Solution Manual for Contemporary Abstract Algebra, 9th Edition

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Complete Solutions Manual to AccompanyContemporary Abstract AlgebraNINTH EDITIONJoseph GallianUniversity of Minnesota Duluth

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iCONTEMPORARY ABSTRACT ALGEBRA 9TH EDITIONSOLUTION MANUALCONTENTSIntegers and Equivalence Relations0 Preliminaries1Groups1 Introduction to Groups72 Groups93 Finite Groups; Subgroups134 Cyclic Groups205 Permutation Groups276 Isomorphisms347 Cosets and Lagrange’s Theorem408 External Direct Products469 Normal Subgroups and Factor Groups5310 Group Homomorphisms5911 Fundamental Theorem of Finite Abelian Groups6512 Introduction to Rings6913 Integral Domains7414 Ideals and Factor Rings8015 Ring Homomorphisms8716 Polynomial Rings9417 Factorization of Polynomials10018 Divisibility in Integral Domains105

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iiFields19 Vector Spaces11020 Extension Fields11421 Algebraic Extensions11822 Finite Fields12323 Geometric Constructions127Special Topics24 Sylow Theorems12925 Finite Simple Groups13526 Generators and Relations14027 Symmetry Groups14428 Frieze Groups and Crystallographic Groups14629 Symmetry and Counting14830 Cayley Digraphs of Groups15131 Introduction to Algebraic Coding Theory15432 An Introduction to Galois Theory15833 Cyclotomic Extensions161

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1CHAPTER 0Preliminaries1.{1,2,3,4};{1,3,5,7};{1,5,7,11};{1,3,7,9,11,13,17,19};{1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24}2.a.2; 10b.4; 40c.4: 120;d.1; 1050e.pq2;p2q33. 12, 2, 2, 10, 1, 0, 4, 5.4.s=−3,t= 2;s= 8,t=−55. By using 0 as an exponent if necessary, we may writea=pm11· · ·pmkkandb=pn11· · ·pnkk, where thep’s are distinct primes and them’s andn’s arenonnegative. Then lcm(a, b) =ps11· · ·pskk, wheresi= max(mi, ni) andgcd(a, b) =pt11· · ·ptkk, whereti= min(mi, ni) Thenlcm(a, b)·gcd(a, b) =pm1+n11· · ·pmk+nkk=ab.6. The first part follows from the Fundamental Theorem of Arithmetic; forthe second part, takea= 4,b= 6,c= 12.7. Writea=nq1+r1andb=nq2+r2, where 0≤r1, r2< n. We mayassume thatr1≥r2. Thena−b=n(q1−q2) + (r1−r2), wherer1−r2≥0. Ifamodn=bmodn, thenr1=r2andndividesa−b. Ifndividesa−b, then by the uniqueness of the remainder, we then haver1−r2= 0. Thus,r1=r2and thereforeamodn=bmodn.8. Writeas+bt=d. Thena′s+b′t= (a/d)s+ (b/d)t= 1.9. By Exercise 7, to prove that (a+b) modn= (a′+b′) modnand(ab) modn= (a′b′) modnit suffices to show thatndivides(a+b)−(a′+b′) andab−a′b′. Sincendivides botha−a′andndividesb−b′, it divides their difference. Becausea=a′modnandb=b′modnthere are integerssandtsuch thata=a′+nsandb=b′+nt. Thusab= (a′+ns)(b′+nt) =a′b′+nsb′+a′nt+nsnt. Thus,ab−a′b′isdivisible byn.10. Writed=au+bv. Sincetdivides bothaandb, it dividesd. Writes=mq+rwhere 0≤r < m. Thenr=s−mqis a common multiple ofbothaandbsor= 0.11. Suppose that there is an integernsuch thatabmodn= 1. Then there isan integerqsuch thatab−nq= 1. Sinceddivides bothaandn,dalsodivides 1. So,d= 1. On the other hand, ifd= 1, then by the corollary ofTheorem 0.2, there are integerssandtsuch thatas+nt= 1. Thus,modulon,as= 1.

