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Herstein Homework Solutions 2.1 - 2.3Solutions Written by Adam Glesser1. In the following determine whether the systems described are groups. Ifthey are not, point out which of the group axioms fail to hold.

(a) G = set of all integers, a b = = a-b.(b) G = set of all positive integers, a -b = ab, the usual product of integers.(c) G = ao, aI, ... ,a6 where

a; . aj = ai+ja; . aj = ai+j-7 if i+ j < 7if i+ j 2 ': 7

(d) G = set of all rational numbers with odd denominators, a . b = = a + b ,the usual addition of rational numbersDetermination: (a) This is not a group. In fact none of the group axiomshold. First, (7 - 5) - 3 = 2 - 3 = -1, while 7 - (5 - 3) = 7 - 2 = 5, so theproduct is not associative. Second, if a E Z and e is the identity element,then a - e = a = e - a. But a - e = a implies e = 0 and then 0 - a = a. Thisdoes not hold for a - I - O. Since there is no identity element, there certainlycan't be inverse elements.

(b) This is not a group. The multiplication is associative, and if e E N isthe identity, then ae = a implies e = 1. But now, for instance, there is nonatural number x such that 2x = 1. Thus G does not contain an inverse of2 (or any other natural number greater that than 1).

(c) This is a group. The identity is ao and the inverse of a; is a7-i.Moreover, one will be able to show later on that this group is (essentially)the same as Zj7Z.

(d) We make the following additional assumption: we consider all frac-tions to be in reduced form, i.e., if ~ E G then (a , b ) = 1. Then G is a group.If ~and ; 'l are in G, then ~ + ; ' l = adb~bc. The denominator is odd and con-sequently if we reduce the fraction, the denominator will remain odd. Thus,G is closed under addition. Since we are dealing with the usual addition, weknow that it is associative. The identity is ~ = 0 and if ~ E G then b a E Gand ~ + -ba = o . 2. Let G be an abelian group. Show that for all a, bEG and nEZ,(a . b)n = an . b",Proof: Let a, bEG, ti E Z. Then (a . b)n = (ab) . (ab) (ab) where thenumber of factors ab is n. Since G is a group, the product is associative and

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so the parenthesis may be dropped. Now as G is abelian, a and b commuteso that we may rewrite (a b ) . (a b ) (a b ) = (a a a ) . ( b b b ) = an . bn .3. If G is a group such that (a . b )2 = a 2 . b2 for all a , bE G, show that Gmust be abelian.Proof: Let a , bEG . Then (ab)2 = a 2 . b2 ====} abab = aabb ====}a -1ababb -1 = a -1aabbb -1 ====} ba = ab o Hence, G is abelian .4. IfG is a group in which (a b )i = a i . b i for three consecutive integers i forall a , bEG , show that G is abelian.Proof: Let iE : : z such that for all a , bE G we have (a b )i-l = a i-I. b i-1 , (a b )i = a i . b i and (a . b )i+l = a i+1 . b i+1 . Then for all a , bEG we have,

a i+1 . b i+1 = (a . b r + 1= (a . b ) . ( a . b r = a . b a i . b i ====} a ' b = b a "anda i . b i = (a b r = (a . b ) . ( a . b r - 1 = a . b a i-I. b i-1 ====} a i-I. b = b . a i-I.Putting these together we get

a i . b = b a i = b a i-I. a = a i-I. b a ====} a b = b a .Since this holds for all a , bEG , the group is abelian .5. Show that the conclusion of Problem 4 does not follow if we assume therelation (a . b )i = a i . b i for just two consecutive integers.Proof: Let G be any group. Then for all a , bEG we have (a b )O = e = a O . b Oand (a b )1 = a - b = a 1. b 1. Thus the relation always holds for two consecutiveintegers but not every group is abelian (e.g., G = 53, the symmetric groupon 3 letters.)

6. In 53 give an example of two elements x,y such that (x . y )2 - I - x2 . y2.Proof: We have ((1,2)(1,2,3))2 = (2,3)2 = e - I - (1,3,2) = (1 ,2 )2 (1 ,2 ,3)2 .7. In 53 show that there are four elements satisfying x2elements satisfying y3 = e. e and three

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Proof: X2 = e: e, (1,2), (1,3), (2,3)y3 = e: e, (1,2,3), (1,3,2) .

