the polynomial in Z Z12 factored form in either ring: x2 –6x – = x – 3 x + 2 But while the equation: x2 –6x – ==x – 3 x + 2 0

has but two solutions in PZx (3 and –2 ), the same equation turns out to have four solutions in P x . The reason, you see, is that while the Z12 ring Z is an integral domain (no zero divisors), the same cannot be said

for Z12 . Indeed there are several pairs on nonzero elements in Z12 with product equal to zero: 26 ====0, 3 4 0 8 3 0 9 4 0 10 6 =0 24 0 mod 12 Your turn: CHECK YOUR UNDERSTANDING 3.16 Solve the equation x2 –6x –0= in: Answer: (a) 3, 6, 7, 10 (b) 3, 6 (a) Z12 (b) Z8 3.3 Integral Domains and Fields 131

DEFINITION 3.8 The characteristic of a ring R is the least nx= x++ x  +x CHARACTERISTIC positive integer n such that nx = 0 for n of them every xR . If no such integer exists, then R is said to have characteristic 0.

The ring Z has characteristic 0, and the cyclic ring Zn has characteristic n. Clearly no finite ring is of characteristic 0. Must every infinite ring have characteristic 0? No: CHECK YOUR UNDERSTANDING 3.17 Answer: See page A-21. Prove that the infinite ring P x has characteristic n. Zn To determine the characteristic of a ring with unity, one need look no further than its unity: THEOREM 3.13 Let R be a ring with unity. If n10 for all nZ + , then R has characteristic 0. If n10= for some nZ + , then the smallest such n is the characteristic of R.

PROOF: If n10 for all nZ + then surely there cannot exist nZ + such that nx = 0 for every xR , and R has characteristic 0. On the other hand, if n10= for some positive integer n, then, for any xR : nx== nx x++ x  +x = 11++ +1 xn =1 x ==0x 0 n of them The smallest such n is then the characteristic of R.

THEOREM 3.14 The characteristic of an integral domain D is either 0 or prime.

PROOF: Assume that D has positive characteristic n, and that n is not prime. Then n can be written as nst= with 1 sn and 1 tn . We then have: 0 ==n1 st 1 =st 12 =s1 t1 Since D has no zero divisors, either s10= or t10= . But this cannot occur, since n is the least positive integer such that n10= . Conclusion: n must be prime.

CHECK YOUR UNDERSTANDING 3.18 Let D be an integral domain of characteristic 3. Show that for every ab  D: ab+ 3 = a3 + b3 . In the exercises you invited to show that for any prime p, if D has characteristic p then: Answer: See page A-21. ab+ p = ap + p . 132 Part 3 From Rings To Fields

PRIME AND MAXIMAL IDEALS

DEFINITION 3.9 Let R be a commutative ring. Note: A proper ideal of R PRIME IDEAL A prime ideal of R is a proper ideal I of R is, by definition, an ideal in for which: R that is distinct for R itself. ab I a I or bI MAXIMAL IDEAL A proper ideal I of R is a maximal ideal if R is the only ideal containing I.

EXAMPLE 3.8 (a) Show that an ideal I in Z is prime if and only if IpZ= , where p is a prime. (b) Show that 5Z is a maximal ideal in Z. SOLUTION: (a) Let p be prime. If ab pZ then, by Theorem 1.9, page 24, pa or pb ; which is to say, that apZ or bpZ . Conversely, assume that InZ= where n  1 is not prime. Let nab= for some positive integers a and b. Then ab I with neither a nor b in I (neither is a multiple of n). (b) If I is an ideal properly containing 5Z , then there must exists aI with a  5Z , i.e. 5 does not divide a. It follows, since 5 is prime, that gcd5 a = 1 . Employing Theorem 1.7, page 23, we have: 15= sat+ for integers s and t. Since 5s and at are both in I: 1  I . It follows, since I is an ideal in Z, that IZ= .

CHECK YOUR UNDERSTANDING 3.19 (a) Show that pZ is a maximal ideal for any prime p. Answer: See page A-21. (b) Prove that I is a maximal ideal in Z if and only if it is prime.

THEOREM 3.15 Let I be an ideal in a commutative ring R The converse of both (a) with unity. and (b) also hold. See Exer- (a) If I is a prime ideal then RI is an integral cise 25 and Exercise 26. domain. (b) If I is a maximal ideal then RI is a field.

PROOF: (a) We need to show that RI has no zero divisors; which We already know that is to say that if ab+ I = I , then either aI+ = I or bI+ = I ; RI is a commutative ring with unity (see The- which is to say that if ab I then either aI or bI . And this is orem 3.6, page 123). so, as I is a prime ideal. (b) Invoking Theorem 3.11, we show that the only ideals of RI are 0 and RI : Let I be a maximal ideal in R, and let K be an ideal in RI . Exercise 18, page 126, assures us that there exists an ideal K in R with IKR such that KKI=  . Since I is a maximal ideal, either KI= , in which case K = 0 , or KR= , in which case KRI=  . 3.3 Integral Domains and Fields 133

FIELDS OF QUOTIENTS Let’s mimic the development in which the integers Z blossom into the field of rational numbers Q, to one that nurtures a general integral domain D into its field of quotients F: From Z to Q From D to the field of quotients F

 Let S = a ab  Z, with b  0 . Z --- Let SD = ab ab  D, with b  0 . b

In Example 1.9, page 29, we demonstrated that Following the procedure of Example 1.9, page a c the relation ---  --- if ad= bc is an equivalence 29, one can show that the relation ab  cd b d if ad= bc is an equivalence relation on SD . relation on SZ . Let Q denote the set of equivalent Let F denote the set of equivalent classes associated with the above equiv- classes associated with the above equivalence relation on SZ . alence relation on SD . Define addition and multiplication in Q as fol- Define addition and multiplication in F as follows: lows: a c ac+ a c ac ab + cd = ad+ cb bd --- + --- = ------and ------= ------b d bd b d bd and ab cd = ac bd You are invited to show in the exercises that the You are invited to show in the exercises that the above operations are well defined; which is to say: above operations are well defined; which is to say: a a c c If ab  a b and cd  c d , then: If ---  ---- and ---  ---- , then: b b d d ad+ cb bd  ad + cb bd ac+ a + c ac ac ------ ------and ------ ------bd bd bd bd and ac bd  ac bd You are also invited to show in the exercises that: You are also invited to show in the exercises that: 0 1 F+ . is a field, with zero 0 a Q+ . is a field, with zero --- and unity --- . 1 1 and unity aa , for any a  0 .

As is the case with the rational numbers, where the equivalence class --a- is simply denoted by the “fraction” --a- , so b b then one generally represents an element ab in the field of quotients F by the two-tuple ab . 134 Part 3 From Rings To Fields

EXERCISES

Exercise 1-6. Find the zero-divisors of the given ring.

1.3Z 2.Z4 3. Z3  Z6

4.ZZ 5 5.Z2  Z5 6. Z4  Z8

Exercise 7-12. Determine the characteristic of the given ring.