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0/Preliminaries212. 7(5n+ 3)−5(7n+ 4) = 113. By the GCD Theorem there are integerssandtsuch thatms+nt= 1.Thenm(sr) +n(tr) =r.14. It suffices to show that (p2+q2+r2) mod 3 = 0. Notice that for anyintegeranot divisible by 3,amod 3 is 1 or 2 and thereforea2mod 3 = 1.So, (p2+q2+r2) mod 3 =p2mod 3 +q2mod 3 +r2mod 3 = 3 mod 3=0.15. Letpbe a prime greater than 3. By the Division Algorithm, we can writepin the form 6n+r, wherersatisfies 0≤r <6. Now observe that6n,6n+ 2,6n+ 3, and 6n+ 4 are not prime.16. By properties of modular arithmetic we have(71000) mod 6 = (7 mod 6)1000= 11000= 1. Similarly,(61001) mod 7 = (6 mod 7)1001=−11001mod 7 =−1 = 6 mod 7.17. Sincestdividesa−b, bothsandtdividea−b. The converse is true whengcd(s, t) = 1.18. Observe that 8402mod 5 = 3402mod 5 and 34mod 5 = 1. Thus, 8402mod5 = (34)10032mod 5 = 4.19. If gcd(a, bc) = 1, then there is no prime that divides bothaandbc. ByEuclid’s Lemma and unique factorization, this means that there is noprime that divides bothaandbor bothaandc. Conversely, if no primedivides bothaandbor bothaandc, then by Euclid’s Lemma, no primedivides bothaandbc.20. If one of the primes did dividek=p1p2· · ·pn+ 1, it would also divide 1.21. Suppose that there are only a finite number of primesp1, p2, . . . , pn. Then,by Exercise 20,p1p2. . . pn+ 1 is not divisible by any prime. This meansthatp1p2. . . pn+ 1, which is larger than any ofp1, p2, . . . , pn, is itselfprime. This contradicts the assumption thatp1, p2, . . . , pnis the list of allprimes.22.−758+358i23.−5+2i4−5i=−5+2i4−5i4+5i4+5i=−3041+−1741i24. Letz1=a+biandz2=c+di. Thenz1z2= (ac−bd) + (ad+bc);|z1|=√a2+b2,|z2|=√c2+d2,|z1z2|=√a2c2+b2d2+a2d2+b2c2=|z1||z2|.25.xNANDyis 1 if and only if both inputs are 0;xXNORyis 1 if and onlyif both inputs are the same.26. Ifx= 1, the output isy, else it isz.

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0/Preliminaries327. LetSbe a set withn+ 1 elements and pick someainS. By induction,Shas 2nsubsets that do not containa. But there is one-to-onecorrespondence between the subsets ofSthat do not containaand thosethat do. So, there are 2·2n= 2n+1subsets in all.28. Use induction and note that2n+132n+2−1 = 18(2n32n)−1 = 18(2n33n−1) + 17.29. Considern= 200! + 2. Then 2 dividesn, 3 dividesn+ 1, 4 dividesn+ 2, . . ., and 202 dividesn+ 200.30. Use induction onn.31. Sayp1p2· · ·pr=q1q2· · ·qs, where thep’s and theq’s are primes. By theGeneralized Euclid’s Lemma,p1divides someqi, sayq1(we may relabeltheq’s if necessary). Thenp1=q1andp2· · ·pr=q2· · ·qs. Repeating thisargument at each step we obtainp2=q2,· · ·, pr=qrandr=s.32. 47. Mimic Example 12.33. Suppose thatSis a set that containsaand whenevern≥abelongs toS,thenn+ 1∈S. We must prove thatScontains all integers greater than orequal toa. LetTbe the set of all integers greater thanathat are not inSand suppose thatTis not empty. Letbbe the smallest integer inT(ifThas no negative integers,bexists because of the Well Ordering Principle; ifThas negative integers, it can have only a finite number of them so thatthere is a smallest one). Thenb−1∈S, and thereforeb= (b−1) + 1∈S.This contradicts our assumption thatbis not inS.34. By the Second Principle of Mathematical Induction,fn=fn−1+fn−2<2n−1+ 2n−2= 2n−2(2 + 1)<2n.35. Forn= 1, observe that 13+ 23+ 33= 36. Assume thatn3+ (n+ 1)3+ (n+ 2)3= 9mfor some integerm. We must prove that(n+ 1)3+ (n+ 2)3+ (n+ 3)3is a multiple of 9. Using the inductionhypothesis we have that(n+ 1)3+ (n+ 2)3+ (n+ 3)3= 9m−n3+ (n+ 3)3=9m−n3+n3+3·n2·3+3·n·9+33= 9m+9n2+27n+27 = 9(m+n2+3n+3).36. You must verify the casesn= 1 andn= 2. This situation arises in caseswhere the arguments that the statement is true fornimplies that it is trueforn+ 2 is different whennis even and whennis odd.37. The statement is true for any divisor of 83−4 = 508.38. One need only verify the equation forn= 0,1,2,3,4,5.Alternatively,observe thatn3−n=n(n−1)(n+ 1).39. Since 3736 mod 24 = 16, it would be 6 p.m.