8. IfG is a finite group, show that there exists a positive integer N such thata N = e for all a E G.Proof: Let a E G . Consider the set M := { a i l O : 0 : ; i : 0 : ; I G I } . Since ev-ery element of M is in G, and there are I G I + 1 powers of a in M, theremust be a repetition. That is, there exists 0 : 0 : ; i< j : 0 : ; I G I such thata i = a j ====} a j-i = e. Therefore, for each a E G, we can find N; E N suchthat a Na = e. Now let N:= lcm{Na la E G} (alternatively one could takeN:= I T Na ) (both of these make sense since G is finite). Then for a E G

aEGwe have aN = ( aNa)N/Na = eN /Na = e.9. Let G be a group.

(a) If G has 3 elements, show that G is abelian.(b) If G has 4 elements, show that G is abelian.(c) If G has 5 elements, show that G is abelian.

Proof: (a) Let G := {e, x , y} where e is the identity element of G. If wedefine x . y = x = x . e, then by Lemma 2.3.2, y = e, a contradiction.Similarly, we cannot define x . y = y. Therefore, we must define x . y = eand this gives X-I. x . Y = X-I ====} Y = X-I ====} y. x = e. Hence, xand y commute. But x certainly commutes with itself, as does y, and so allelements of G commute. Therefore, G is abelian.

(b) Let G:= {e, Xl, X2, X3} where e is the identity element of G. As in (a),we will show that Xl, X2, and X3 all commute. This will show that G is abelian.Let i, j E {I, 2, 3}. As in the proof of (a), we must have Xi Xj E {e, Xk} wherekisdifferentfromiandj. Ifxixj=e,thenxj=xil ====} XjXi=e ====} Xiand Xj commute. If Xi Xj = Xk, then Xj . Xi must also equal Xk (else it wouldequal e and then Xi . Xj would equal e as well.) Thus, Xi and Xj alwayscommute. But we chose i and j randomly, so G is abelian.

(c) Let G:= {e,xI,x2,X3,X4} where e is the identity element of G. Leti, j E {I, 2, 3, 4}. We will show that Xi . Xj = Xj . Xi. To this end, letk, l E {I, 2, 3, 4} such that Xi Xj = Xk and Xj Xi = Xl. By the argument usedin (a), we may assume that k, l ~ {i, j} . We now consider the inverse of Xi.As Xi - I - e, xiI - I - e. If xiI = Xj, then Xi and Xj commute and we are done.If xiI = Xk, then Xj . Xi = Xl ====} Xj Xi . Xk = Xl Xk ====} Xj = Xl Xk. Nowusing the other equation, we get Xk = Xi . Xj = Xi . Xl . Xk. By Lemma 2.3.2,

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e = Xi . Xl ====} Xl = xiI ====} l = k and so again Xi and Xj commute. Weproceed similarly if xiI = Xl.The only case remaining then is if xiI = Xi. But for the exact samereasons we may assume that also xjl = Xj. Lemma 2.3.1(d) implies that-1 ( ) 1 -1 -1 W h t txk = Xi Xj - = Xj . Xi = Xj . Xi = Xl. e now WIS 0 compu e Xk . Xi.

By the arguments used in (a), Xk . Xi E {e, Xj, Xl}. Since l - I - iand X;;1 = Xl,Xk . Xi - I - e. If Xl = Xk . Xi = Xi Xj . Xi = Xl Xi, then by Lemma 2.3.2, Xi = ea contradiction. Thus, Xk . Xi = Xj ====} Xj Xi = Xk . Xi . Xi = Xk ====} k = l.This shows that Xi and Xj must always commute. Since these were chosenarbitrarily, G is abelian.

[Note: This problem (especially parts (a) and (c)) can be solved in a moreelegant fashion using Lagrange's Theorem and Corollary 5 from 2.4. In fact,Corollary 5 will imply that there is "only" one group of order 3 and" only"one group of order 5 (where" only" will be explained more fully in 2.7.) Inthe case of a group of order 4, there will turn out to be 2 possible groups.]

10. Show that if every element of the group G is its own inverse, then G isabelian.Proof: Let x, y E G. Then xy = (xy)-1 = y-lx-l = yX .11. If G is a group of even order, prove it has an element a - I - e satisfyinga2 = e.Proof: We define an equivalence relation on G, namely a rv b if either b = aor b = a-I. Each equivalence class clearly has either 1 or 2 elements. Ifa E G such that cl(a) = {a } then a = a -I and so a2 = e. So it will sufficeto show that there is a class of an element other than e with only 1 element.But the relation partitions G into disjoint subsets and since c l ( e) = {e}, theremaining elements form a subset of odd order. A set with odd order cannotbe a disjoint union of sets with even order, hence one of the classes musthave only one element and the result is proved .