7.3Z 8.Z4 9. Z3  Z6

10.ZZ 5 11.Z2  Z5 12. Z4  Z8

Exercise 13-15. Solve the equation x2 –65x +0= in:

13.Z2 14.Z5 15. Z12

Exercise 16-18. Solve the equation x3 –43x –0= in:

16.Z2 17.Z5 18. Z12

19. Show that Zp has no zero divisors for any prime p.

20. Show that the zero divisors of Zn are the nonzero elements that are not relatively prime to n.

21. Show that every nonzero element in Zn is a unit or a zero-divisor.

22. Let R be a finite commutative ring with unity. Prove that every nonzero element in Zn is a unit or a zero-divisor. 23. Give an example of a ring R that contains a nonzero element that is neither a zero-divisor nor a unit. 24. Show that any nonzero element a in a commutative ring R is a zero-divisor if and only if a2b = 0 for some b  0 . 25. Let R and S be nonzero rings. Can RS be an integral domain? 26. Give an example of a commutative ring R without zero-divisors that is not an integral domain. 27. A nonempty subset S of an integral domain D is called a subdomain of D if it is an integral domain under the operations of D. Prove that a nonempty subset of D is a subdomain of D if and only if S is a subring of D that contains the unity of D. 28. Prove that the intersection of two subdomains of an integral domain D is also a subdomain of D. (See Exercise 18.) 29. Find all subdomains of Z. (See Exercise 27.) 3.3 Division Rings and Fields 135

30. Show that the only subdomain of Zp , for p prime, is Zp . (See Exercise 18.) 31. Let D be an integral domain of prime characteristic p. Show that for every ab  D : ab+ p = ap + bp 32. Prove that every maximal ideal in a commutative ring with identity is a prime ideal. 33. Let I be an ideal in a commutative ring R with unity. Prove that I is a prime ideal of R if and only if RI is an integral domain. [See Theorem 3.15(a).] 34. Let I be an ideal in a commutative ring R with unity. Prove that I is maximal in R if and only if RI is afield. [See Theorem 3.15(b).] 35. Prove that every proper ideal on a ring with unity is contained in a maximal ideal. 36. Let R be a commutative ring. Prove that if P is a prime ideal of R that contains no zero-divisors, then R is an integral domain. 37. Let R be a commutative ring. Let I and J be ideals of R. Show that if P is a prime ideal of R that contains IJ , then either I or J is contained in P.

38. Show that the subset S = 03 is an ideal in Z6 . Show that while S is not an integral domain, Z6  S is a field. 39. Show that any ring homomorphism : FR from a field F to a ring R  0 is one-to-one. 40. Let R be a commutative ring. Prove that R is a field if and only if 0 is a maximal ideal. 41. Referring to the “From D to the field of quotients F” development on page 133,verify that the operations: ab + cd = ad+ cb bd and ab cd = ac bd are well defined. 42. Referring to the “From Z to Q” development on page 133,verify that the operations: a c ac+ a c ac --- + --- = ------and ------= ------b d bd b d bd are well defined.

43. Referring to the “From Z to Q” development on page 133,verify that Q+ . is a field, with 0 1 zero --- and unity --- . 1 1 44. Referring to the “From D to the field of quotients F” development on page 133,verify that F+ . is a field, with zero 0 a and unity aa , for any a  0 . 45. Establish Fermat’s Little Theorem: If aZ and if p is a prime not dividing p, then: ap – 1  1 (mod p  136 Part 3 From Rings To Fields

46. Show that for any prime p and any aZ : ap  a (mod p 

PROVE OR GIVE A COUNTEREXAMPLE

47. The intersection of subdomains of an integral domain D is a subdomain of D. (See Exercise 18.) 48. If : DR is a homomorphism from the integral domain D to a ring R, then D is a n integral domain. 49. Let R be a commutative ring with unity. If P is a prime ideal of R and if J is a subring of R, then PJ is a prime ideal of R. 50. Let R be a commutative ring with unity. If P is a prime ideal of R and if I is an ideal of R, then PJ is a prime ideal of R. CYU SOLUTIONS A-1 APPENDIX A CHECK YOUR UNDERSTANDING SOLUTIONS PART 1 PRELIMINARIES 1.1 Functions  ab 2 CYU 1.1 For f: M22   , given by f = ad+ , and g:   R given by cd gx= 2xx 2 , we have:

13 13 (a) g f =====gfg14+ g5 25  52 10 25  24 24

ab ab (b) g f ===gfga+ d 2ad+  ad+ 2  cd cd = 2a + 2da 2 ++2ad b2

 4 ab CYU 1.2 (a) Let f: M22  R be given by f = dc –  3ab cd

ab ab One-to-one: f ==f  dc –  3ab dc –  3ab cd cd dd=  dd=    –c = –c  cc=  ab ab   = 3a = 3a  aa=  cd cd   bb=  bb= 

ab ab Onto: For given xyzw , we find such that f = xyzw : cd cd

dx=  dx=    ab –c = y  cy= –  f ==xyzw  dc–3ab xyzw   cd 3az=  az=  3    bw=  bw= 

z  3 w Hence: f = xyzw . –y x A- 2 APPENDIX A

ab ba 00 00 00 (b) f = is not one-to one: f ==f  . cd cd+2b 00 10 00

ab 10 ab b a 1 0 f is not onto, since no element is mapped to : f =  . cd 03 cd cd+ 2b 0 3

CYU 1.3 Let yY . Since yf –1y  f –1 , f –1y  y  f , which is to say: ff–1y = y .

 4 ab CYU 1.4 The function f: M22  R given by f = dc –  3ab is a bijection [see cd ab ab CYU 1.2(a)]. To find its inverse we determine for which f = xyzw : cd cd

dx=  az=  3    ab –c = y  bw=  f ==xyzw  dc –  3ab xyzw   cd 3az=  cy= –    bw=  dx= 

Conclusion: f –1xyzw = z  3 w –y x · –1 z  3 w z Moreover: ffxyzw ==f xy–3– ---  w =xyzw –y x 3

–1 ab –1 3a  3 b ab and: f f ===f dc –  3ab cd ––c d cd

–2y x ab CYU 1.5 From fxyzw = and g = dc –  3ab we have: 3wz cd

–2y x gf xyzw ===gfxyzw gz –3w –3y 2x 3wz

To find its inverse of gf: 4  4 we start with xyzw  4 on the right side of gf: 4  4 and find abcd (on the left side) for which gf abcd = xyzw –1 (we will then turn things around to arrive at gf ). Let’s do it: CYU SOLUTIONS A-3

gf abcd ==xyzw  gfabcd xyzw

–2b a  g==xyzw  c –3d –3b 2a xyzw 3dc cx=  aw=  2    –3d = y  bz= –3    –3bz=  cx=    2aw=  dy= –3  w z y At this point we have g f ----–--- x –--- = xyzw ; and, consequently:  2 3 3 –1 w z y g f xyzw = ----–--- x –---  2 3 3 –1 w z y We now verify that f g–1 xyzw also equals ----–--- x –--- , where  2 3 3

z  3 w –1ab b c g–1 xyzw = and f = ---–a d --- : –y x cd 2 3  –1 –1 –1 –1 –1 z  3 w w z y f g xyzw ===f g xyzw f ----–--- x –--- –y x 2 3 3

1.2 Principle of Mathematical Induction CYU 1.6 (a) The equation 24+= 1234+++–  13+ illustrate that the sum of the first two even integers can be expressed as the sum of the first four integers minus the sum of the first two odd integer. Generalizing, we anticipate that the sum of the first n even integers is the sum of the first 2n integers minus the sum of the first n odd integers; leading us to the conjecture that the sum of the first n even integers equals n2 + n : 2n2n + 1 ------–2n2 ==n2 + nn– 2 n2 + n 2 sum or first 2n integers sum of first n odd integers (Eample 1.16) (page 34) (b) Let Pn be the proposition that the sum of the first n even integers equals n2 + n . I. Since the sum of the first 1 even integers is 2, P1 ==12 +21 is true. II. Assume Pk is true; that is: 246+++ +2k = k2 + k . III. We complete the proof by verifying that Pk+ 1 is true; which is to say, that 246+++ +2k +2k + 1 = k + 1 2 + k + 1 :

by II 246+++ +2k +2k + 1 = k2 ++k 2k + 1 ==k2 ++2k 1 + k + 1 k + 1 2 + k + 1 A- 4 APPENDIX A

CYU 1.7 (a) False — a counterexample: 43+ 1 , and 4 divides neither 3 nor 1. (b) True: Since ab , there exists h such that: (1) bah= . Since ab+ c , there exists k such that: (2) bc+ = ak . From (2): cakb= – . From (1): cakah==– ak– h . Since cat= (where tkh= – ): ac .