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0/Preliminaries440. 541. Observe that the number with the decimal representationa9a8. . . a1a0isa9109+a8108+· · ·+a110 +a0. From Exercise 9 and the fact thatai10imod 9 =aimod 9 we deduce that the check digit is(a9+a8+· · ·+a1+a0) mod 9. So, substituting 0 for 9 or vice versa foranyaidoes not change the value of (a9+a8+· · ·+a1+a0) mod 9.42. No43. For the case in which the check digit is not involved, the argument givenExercise 41 applies to transposition errors. Denote the money ordernumber bya9a8. . . a1a0cwherecis the check digit. For a transpositioninvolving the check digitc= (a9+a8+· · ·+a0) mod 9 to go undetected,we must havea0= (a9+a8+· · ·+a1+c) mod 9. Substituting forcyields2(a9+a8+· · ·+a0) mod 9 =a0. Then cancelling thea0, multiplying bysides by 5, and reducing module 9, we have10(a9+a8+· · ·+a1) =a9+a8+· · ·+a1= 0. It follows thatc=a9+a8· · ·+a1+a0=a0. In this case the transposition does not yieldan error.44. 445. Say the number isa8a7. . . a1a0=a8108+a7107+· · ·+a110 +a0. Thenthe error is undetected if and only if (ai10i−a′i10i) mod 7 = 0.Multiplying both sides by 5iand noting that 50 mod 7 = 1, we obtain(ai−a′i) mod 7 = 0.46. All except those involvingaandbwith|a−b|= 7.47. 448. Observe that for any integerkbetween 0 and 8,k÷9 =.kkk . . . .50. 751. Say that the weight foraisi. Then an error is undetected if modulo 11,ai+b(i−1) +c(i−2) =bi+c(i−1) +a(i−2). This reduces to the caseswhere (2a−b−c) mod 11 = 0.52. Say the valid number isa1a2. . . a10andaiandai+1were transposed.Then, modulo 11, 10a1+ 9a2+· · ·+a10= 0 and10a1+· · ·+ (11−i)ai+1+ (11−(i+ 1))ai+· · ·+a10= 5. Thus, 5 = 5−0 =(10a1+· · ·+ (11−i)ai+1+ (11−(i+ 1))ai+a10)−(10a1+ 9a2+· · ·+a10).It follows that (ai+1−ai) mod 11 = 5. Now look for adjacent digitsxandyin the invalid number so that (x−y) mod 11 = 5. Since the only pair is39, the correct number is 0-669-09325-4.

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0/Preliminaries553. Since 10a1+ 9a2+· · ·+a10= 0 mod 11 if and only if0 = (−10a1−9a2− · · · −10a10) mod 11 = (a1+ 2a2+· · ·+ 10a10) mod 11,the check digit would be the same.54. 734458606155. First note that the sum of the digits modulo 11 is 2. So, some digit is 2too large. Say the error is in positioni. Then10 = (4,3,0,2,5,1,1,5,6,8)·(1,2,3,4,5,6,7,8,9,10) mod 11 = 2i. Thus,the digit in position 5 to 2 too large. So, the correct number is 4302311568.56. An error in an even numbered position changes the value of the sum by aneven amount. However,(9·1 + 8·4 + 7·9 + 6·1 + 5·0 + 4·5 + 3·2 + 2·6 + 7) mod 10 = 5.57. 2. Sinceβis one-to-one,β(α(a1)) =β(α(a2)) implies thatα(a1) =α(a2)and sinceαis one-to-one,a1=a2.3. Letc∈C. There is abinBsuch thatβ(b) =cand anainAsuch thatα(a) =b. Thus, (βα)(a) =β(α(a)) =β(b) =c.4. Sinceαis one-to-one and onto we may defineα−1(x) =yif and only ifα(y) =x. Thenα−1(α(a)) =aandα(α−1(b)) =b.58.a−a= 0; ifa−bis an integerkthenb−ais the integer−k; ifa−bisthe integernandb−cis the integerm, thena−c= (a−b) + (b−c) isthe integern+m. The set of equivalence classes is{[k]|0≤k <1,kis real}. The equivalence classes can be represented bythe real numbers in the interval [0,1). For any real numbera, [a] ={a+k|wherekranges over all integers}.59. No. (1,0)∈Rand (0,−1)∈Rbut (1,−1)6∈R.60. Obviously,a+a= 2ais even anda+bis even impliesb+ais even. Ifa+bandb+care even, thena+c= (a+b) + (b+c)−2bis also even. Theequivalence classes are the set of even integers and the set of odd integers.61.abelongs to the same subset asa. Ifaandbbelong to the subsetAandbandcbelong to the subsetB, thenA=B, since the distinct subsets ofPare disjoint. So,aandcbelong toA.62. Suppose thatnis odd prime greater than 3 andn+ 2 andn+ 4 are alsoprime. Thennmod 3 = 1 ornmod 3 = 2. Ifnmod 3 = 1 thenn+ 2 mod 3 = 0 and so is not prime. Ifnmod 3 = 2 thenn+ 4 mod 3 = 0and so is not prime.