12. Let G be a nonempty set closed under an associative product, which inaddition satisfies:

(a) There exists an e E G such that a e = a for all a E G.(b) Given a E G, there exists an element y ( a ) E G such that a . y ( a ) = e.Prove that G must be a group under this product.

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Proof: It will suffice to show that e . a = e and y( a ) . a = e for all a E G.But y(a)a = y(a)ae = y(a)ay(a)y(y(a) ) = y(a)ey(y(a)) = y(a)y(y(a) ) = eand then ea = ay(a)a = a e = a .13. Prove, by an example, that the conclusion of Problem 12 is false if weassumed instead:

(a') There exists an e E G such that a . e = a for all a E G.(b') Given a E G, there exists y( a ) E G such that y( a ) . a = e.

Proof: Let G := {e, a } where multiplication is given by:e e e e a = ea e a a a = a .

Note that under this multiplication, the product is determined completely bythe first factor. One, therefore, sees that this product is associative. More-over, e satisfies property (a'), while y ( e) = e and y ( a ) = e and so G satisfiesboth properties. But G is not a group since there is no identity element .14. Suppose a finite set G is closed under an associative product and thatboth cancellation laws hold in G. Prove that G must be a group.Proof: Let G := {gl, ... ,gn} and let g be an arbitrary element in G. Notethat for i, j E {I, ... , n}, if 9 . g i = 9 . g j, then the cancellation law impliesthat g i = g j. Therefore, {g . g i 1 1 : 0 : ; i 0 : ; n} = G. In particular, there exists isuch that 9 . g i = g . Let x E G. Then we may proceed as above and find ag j E G such that x = g j. g . Then X g i = (g j. g ). g i = g j. (g . g i) = g j . 9 = x.Set e :=g i. Then we have proved that X e = x for all x E G. Moreover, sincee E G, our argument above shows that for all x E G, there exists y(x) E Gsuch that x . y ( x ) = e. By Problem 12, G is a group .15. (a) Using the result of Problem 14, prove that the nonzero integersmodulo p, p a prime number, form a group under multiplication mod p.

(b) Do part (a) for the nonzero integers relatively prime to n under mul-tiplication mod n.Proof: (a) By Problem 14 it will suffice to show that the cancellation lawshold, that is, if a , b , x E : : z such that p t a b x (this is shorthand for saying pdoes not divide a, b, or x), then a x = = b x (mod p) ====} a = = b (mod p) andxa = = xb (mod p) ====} a = = b (mod p). So let a , b , x E : : z such that p t a bxand a x = = b x (mod p). Then pla x - b x = (a - b )x . Since p t x , we havep ia - b ====} a = = b (mod p). We proceed similarly if xa = = xb (mod p).

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(b) The proof is similar to that in (a), usmg Lemma 1.3.2 to see thatnl(a-b)x,(n,x)=l ===? nla -b .16. In Problem 14 show by an example that if one just assumed one of thecancellation laws, then the conclusion need not follow.Proof:

17. Prove that in Problem 14 infinite examples exist, satisfying the condi-tions, which are not groups.Proof:

18. For any ti > 2 construct a non-abelian group of order 2n.Proof: For ti > 2 we will a construct a non-abelian group G of order 2n inthe following way: Let G be defined as all formal symbols xiyj, i= 0, 1 andj = 0,1,2, ... ,n - 1 where we assume

., .,x' y J if and only if . f fz=z,)=),

xyyn = e,-1y x.

One can easily see that G is a group with identity xOyO and (xiyj)-l =xiy(-l)i+lj and since x and y do not commute, G is non-abelian. Finally, therelations imply that G has 2n elements .

19. If S is a set closed under an associative operation, prove that no matterhow you bracket a1a2 ... an, retaining the order of the elements, you get thesame element in S (e.g., (a1 . a 2) . (a3 a4 ) = a1 . (a2 (a3 a4 )); use inductionon n).Proof:

20. Let G be the set of all real 2 X 2 matrices (~ ~), where ad - b e - I - 0 isa rational number. Prove that G forms a group under matrix multiplication.Proof: Note that a d - b e is the determinant of the matrix and since thedeterminant is multiplicative, the operation is closed. It is also associative.We may set e := (~ ~) and this is the identity of G. If (~ ~) E G, thenad~be E Q - {O} ===? (ad~ge - a~~be) E G and multiplication shows that

- ad-be ad-be

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this is (~ ~) -1. Thus, G is a group .