CYU 1.8 (a) Let Pn be the proposition n!  n2 : I.P4 is true: 4!1234== 24 42 . II. Assume Pk is true: k!  k2 (for k  4 ) III. We show Pk+ 1 is true; namely, that k + 1 !  k + 1 2 : k + 1 ! = k!k + 1  k2k + 1 II Now what? Well, if we can show that k2k + 1  k + 1 2 , then we will be done. Let’s do it: Since k  4 , k  2 , and therefore k2 = kk  2kk + 1 . Multiplying both sides by the positive number k + 1 : k2k + 1  k + 1 2 .

(b) Let Pn be the proposition that 6 n3 + 5n for all integers n  1 . I. True at n = 1 : 613 + 51 . II. Assume Pk is true; that is: 6 k3 + 5k . III. To establish that 6 k + 1 3 + 5k + 1 , we begin by noting that k + 1 3 + 5k + 1 = k3 ++3k2 8k + 6 and then set our sights on showing that 6 k3 ++3k2 8k (for clearly 66 ). Wanting to get II into play we rewrite k3 ++3k2 8k in the form k3 + 5k + 3k2 + 3k . Our induction hypothesis allows us to assume that 6 k3 + 5k . If we can show that 6 3k2 + 3k , then we will be done, by virtue of Theorem 1.6(b), page 28. Let’s do it: Since 3k2 +33k = kk+ 1 , and since either k or k + 1 is even: 6 is a factor of 3x2 + 3k .

CYU 1.9 Let Pn be a proposition for which P1 is True and for which the validity at k implies the validity at k + 1 . We are to show, using the Well-Ordering Principle, that Pn is True for all n. Suppose not (we will arrive at a contradiction): Let SnZ=  + Pn is False . Since P1 is True, S   . The Well-Ordering

Principle tells us that S contains a least element, n0 . But since the validity at n0 – 1 implies the validity at n0 , n0 – 1 must be in S — contradicting the minimality of n0 . CYU SOLUTIONS A-5

1.3 The Division Algorithm and Beyond CYU 1.10 The division algorithm tells us that n must be of the form 3m , or 3m + 1 , or 3m + 2 , for some integer m. We show that, in each case, n2 = 3q or n2 = 3q + 1 for some integer q: If n = 3m , then n2 ==9m2 3q with q = 3m2 . If n = 3m + 1 , then n2 ==9m2 ++6m 1 33m2 + 2m +31 =q + 1 . If n = 3m + 2 , then n2 ==9m2 ++312m 4 3m2 ++4m 1 +31 =q + 1 .

CYU 1.11 We simply show that c  0 divides nZ if and only if cn : cknc==== kn c kn c hn where hk = since c  0 CYU 1.12 Proof by contradiction: Assume that gcd a c = 1 . From Theorem 1.9: if abc , and if gcd a c = 1 , then ab — contradicting the given condition that ab .

 CYU 1.13 Let Pn be the proposition that if pa1a2 an , then pai for some 1 in . I.P1 is trivially True.  II. Assume Pk is True: If pa1a2 ak , then pai for some 1 ik .   III. Suppose pa1a2 akak + 1 ; or, to write it another way: pa1a2 ak ak + 1 .  If pak + 1 then we are done. If not, then by Theorem 1.8: pa1a2 ak . Invoking II we conclude that pai for some 1 ik .

r1 r2 rs m1 m2 mt CYU 1.14 Let ap=1 p2 ps  b = q1 q2 qt be the prime decompositions of a and b, with distinct primes p1p2  ps , and distinct primes q1q2  qt .

r1 r2 rs ri Since an : nakp==1 p2 ps  k . It follows that and each pi must appear in the prime decomposition of n, for 1 is (with possibly additional pi ’s appearing

mi in the prime decomposition of k). Similarly, since bn , each qi must appear in the prime decomposition of n, for 1 it .

Since a and b are relatively prime, none of the pis is equal to any of the qis . It fol-

r1 r2 rs m1 m2 mt lows that p1 p2 ps q1 q2 qt appears in the prime decomposition of n, and

r1 r2 rs m1 m2 mt that therefore ab= p1 p2 ps q1 q2 qt divides n. A- 6 APPENDIX A

1.4 Equivalence Relations CYU 1.15 Reflexive: Let AS . Since I: AA , given by Ia= a aA is a bijection, Aa . Symmetric: If A~B for AB  S , then there exists a bijection f: AB . Theorem 1.1(a), page 5, tells us that f –1: BA is a bijection. Hence, BA . Transitive: If A~B and B~C with ABC   S , then there exists bijections f: AB and g: BC . Theorem 1.2(c), page 7, tells us that gf: AC is a bijection. Hence, A~C .

CYU 1.16 (a) No: 12 23  . (b) Yes: Every element of  is either an integer or is contained in some ii + 1 for some integer i  0 or in some –i –1i – for some i  1 . Moreover the sets in   n nZ ii + 1 i = 0  –i –1i – i = 1 are mutually disjoint.

CYU 1.17 Let ad= anr+ a and bd= bnr+ b with 0  ra  n and 0  rb  n .

If ra = rb , then ab– ==dand– bn nda – db . Since na– b , ab mod n .

Conversely, assume that ra  rb , say: 0  rb ra n .Then:

ab– ==danr+ a – dbnr+ b da – db nr+ a – rb

Assume that na– b . Since ra – rb = ab– – da – db n , n would have to divide

ra – rb ; which it cant, sine 0 ra – rb n . It follows that ra  rb  ab mod n .

CYU 1.18 (a) a n ==a n  aa– hn and b n ==b n  bb– kn , for hk  Z . Then: ab– ab ===hn+ a babkn– – hbn++ ab– ab kan hb+ ka n

Since nabab– , ab n = ab n .

(b) a nb nc n ==a nbc n abcn ==ab c n ab nc n =a nb n c n

==ab+ ac n ab n + ac n =a nb n + a nc n

CYU SOLUTIONS A-7

PART 2 GROUPS 2.1 Definitions and Examples CYU 2.1 (a) Closure: The sum of two n-tuples is again an n-tuple.

Associative: a1a2  an + b1b2  b + c1c2  cn = a1 + b1a2 + b2  an + bn + c1c2  cn

= a1 + b1a2 + b2  an + bn + c1c2  cn

= a1 + b1 + c1a2 + b2 + c2  an + bn + cn

= a1 + b1 + c1  a2 + b2 + c2 an + bn + cn

= a1a2  an + b1b2  b + c1c2  cn

Identity: a1a2  an + 00 0 ==a1 + 0a2 + 0  an + 0 a1a2  an

Invere: a1a2  an + –a1–a2  –an = a1 + –a1 a2 + –a  an + –an = 00 0 (b) Closure: The sum of 2 two-by-two matrices is again a two-by-two matrix.  a1 b1 a2 b2 a3 b3 a1 +a2 b1 + b2 a3 b3 a1 + a2 +a3 b1 + b2 + b3 Associative: + + ==+ c1 d1 c2 d2 c3 d3 c1 +c2 d1 + d2 c3 d3 c1 + c2 +c3 d1 + d2 + d3  a +a + a b + b + b = 1 2 3 1 2 3 c1 +cc+ 3 d1 + d2 + d3

 a1 b1 a2 b2 a3 b3 = + + c1 d1 c2 d2 c3 d3 

Identity: ab + 00 ==a +0 b + 0 ab cd 00 c +0 d + 0 cd

Inverse: ab + –a –b ==aa+– bb+ – 00 cd –c –d cc+–  dd+ – 00

CYU 2.2 The values in column a follow from the observation that 0+n nn= for 0 n 3 . As for column b, row 3: 3+ 1 = 0 , since 31+414== + 0 4 a b c d As for column c, rows 2 and 3: 2+42 = 0 and 3+42 = 1 , since: + 4 0123 22+414== + 0 and 32+514== + 1 . 0 0123 As for column d, rows 1, 2, and 3: 1+ 3 = 0 , 2+ 3 = 1 , 4 4 1 1230 and 3+ 3 = 2 , since: 13+414== + 0 , 4 22301 23+514== + 1 , 33+614== + 2 . 33012 A- 8 APPENDIX A

th CYU 2.3 Let Gea= 1 a2  an – 1 . By construction, the i column of G’s group table is precisely eaia1ai a2ai  an – 1ai . The fact that every element of G appears exactly one time in that row is a consequence of Exercise 37, which asserts that the function k : GG given by ai k g = ga is a bijection. ai i

1 2 3 4 5 1 2 3 4 5 CYU 2.4 From  =  and  = we have: 1 5 2 3 4 5 3 2 1 4 1 2 3 4 5  1 2 3 4 5  1 5 2 3 4  :  5 4 3 2 1 5 4 3 2 1

1 2 3 4 5  1 2 3 4 5  5 3 2 1 4  :  4 2 5 1 3 4 2 5 1 3

CYU 2.5 (a) We know that 1 and 5 are generators of Z6 [Example 2.2(a)]. The remaining 4 ele-

ments in Z6 are not:

2+ 24= 4+ 42= 0+600= 6 3+630= 6 2+62+620= 4+64+640=

0 1 11 12 121 (b)  is cyclic, with generator  :   == . S2:  1 1 1 0 22 21 212

(c) For n  2 , consider the following bijections   Sn : 1 ==2 2 = 1 and i i for 3 in 2 ==3 3 = 2 and i i for i not equal to 2 or 3

Since  1 === 1 2 3 and  1 === 1 1 2 ,

Sn is not abelian, and therefore not cyclic. CYU SOLUTIONS A-9

2.2 Elementary Properties of Groups CYU 2.6 Since bca bca ===bc abc abcea bca : bca= e (Theorem 2.6).

12 12 13

CYU 2. 7 (a) False: For    S3 given by : 23 , : 21 and : 22 we have: 31 33 31     121 131 : 233 and : 223 as well. 312 312

(b) True: ab+ = bc+  ba+ ==bc+  a c commutativity Theorem 2.9

CYU 2.8 We show that the equation ax+ = b has a unique solution in  + : Existence: ax+ = b – a + ax+ = –ab+ – a + a + x = – a + b  0 + x ==– a + bx – a + b Uniqueness: If x and x then: ax+ = b and ax+ = b  ax+ ==ax+  x x Theorem 2.8

CYU 2.9 We know that we have to consider a non-abelian group, and turn to our friend S3 . Spe- 13 13 13 –1 cifically for   S3 given by : 22 and  = 21 we have:  = 22 , 31 32 31   12 132 12 –1  = 23 , and  ==221 21 , so that: 31 313 33 –1 –1 12 131 11 –1 –1 –1  ===21 while   223 23 33 312 32

–1 –1 –1 –1 CYU 2.10 I. ana2a1 = a1 a2 an clearly holds for n = 1 .  –1 –1 –1 –1 II. Assume ak a2a1 = a1 a2 ak . Then: III.  –1  –1 ak + 1  ak a2a1 = ak + 1  ak a2a1 –1 –1 Theorem 2.12:  –1 –1 –1 –1 ==ak a2a1 ak + 1 a1 a2 ak  ak + 1 II A- 10 APPENDIX A

CYU 2.11 (a) 1 2 3 4 (b) 14= 4  2 3 4 1 24==4+2448 2  34==8+ 412 3 4 1 3 e 24 3  44==12+ 416  4 1 2 3 24  54==16+ 420 4 24 1 2 3 4 64==20+244 0: 4 has order 6  has order 4

(c) Let gcd a n = d . We show that d is the smallest positive integer m such that ma= kn for some k : n k--- kn d ma=== kn m ------(*) a --a- d n a Since gcd a n = d : --- and --- are relatively prime. It follows, from (*), that --a- k . d d d Turning to m = -----kn- we see that m will be smallest when k is smallest; which is to say, a a ---  n n when k = --a- . Hence, the smallest m turns out to be -----kn- ==------d --- . d a a d 2.3 Subgroups CYU 2.12 We already know that 6Z is a subgroup of Z. To show that it is a subgroup of 3Z we need but observe that 6Z  3Z : n  6Z  n = 6m for mZ  n = 32m  n  3Z

–1 h2 h1 ==eh 1 h2 CYU 2.13 h k = h k  h–1h = k k–1  1 1 2 2 2 1 2 1  –1  k2k1 ==ek 1 k2

both in H and K --so

CYU 2.14 We show that 3 ==Z8 01234567 by demonstrating that every element of Z8 is a multiple of 3:

13 ======3 2 3 3 +836 33 3 +83 +831 43 3 +83 +83 +834 =

53 =====3 +83 +83 +83 +837 63 2 7 3 5 8 3 0

Claim: 4 = 04 : 14 ===4 2 4 4 +840 . Fine, but can we pick up other elements of Z8 by taking additional multiples of 4? No: The division algorithm assures us that nq= 4 + r for any nZ , with 0  r  4 . From the above we know that 14 and 24 are in 04 , and surely 04   04 . The only possible loose end is 34 . Let’s tie it up:

34 ===4 +84 +840 +844 CYU SOLUTIONS A-11

CYU 2.15 A direct consequence of CYU 2.11(c), page 57.

CYU 2.16 ii  i : Let S be the intersection of all subgroups of G that containing A. Since A is a subgroup of G containing A that is contained in every subgroup of G that contains A: SA= .

CYU 2.17 We show that 2 3 = S3 by observing that every element in  1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3  S = e = ,  = ,  =   = ,  = ,  =  3  1 2 3 1 2 3 1 2 3 1 2 3 1 3 2 4 3 2 1 5 2 1 3   

can be expressed a a product of the permutations 2 3 (details omitted): 2 2 e = 3 , 1 = 2 , 2 = 2 , 3 = 3 , 4 = 23 , 5 = 32

2.4 Homomorphisms and Isomorphisms CYU 2.18 Let : G  G be given by a = e . Since for every ab  G , ab ==eee =a b ,  is a homomorphism.

CYU 2.19 For ab  G we have:  ab ==ab a b =a b

=  a  b

CYU 2.20 Homomorphism:

2n1 + 2n2 ====2n1 + n2 8n1 + n2 8n1 + 8n2 2n1 + 2n1 Ker ===2n 2n = 0 2n 8n = 0 0 . Im === 2n 8n 8Z

CYU 2.21 We are to show that for any ab  G , a == b  a b . Let’s do it:

a == b  ca = cb  c a c b c = c b a –1 c = c b a–1 c ==cba–1  c cba–1  c–1c = c–1cba–1  e ==ba–1  a b A- 12 APPENDIX A

CYU 2.22 (a) We show that the relation  given by GG  if G is isomorphic to G is an equivalence relation: Reflexive GG since the identity map Ig= g is clearly an isomorphism. Symmetric GG   G  G : Let : GG  be an isomorphism. Theorem 1.1(a), page 5, assures us that the map –1: G  G is a bijection. We show that it is also a homomorphism: For a b  G –1ab = –1a –1b since let ab  G be such that a = a and b = b . Since ab = ab : –1ab ==ab –1a –1b .

Transitive G1  G2 and G2  G3  G1  G3 : Follows from Theorem 1.2(c), page 7, and CYU 2.19. (b) We show that the map : nZ mZ given by nz = mz is an isomorphism: One-to one: nz = nz  mz = mz  z = z Onto: For given mz mZ : nz = mz . Homomorphism: nz+ nz ====nz+ z mz+ z mz+ mz nz + nz

(c) Let gG . The map ig: GG is a bijection: –1 –1 –1 –1 –1 –1 One-to one: iga = igb  gag = gbg  g gag g ==g gbg ga b –1 –1 –1 Onto: For aG , igg ag ==ggag g a –1 –1 –1 Homomorphism: igab ==gabg gag gbg =iga igb

CYU 2.23 We show that : a  Z given by an = n is an isomorphism: One-to-one: an === am  n ma n am Onto: For nZ , an = n . Homomorphism: anam ===anm+ nm+ an + am .