21. Let G be the set of all real 2 X 2 matrices (~ ~), where ad - I - o . Provethat G forms a group under matrix multiplication. Is G abelian?Proof: Let ( ~ 1 ~ : ) , ( ~ 2 ~ : ) E G. ThenThus, matrix multiplication is a binary operation on G. Since matrix mul-tiplication is associative, it remains to show that G contains an identityelement and inverses. But clearly (~ ~) E G and if (~ ~) E G then

( ~ - : d ) E G and this latter matrix is the inverse of the former. Thus Go 1dIS a group. It is not abelian however since, for example,

(~ ~). (~ ~) = (~ ~)and (~~)22. Let G be the set of all real 2 X 2 matrices (~ a~I ) ' where a - I - o . Provethat G forms an abelian group under matrix multiplication.Proof: Let (~ a~I)' (~ b~l) E G. Then

(a 0 ) (b 0 ) (a b 0 ) (a b 0 )o a -I . 0 b -1 = 0 a -1b -1 = 0 (ab) -1 E G ,(where we note that a b = b a since a and b are real numbers). Thus G isclosed under multiplication. One now easily sees that (~ ~) is the identity

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of G and (~ a~1 ) -1 (a ;1 ~ ). Therefore G is a group. Finally,

(b a 0 )o (b a )-1 (a b 0 )o (a b )-1 ,and so G is abelian. 23. Construct in the G of Problem 21 a subgroup of order 4.Proof: [Note: See 2.4 for more about the notion of a subgroup] A subgroupof G is a subset of G that, under the same multiplication, also forms a group.One can check that the subset

satisfies the group axioms .24. Let G be the set of all 2 X 2 matrices (~ ~), where a.b.c.d are integersmodulo 2, such that ad - b e - I - O . Using matrix multiplication as the operationin G, prove that G is a group of order 6.Proof: The same proof used in Problem 20 shows that G is a group (notethat the inverse is easier to write down since a d - b e - I - 0 ====} a d - b e = = 1(mod 2).) To help calculate the order of G, we make the following observa-tion: If (a , e) = (0,0) or (b , d) = (0,0 or (a , e) = (b , d) then a d - b e = o .Therefore, we have only 3 choices for (a,c), namely, (1,0), (0,1) and (1,1).Once we have made this choice, only two choices remain for (b,d) (since wehave already established that (a , e) - I - ( b , d ) - I - (0,0).) Therefore, G has nomore than 3 . 2 = 6 elements. On the other hand, one can easily see thata d - b e - I - 0 for all 6 of the matrices we have produced. Thus, IG I = 6.25. (a) Let G be the group ofall2x2 matrices (~ ~), such that ad - b e - I - 0and a.b.c.d are integers modulo 3, relative to matrix multiplication. Showthat IG I = 48.(b) Ifwe modify the example of G in part (a) by insisting that a d - b e = 1,then what is I G I ?Proof: (a) We first recall some facts from linear algebra. A matrix is in-vertible if and only if the determinant is non-zero. This is precisely what the

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condition a d - b e - I - 0 is guaranteeing. Also, a matrix is invertible if and onlyif the columns are linearly independent. Two vectors are linearly indepen-dent if and only if they are not scalar multiples of each other. Putting thisall together, we compute the order of G.

Let A = (~ ~) E G. We may put any non-zero vector in the first columnof A, so there are 32 - 1 possibilities for (a,c). Now of the 32 possibilitiesfor a 2-vector over the integers modulo 3 , exactly 3 will be scalar multiplesof (a,c). Therefore, there will be 32 - 3 possibilities for (b,d) once (a,c) ischosen. Consequently, G has (32 - 1)(32 - 3) = 48 elements.

(b) Let H 2 be the subset of G consisting of all matrices in G such thata d - b e = 2. Then there is a bijection between Hand H 2 given by A f - - - 7 2 Afor A E H. Thus, IHI = IGI/2 = 48.26. (a) Let G be the group of all 2 X 2 matrices (~ ~), such that a d - b e - I - 0and a.b.c.d are integers modulo p, p a prime number. G forms a group relativeto matrix multiplication. What is IGI?

(b) Let H be the subgroup of the group G consisting of all matrices in Gsuch that a d - b e = 1. What is IHI?Proof: (a) Using the same proof as in Problem 25( a), with 3 replaced by p,we get IG I = ( p 2 - 1) ( p 2 - p ) .

(b) Let Hi be the subset of G consisting of all matrices in G such thata d - b e = i, for i = 2, ... p - 1. Then for each i, there is a bijection betweenH and Hi given by A f - - - 7 i.A for A E H. Thus, IHI = IGI/(p - 1) =( p 2 _ 1) ( p 2 - p ) / ( p - 1).

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