CYU 2.24 Let : GG  be an isomorphism. For a b  G let ab  G be such that a = a and b = b . Then: ab=====a b ab ba b a ba CYU SOLUTIONS A-13

2.5 Symmetric Groups

12345678910 CYU 2.25 (a) For  =  we have: 39415627810 1 =====321 3 431 4 1  134 is a cycle. Picking the first element not moved by the above cycle; namely 2, we have: 2 ======922 9 832 8 742 7 2  2987 a cycle Since the elements not contained in either of the above two cycles are stationary under  :  = 134 2987 (b) One possible answer: 12 34 56789 .

12345678 CYU 2.26 For  =  we have: 38674152

12345678 12345678 12345678 2 = , 3 = , 4 =  62157348 18345672 32674158

12345678 12345678 Since 6 is the smallest n for 5 = , 6 =  68157342 12345678 which n ==e, o( 6

CYU 2.27 (a) Following the construction above Theorem 2.32 we have: 3251 = 31 35 32

1234 5 678910 = 124 5108679 (b) 24311059678 = 14 12 56 58 510 79 (c) Since ij ij = ij , ij –1 = ij

CYU 2.28 (a) Since ei= i for every 1 in , e can be expressed as a product of 0 transpositions, and 0 is certainly an even number. (In the event that n  1 , you also have: e = 12 12 )

1234 5 678910 (b) Since = 14 12 56 58 510 79 , the 24311059678 transposition is even.

CYU 2.29 As you can easily check, e1 2 are even and the rest are odd. Consequently:

A3 = e1 2   A- 14 APPENDIX A

2.6 Normal Subgroups and Factor Groups CYU 2.30 We show that the function f: HaH given by fh= ah is a bijection:

One-to-one: fh1 ===fh2  ah1 ah2  h1 h2 . Theorem 2.10, page 56 Onto: For ah aH, fh= ah

CYU 2.31 Consider the homomorphism : Z2 +2  S3 given by: 123 123 0 =  and 1 =  123 132

123 123 While Z2 is normal in Z2 , Z2 =   fails to be a normal sub- 123 132 group of S3 (see Example 2.10). CYU 2.32 (a) Follows from CYU 2.31. (b) Let Ga=  and let N  G (actually every subgroup of G is normal). We show that GN = aN : Let gN G N . Since ga  , there exists m such that ga= m , We then have: gN== amNaNaNaN = aN m  gN  aN m times The above argument shows that GN  aN . Clearly: aN  GN . CYU 2.33 Since, for any aG , ag= ga for every gG : ZG==aGag = ga gG G Since for any ab  G , aba–1b–1 ==aa–1bb–1 e : CG===aba–1b–1 ab  G e e

CYU 2.34 We first show that the function : Sn  –11 given by:  1 if  is an even permutation =  is a homomorphism by considering four cases: –1 if  is an odd permutation If  and  are even, then so is  , and we have: ==111 = If  and  are odd, then  is even, and we have: ==11– –1 =  If  is even and  is odd, then  is odd, and we have: ==–11 –1 =  If  is odd and  is even, then  is odd, and we have: ==–11 – 1 = 

Since 1 is the identity in the group –11 , Ker = An . Invoking the First Isomorphism Theorem, we have: GS n  An . CYU 2.35 Employing Theorem 2.42 to the homomorphism : GG  we have: G  GK . Restricting  to the group N we arrive at a homomorphism N: NN  . In this setting, Theorem 2.42 tels us that N  NK . Consequently: GN  GK  NK CYU SOLUTIONS A-15

2.7 Direct Products CYU 2.36 (a) Associativity:  For a1a2  an b1b2  bn c1c2  cn  G1  G2   Gn :

a1a2  an b1b2  bn c1c2  cn = a1b1 c1 a2b2 c2 anbn cn

= a1b1c1  a2b2c2 anbncn

= a1a2  an b1b2  bn c1c2  cn

Identity: Letting ei denote the identity in Gi we have:

a1a2  an e1e2  en = a1e1a2e2  anen a1a2  an –1 –1 –1 Inverses: a1a2  an a1 a2  an = e1e2  en

(b) If each Gi is abelian, then:

a1a2  an b1b2  bn = a1b1a2b2  anbn

==b1a1b2a2  bnan b1b2  bn a1a2  an

Conversely, assume that not all of the Gi are abelian. For definiteness, assume that G1 is not abelian, with ab ba . We then have:

ae2  en be2  en = ab e2  en  ba e2  en = be2  en ae2  en

CYU 2.37 Noting that 3 has order 2 in Z6 and is of order 4 in Z4 , and that 4 has order 4 in 16 , we conclude that 334 has order lcm244 = 4 in Z30 .

CYU 2.38 Using Induction on s we show that if n1n2  ns are relatively prime, then the group Z  Z    Z is cyclic and isomorphic to Z : n1 n2 ns n1n2ns I. True if s = 2 , by Theorem 2.44. II. Assume True for sk= ; i.e: Z  Z    Z is cyclic and isomorphic to Z . n1 n2 nk n1n2nk III. We establish validity for sk= + 1 ; i.e, that Z  Z    Z  Z is cyclic and isomorphic to Z : n1 n2 nk nk + 1 n1n2nknk + 1 Z  Z    Z  Z  Z  Z    Z  Z n1 n2 nk nk + 1 n1 n2 nk nk + 1

by II:  Z   Z (with Z  cyclic) n1n2 nk nk + 1 n1n2 nk by I: Z Z n1n2nk nk + 1 n1n2nknk + 1  note that the two number n1n2 nk and nk + 1 are relatively prime In the event that n n  n are not relatively prime, Z  Z    Z is not iso- 1 2 s n1 n2 ns morphic to Z since no element in Z  Z    Z has order n n n . n1n2ns n1 n2 ns 1 2 s A- 16 APPENDIX A

CYU 2.39 We are given that GHK= with every aG having a unique representation of the form hk . Suppose that aHK  , with ahk= . But aae= is also a representation of a, where aH and eK . It follows, from the unique representation condition, that ke= . Similarly, since aea= : he= . Consequently: ae= .

CYU 2.40 G3 has an element of order 8 while G6 does not.

PART 3 From Rings To Fields 3.1 Definitions and Examples

CYU 3.1 (a) Since S3 = S3  is not an abelian group, it cannot be turned into a ring by imposing any additional operator “*”. (b) Let G + be an abelian group. By defining a*b = 0 for every ab  G , we arrive at a ring G+ * .

CYU 3.2 We already know that nZ + is an abelian group (Example 2.4, page 62). In addition, nZ is closed under multiplication: na nb = nanb . Moreover, since the associative and distributive properties hold for all in integers, they will surely hold for the integers in nZ .

CYU 3.3 Using induction we first show that nab==na banb for n  0 : I. nab==na banb for n = 0 . II. Assume kab==ka bakb for given k  0 III. We show k + 1 ab = ak+ 1 b (A similar argument can be used to show that k + 1 a bak= + 1 b ): k + 1 ab ====kab+ ab akb+ ab akb+ b ak+ 1 b by II

In the event that n  0 we have: nab==anb –nab –anb –nab =anb– by Theorem 3.1(b) –0n  CYU SOLUTIONS A-17

CYU 3.4 (a) We first verify that the nonempty set M22 = M22  + satisfies the three properties of Definition 2.1, page 41:

a b a b a b a b a +a b + b a +a + a b + b + b 1. 1 1 + 2 2 + 3 3 ==1 1 + 2 3 2 3 1 2 3 1 2 3  c1 d1 c2 d2 c3 d3 c1 d1 c2 +c3 d2 + d3 c1 +c2 + c3 d1 + d2 + d3 a + a +a b + b + b = 1 2 3 1 2 3 c1 + c2 +c3 d1 + d2 + d3

a b a b a b = 1 1 + 2 2 + 3 3  c1 d1 c2 d2 c3 d3

ab ab 00 a + 0 b + 0 ab 2. For every  M22 : + == cd cd 00 c + 0 d + 0 cd

ab ab –a –b aa– bb– 00 3. For given  M22 : + == cd cd –c –d cc– dd– 00

Moreover, the group M22  + is abelian:

a b a b a +a b + b a +a b + b a b a b 1 1 + 2 2 ===1 2 1 2 2 1 2 1 2 2 + 1 1 c1 d1 c2 d2 c1 +c2 d1 + d2 c2 +c1 d2 + d1 c2 d2 c1 d1

Properties 2 and 3 of Definition 3.1 are also satisfied:

a b a b a b a b a a +b c a b + b d 1 1 2 2 3 3 = 1 1 2 3 2 3 2 3 2 3 2.  c1 d1 c2 d2 c3 d3 c1 d1 c2a3 +d2c3 c2b3 + d2d3

a a a + b c +b c a + d c a a b + b d + b c b + d d = 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 1 2 3 2 3 c1a2a3 + b2c3 +d1c2a3 + d2c3 c1a2b3 + b2d3 + d1c2b3 + d2d3

a a + b c a +a b + b d c a a + b c b + a b + b d d = 1 2 1 2 3 1 2 1 2 3 1 2 1 2 3 1 2 1 2 3 c1a2 + d1c2 a3 + c1b2 + d1d2 c3 c1a2 + d1c2 b3 + c1b2 + d1d2 d3

a b a b a b = 1 1 2 2 3 3  c1 d1 c2 d2 c3 d3

a b a b a b a b a +a b + b 1 1 2 2 + 3 3 = 1 1 2 3 2 3 3. c1 d1 c2 d2 c3 d3 c1 d1 c2 +c3 d2 + d3

a a + a +b c + c a b + b + b d + d = 1 2 3 1 2 3 1 2 3 1 2 3 c1a2 + a3 +d1c2 + c3 c1b2 + b3 + d1d2 + d3

a a + b c +a a + b c a b + b d + a b + b d = 1 2 1 2 1 3 1 3 1 2 1 2 1 3 1 3 c a + d c +c a + d c c b + d d + c b + d d 1 2 1 2 1 3 1 3 1 2 1 2 1 3 1 3

a b a b a b a b = 1 1 2 2 + 1 1 3 3 c1 d1 c2 d2 c1 d1 c3 d3

a b a b a b a b a b a b a b 1 1 + 2 2 3 3 = 1 1 3 3 + 2 2 3 3 In a similar fashion one can show that:  c1 d1 c2 d2 c3 d3 c1 d1 c3 d3 c2 d2 c3 d3 A- 18 APPENDIX A

10 It is easy to show that is the unity in M22 : 01

10 ab ==ab 10 ab 01 cd cd 01 cd (b) If 1 and 1 are unities in a ring R, then: 11==1 1

CYU 3.5 Does there exist ab such that ab 23 ==2a –4b 3a – 6b 10 ? cd cd –64 – 2c –4d 3c – 6d 01 14+ b If so, then: 2a –14b = and 3a –06b = , or: a = ------and a = 2b , or: 2 14+ b ------=== 2 b  14+ b 4b  1 0 ! 2

CYU 3.6 (a) Challenging each element in Z6 = 012345 we find that, apart from 1, only 5 has a multiplicative inverses: multiplying 2345 mod 6 24024 35303 42042 54321 Ah! 55 = 1

(b) If m and n are relatively prime then, by Theorem 1.7, page 23: 1 = sm+ tn for some integers s and t. It follows that sm = 1 – tn which says that sm is congruent to 1 modulo n, and that m is a unit. If gcd m n = d  1 . Then, by Theorem 1.6, page 22: dsmtn= + for some integers s and t. It follows that sm= d– tn which shows that 1 is not congruent to sm modulo n (it is congruent to d modulo n, with 1  dn ). It follows that m is not a unit.

CYU 3.7 Expressing Exercise 38 (page 70) in additive form we have: A (nonempty) subset S of a group G is a subgroup of G if and only if ss  S s– s S . It follows that property (i) of Theorem 3.2: S, + is a subgroup of R, + can be replace d with: ss  S ss–1  S . CYU 3.8 Employing CYU 3.7:

(a)00 – 00 = 00 H and 00 00 = 00 H . ab cd ac– bd– ab cd bc bd

(b) Let xy  Sa . Since ax– y ===ax–00 ay –0 : xy–  Sa .

Since axy===ax y 0y 0 : xy Sa CYU SOLUTIONS A-19

3.2 Homomorphisms, and Quotient Rings CYU 3.9 (a) Let ab  –1H . To say that ab– and ab are contained in –1H is to say that ab– and ab are contained in –1H ; and they are: ab– ==a – b  H and ab a b  H (since H is a subring of G ) (b) Assume there exists an isomorphism : 3Z  5Z . If so, then: 9 == 33 3 3 and 9 == 333++ 33 This implies that 3 3 = 33 or that 3 – 3 3 = 0 . Since 3 can’t be zero (if it were, then  would map everything to zero), 3 must equal 3 — a contradiction since 35 Z .

CYU 3.10 (a) We already know that InZ= is a subring of Z [Example 3.4(a), page 117]. It is an ideal since, for any mZ and ns nZ : mns= nms nZ . The converse follows from the fact that all subgroups of Z are of the form nZ (Exercise 35, page 70). (b) We already know that I is a subring of R [Theorem 3.4(a)]. It is an ideal: For x  R , choose xR such that x = x . Then, for any a  I we have: xa == x a xa  I and a x==a x ax  I since I is an ideal

CYU 3.11 Theorem 2.23(d), page 73, assures us that I = –1I is an additive subgroup of R. As for the second part of Definition 3.6: Let iI and rR . To show that ri I we need but verify that ri  I . Easy enough. Since i  I and since I is an ideal in R : ri = r i  I A similar argument can be used to show that ir I .

As 0  I , KI . Noting that the function I: II  given by Ii = i is an onto homomorphism with kernel K, we have: IK  I .

CYU 3.12 For ab  Z, let:

( 1 ) aq=anr+ a or ra = aq– an where 0  ra  n . S o : a = ra .

(2) bq=bnr+ b or rb = bq– bn where 0  rb  n . So: b = rb . (3) ab= qn+ r or r = ab– qn where 0  rn . So ab = r .

We compete the proof by showing that rr–arb  0 mod n : 3 1 and 2 rr– arb ==ab– qn – rarb ab– qn – aq– an bq– bn

= ab– qn – ab– bqan – aqbnq+ anqbn

= – qn ++bqanaqbnq– anqbn

= – q +++bqa aqb qanqb n A- 20 APPENDIX A

3.3 Integral Domains and Fields

CYU 3.13 (a) Z5 = 01234 is a commutative with unity 1. It is easy to see that it has no zero divisors and that every nonzero element has a multiplicative inverse: 11 ====1 2 3 32 1 4 4 1

Conclusion: Z5 is a field.

(b)Z15 = 012 14 is a commutative ring with unity1. It fails to be an integral domain as it has zero divisors (35 = 0 ).

CYU 3.14 (a) Let R be a commutative ring with unity in which the cancellation property holds. We show that R has no zero divisors (i.e, that R is an integral domain): Let ab = 0 . Since a00= : ab= a0 . “Canceling the a,” we have b = 0 . On the other hand, if R is an integral domain then, by its very definition, it has no zero divisors.

(b)fax = fay  ax ==ay x y .

CYU 3.15 By CYU 3.6(b), page 116, every nonzero element in Zp = 012 p – 1 is a unit. It follows that Zp is an integral domain, and therefore a field (Theorem 3.12).

CYU 3.16 (a) Consider the factored form x2 –6x – ==x – 3 x + 2 0 . Clearly x = 3 is a solution. Not quite as clear is that 10 is also a solution, as 10+0 2 = . From the discussion preceding this CYU we know that there are five pairs of numbers in Z12 = 01234567891011  (not involving 0) whose product equals 0 (in Z12 )— specifically: 26 34 83 94 and 10 6 . The plan, now, is to find those elements x in Z12 for which the product x – 3 x + 2 turns out to involve any or the above five pairs. A direct calculation shows that 6 and 7 are the only winners: For x ===6: x – 3 x + 2 63– 62+ 38 For x ===7: x – 3 x + 2 73– 72+ 49 2 Conclusion: 3, 10, 6, and 7 are the solutions of x –6x –0= in Z12 . 2 (b) In Z8 = 01234567 the equation x –6x – ==x – 3 x + 2 0 is seen to have solutions 3 and 6 (since 62+0= ). Here are the only pairs (not involving 0) with product equal to 0 (in Z8 ): 24 and 46 . A direct calculation shows that for no x in Z8 does the product x – 3 x + 2 involve either 24 or 46 . For example, if x = 5 , then x – 3 x + 2 involves the pair 27 . 2 Conclusion: 3 and 6 are the solutions of x –6x –0= in Z8 . CYU SOLUTIONS A-21

CYU 3.17 We first acknowledge the fact that the ring P x is infinite, for it contains the infinite Zn set of polynomials xn  . As the coefficients of any polynomial px in P x n = 1 Zn are elements in Z , np x = 0 for every px P x . Moreover, for no 0 mn n Zn is it true that np x = 0 , where px is the constant polynomial 1. It follows that P x has characteristic n. Zn

CYU 3.18 Expanding ab+ 3 we find that ab+ 3 = a3 +++3a2b 3ab2 b3 . Since the given domain D had characteristic 3, the terms 3a2b and 3ab2 are zero. Consequently: ab+ 3 = a3 + b3 .

CYU 3.19 (a) If pZ is not a maximal ideal, then pZ nZ for some n  1 [see CYU 3.10(a), page 124]. Consequently pnm= for some mZ — contradicting the given condition that p is prime. (b) Assume that ImZ= is a maximal ideal. If m is not prime, then mab= with neither a nor b equal to 1. But then: mZ aZ Z

mk aZ kZ since mk = abk 1  aZ Contradicting the given condition that mZ is a maximal ideal in Z. So, every maximal ideal in Z is of the form pZ for p prime. As such, it is a prime ideal, for: ab pZ p ab p a or bb a pZ or bpZ Theorem 1.9, page 24 Conversely, assume that ImZ= is a prime ideal in Z. If m is not prime, then mab= with neither a nor b equal to 1. But then I is not a prime ideal, since ab mZ with neither a nor b contained in mZ . So: ImZ= prime  m prime  ImZ= maximal (a) A- 22 APPENDIX A Determinants B-1

1 APPENDIX B

We offer Professor Goldberg’s proof that the groups Z4 and K appearing in Figure 2.1, page 43, are the only groups of order 4. PROPOSITION: Let S be a group of order 4, with identity e. Then, for every aS , there exists a positive integer d so that: •;ad = e •ak  e , for any positive integer k smaller than d; and • d = 1 2 or 4. PROOF: Choose an arbitrary element a of S. Consider the following elements of S: eaa2 a3 . Since S has 4 elements, by an elementary application of the pigeonhole principle, either: case 1. ai ==e i 123or 4 and/or case 2. ai = aj , some integers ij , 1  ji  4 . In either case (using inverses for case 2), we obtain that ak = e for some integer k  1234 : for case 1, ki= ; for case 2 kij= – . Hence the set kZ + ak = e is not empty. Let d be the minimum of this set. From the above, 1 d 4 . To finish the proof of the Proposition, it remains to show that d cannot be 3: If d = 3 , the elements eaa2 are distinct. Since S is of order 4, bS with beaa 2 . So, eaa2 b = S . By closure, ab S . But note that: ab b , since ab== b a e , impossible; ab a2 , since ab== a2  b a , impossible; ab a , since ab== a b e , impossible; ab e, since ab== e a2ab a2 (using that a3 = e ), impossible. Hence S contains at least 5 distinct elements, contradicting that it has order 4. So d cannot be 3. Using the above Proposition, it is easy to classify groups of order 4. By the Proposition, there are 2 cases: case 1. aS with aea  2  ea 3  e and a4  e , or case 2. aS , a2 = e . In Case 1, we have Seaa= 2 a3 and a4 = e Up to “renaming”,

S is Z4 . In Case 2, pick an element aS with ae . Next, pick an element be with ba . Since we are in Case 2, a2 ==b2 e , and eab==2 abab. Multiplying on the left by a and on the right by b we obtain ab= ba . Hence, S=  eabab (it is easy to see that ab does not equal e, a, or b), each element is of order 2, and S is commutative. B-2 Determinants DETERMINANTS We define a function that assigns to each square matrix a (real) num- jthcolumn ber: 1

st out row DETERMINANT For A = ab: out cd

det ab = ad– bc nn cd

For AM nn , with n  2 , let A1j denote the resulting in a n – 1  n – 1 matrix obtained by deleting asquare matrix th of dimension the first row and j column of the matrix A (see n – 1 margin). Then: n 1 + j detA =  –1 a1j detA1j j = 1 The above definition defines the determinant of a matrix by an expansion process involving the first row of the given matrix. The following theorem (proof omitted), known as the Laplace Expansion Theorem, enables one to expand along any row or column of the matrix.

THEOREM 1 For given AM nn , Aij will denote the n – 1  n – 1 submatrix of A obtained th th Note that the sign of the by deleting the i row and j column of A. ij+ We then have: –1 has an alternat- n ing checkerboard pattern EXPANDING ALONG i + j + _ + _ + _ + _ THE ith ROW detA = –1 aij detAij _ _ _ _  + + + + + _ + _ + _ + _ j = 1 _ + _ + _ + _ + _ _ _ _ and: + + + + EXPANDING ALONG _ + _ + _ + _ + _ _ _ _ n + + + + THE jth COLUMN _ + _ + _ + _ + i + j detA =  –1 aij detAij i = 1

THEOREM 2 The determinant of the identity matrix In  Mnn is 1.

PROOF: By induction on the dimension, n, of Mnn .

I. Holds for n = 2 : det 10 = 1 . 01 Determinants B-3

II. Assume claim holds for nk= : detIk = 1

1 0000 III. We establish validity at nk= + 1 : detIk + 1 = 1 01000 00100 Ik + 1 Expanding across the first row of Ik + 1 (see margin) we have: I 00010 k detIk + 1 ==detIk 1 00001

THEOREM 3 If two rows of AM nn are interchanged, then the determinant of the resulting matrix is –detA .

PROOF: By induction on the dimension of the matrix A. For n = 2 :

det ab ==ad– bc and det cd cb– da cd ab negative of each other Assume the claim holds for matrices of dimension k  2 (the induction hypothesis).

Let Aa= ij be a matrix of dimension k + 1 , and let Bb= ij denote the matrix obtained by interchanging rows p and q of A. Let i be the index of a row other than p and q. Expanding about row i we have: k + 1 k + 1 ij+ ij+ detA =det –1 aij Aij and det B =det –1 bij Bij j = 1 j = 1 Since rows p and q were switched to go from A to B, row i of B still equals that of A, and therefore: bij = aij . Since Bij is the matrix Aij with two of its rows interchanged, and since those matrices are of dimension k, we have: detBij = –detAij (the induction hypothesis). Consequently: k + 1 k + 1 detB = –1 ij+ b detB = –1 ij+ a –detA  ij ij  ij ij j = 1 j = 1

k + 1 ij+ ==–  –1 aijdetAij –detA j = 1 B-4 Determinants ANSWERS C-1

Appendix C Answers to Selected Exercises Part 1 Preliminaries 1.1 Functions. 1. U 3. 15nn U 5. B 7. D 9. U 11.  13. F 15. D 17. 13579111315   19. E 27. One-to-one and onto 29. Not one-to-one, not onto 31. One-to-one not onto 33. Not one-to-one, not onto 35. One-to-one and onto 37. Not one-to-one, not onto 39. Not one-to-one, not onto –1 y + 2 –1 y –1 a 41. f y = ------43. f y = ------45. f ab = --- b – 3 3 2 – y 5

--a- cd– 2 –1 ab –1 1 47. f = b 49. f abc = ---s – 2b cd a --- 2 2 1 ---–2a + b – 2c 2

1.2 Principle of Mathematical Induction. Each exercise calls for a verification or proof.

1.3 The Division Algorithm and Beyond.

1. qr==0 3. q ==–127 r 5. 5 7. 60 9. 90

1.4 Equivalence Relations.

23. Yes 25. No 27. Yes 29. No 31. Yes 33. Yes

43. n = n + 5kk Z 45. x = –xx 47. x0 y0 = x0 y y   ANSWERS C-2

Part 2 Groups 2.1 Definitions and Examples. 1. A cyclic group with generator 2. 2. Not a group. It does not contain an identity. 3. Not a group. It does not contain an identity. 5. Not a group. It does not contain an identity. 7. Not a group. 1 is the identity, but 2 has no inverse. 9. Abelian group. Not cyclic. 11. Abelian group. Not cyclic.

 e if n  0 mod 3 2 3 n  –n  e if n is even 13. 3 = e , 3 = 3 15. 1 =  1 if n  1 mod 3 17. 3 =    3 if n is odd  2 if n  2 mod 3

 e if n  0 mod 3  e if n  0 mod 3 2 3 n  –n  19. 2 = 1 , 2 = e 21. 2 =  2 if n  1 mod 3 23. 2 =  1 if n  1 mod 3    1 if n  2 mod 3  2 if n  2 mod 3

123456 123456 123456 25.  =  27.  =  29. 5 ==–1  143652 563412 612345

123456 123456 31. 101 ==5 –1 = 33. 101 ==5 –1 = 612345 214365 35. An abelian non-cyclic group. 37. Not a group 39. A cyclic abelian group 2.2 Elementary Properties of Groups. 1.(a) e (b) a (c) a–1cb–1 (d) aba–1

2.3 Subgroups. 1. Yes 3. No 5. Yes 7. No 9. Yes 11. Yes 13. No 15. Yes 17. No 19. Yes 21. Yes 23. Yes 25. No 27. Yes 29. No 31. Yes 33. Yes

2.4 Homomorphisms and Isomorphisms. 1. Yes 3. No 5. Yes 7. Yes 9. Yes 11. Yes 13. Yes ANSWERS C-3

2.5 Symmetric Groups. 1. 25 134 = 25 14 13 3. 152 = 12 15

5. 2564 = 24 26 25 7. 1634 25 = 14 13 16 25

9. 12 34 11. 3 13. 4 15. 7 17.  = 12453 19.  = 1342

2.6 Normal Subgroups and Factor Groups 1. No 3. Yes

2.7 Direct Products. 1. 36 3. 36 5. 36

7. Order 1: 00 , order 2: 10 , order 3: 01 , 02 , order 4: 11 , 12

9. The element 00e has order 1. The remaining seven elements have order 2.

11. 00  00 01 02

12 13. Here are the proper subgroups of Z2  Z2  S2 , where S2 = e  with  =  : 21 00e  00e  10e 00e  01e 00e  00 00e  11e  00e  01 00e  10  00e  11 00e 01e 10 11  00e 10e 01 11 00e 00 11e 11

Z  Z  Z  Z  Z 15. Z2  Z2  Z3  Z3 17. 2 2 5 3 3 19. 10 Z  Z  Z  Z Z4  Z2  Z3 4 5 3 3 Z  Z  Z  Z Z2  Z2  Z9 2 2 5 9 Z  Z  Z Z4  Z9 4 5 9 ANSWERS C-4

Part 3 From Rings to Fields

3.1 Definitions and Examples. 1. Yes 3. No 5. Yes 7. Yes 9. Yes 11. No 13. No 15. Yes

17. No 19. Yes 21. –11 23. 1234 25. 11 –11 11 – –11 –

27. 11 12 14 15 17 18 51 52 54 55 57  58

3.2 Homomorphisms and Quotient Rings. 1. Yes 3. No 5. No 7. No 9. Yes

19. n = nn is the only homomorphism from Z to ZZ .

3.3 Integral Domains and Fields.

1. None 3. 02  03 5. None 7. 0 9. 6 11. 10 Index I-1

A Function, 2 Bijection, 4 Abelian Group, 43 Domain, 2 Addition Modulo n, 42 Composition, 3 Alternating Group, 88 Inverse, 5 Alternate Principle of Induction, 17 One-to-One, 4 Automorphism, 76 Onto, 4 Range, 2 Fundamental Theorem of B Finitely Generated Abelian Groups, 107 Bijection, 4 Fundamental Counting Principle, 107 G C Generator, 48 Cardinality, 31 Generated Subgroup, 65 Cancellation Law 55 Greatest Common Divisor, 22 Cartesian Product, 2, 103 Group, 41 Cayley’s Theorem, 78 Abelian, 43 Center Subgroup, 96 Alternating, 88 Characteristic, 131 Cyclic, 48 Commutative Ring, 111 Generator, 48 Commutator Subgroup, 96 Factor, 95 Klein, 43 Composition, 3 Order, 43 Congruence Modulo n, 33 Symmetric, 84 Cycle, 84 Table, 43 Decomposition, 84 Cyclic Group, 48 H Generator, 48 Homomorphism (Group), 72 Image, 75 D Kernel, 75 Direct Product, 103 Homomorphism (Ring), 121 External, 103 Internal, 105 I Division Algorithm, 21 Ideal, 123 Divisibility, 15 Maximal, 132 Domain, 2 Prime, 132 Image, 75 E Index of a subgroup, 93 Equivalence Class, 31 Induction, 13 Equivalence Relation, 29 Integral Domain, 128 Even Integer, 15 Inverse Function, 5 Even Permutation, 88 Isomorphism (Group), 75 Theorems, 98 First, 98 Second, 99 F Third, 99 Factor Group, 95 Isomorphism (Ring), 116 Field, 128 Theorems, 124 of Quotients, 133 First, 124 I-2 Index

Second, 125 Quotient Ring, 123 Third, 125 Inverse Function, 5 R Range, 2 K Relation, 29 Kernel, 74 Equivalence, 29 Klein Group, 43 Reflexive, 29 Symmetric, 29 L Relatively Prime, 23 Lagrange’s Theorem, 65 Ring, 111 Least Common Multiple, 104 characteristic, 131 Commutative, 113 With Unity, 113 M S Mathematical Induction 13, 16 Maximal Ideal, 132 Set, 1 Complement, 1 Disjoint, 1 N Equality, 1 Normal Subgroup, 93 Intersection, 1 Subset, 1 Proper, 1 O Union, 1 Odd Integer, 15 Subgroup, 62 Odd permutation, 88 Center, 96 One-to-One, 4 Commutator, 96 Onto, 4 Generated by a, 63 Order of an Element, 58 Generated by subsets, 65 Order of a Group, 43 Normal, 93 Order of a Permutation, 86 Subring, 116 Symmetric Group, 46, 84 P T Partition, 32 Transposition, 85 Permutation, 46 Even, 88 U Odd, 88 Unit, 114 Prime number, 24 Unity, 113 Prime Ideal, 132 Principle of Mathematical Induction, 13,16 W Q Well Ordering Principle, 18 Quotient Group, 95 Z Quotient Field, 133 Zero Divisor, 128 Index I-3