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0/Preliminaries663. The last digit of 3100is the value of 3100mod 10. Observe that 3100mod10 is the same as ((34mod 10)25mod 10 and 34mod 10 = 1. Similarly,the last digit of 2100is the value of 2100mod 10. Observe that 25mod 10= 2 so that 2100mod 10 is the same as(25mod 10)20mod 10 = 220mod 10 = (25)4mod 10 = 24mod 10 = 6.64. Suppose that there are integersa, b, c,anddwith gcd(a, b) = 1 andgcd(c, d) = 1 such thata2/b2−c2/d2= 1002. Thena2d2−c2b2= 1002b2d2. If bothbanddare odd, then modulo 4,b2=d2= 1 anda2/b2−c2/d2= 1002 reduces toa2−c2= 2. This case ishandled in Example 7. If 2i(i >0) dividesb, thenais odd anda2d2−c2b2= 1002b2d2implies that 2idividesdalso. It follows that if 2nis the highest power of 2 that divides one ofbord, then 2nis the highestpower of 2 that divides the other. So dividing both sides ofa2d2−c2b2= 1002b2d2by 2nwe get an equation of the same form wherebothbanddare odd. Taking both sides modulo 4 and recalling that foroddx, x2mod 4 = 1 we have thata2d2−c2b2= 1002b2d2reducesa2−c2= 2, which was done in Example 7.65. Applyγ−1to both sides ofαγ=βγ.

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7CHAPTER 1Introduction to Groups1. Three rotations: 0◦, 120◦, 240◦, and three reflections across lines fromvertices to midpoints of opposite sides.2. LetR=R120,R2=R240,Fa reflection across a vertical axis,F′=RFandF′′=R2FR0RR2FF′F′′R0R0RR2FF′F′′RRR2R0F′F′′FR2R2R0RF′′FF′FFF′′F′R0R2RF′F′FF′′RR0R2F′′F′′F′FR2RR03.a.Vb.R270c.R0d.R0, R180, H, V, D, D′e.none4. Five rotations: 0◦, 72◦, 144◦, 216◦, 288◦, and five reflections across linesfrom vertices to midpoints of opposite sides.5.Dnhasnrotations of the formk(360◦/n), wherek= 0, . . . , n−1. Inaddition,Dnhasnreflections. Whennis odd, the axes of reflection arethe lines from the vertices to the midpoints of the opposite sides. Whennis even, half of the axes of reflection are obtained by joining oppositevertices; the other half, by joining midpoints of opposite sides.6. A nonidentity rotation leaves only one point fixed – the center of rotation.A reflection leaves the axis of reflection fixed. A reflection followed by adifferent reflection would leave only one point fixed (the intersection of thetwo axes of reflection) so it must be a rotation.7. A rotation followed by a rotation either fixes every point (and so is theidentity) or fixes only the center of rotation. However, a reflection fixes aline.8. In either case, the set of points fixed is some axis of reflection.9. Observe that 1·1 = 1; 1(−1) =−1; (−1)1 =−1; (−1)(−1) = 1. Theserelationships also hold when 1 is replaced by a “rotation” and−1 isreplaced by a “reflection.”10. reflection.

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1/Introduction to Groups811. Thinking geometrically and observing that even powers of elements of adihedral group do not change orentation we note that each ofa, bandcappears an even number of times in the expression. So, there is no changein orentation. Thus, the expression is a rotation. Alternatively, as inExercise 9, we associate each ofa, bandcwith 1 if they are rotations and−1 if they are reflections and we observe that in the producta2b4ac5a3cthe terms involvingarepresents six 1s or six−1s, the termb4representsfour 1s or four−1s, and the terms involvingcrepresents six 1s or six−1s.Thus the product of all the 1s and−1s is 1. So the expression is a rotation.12. H, I, O, X. Rotations of 0◦,180◦, horizontal reflection, and verticalreflection.13. InD4, HD=DVbutH6=V.14.Dnis not commutative.15.R0,R180,H,V16. Rotations of 0◦and 180◦; Rotations of 0◦and 180◦and reflections aboutthe diagonals.17.R0,R180,H,V18. Let the distance from a point on oneHto the corresponding point on anadjacentHbe one unit. Then translations of any number of units to theright or left are symmetries; reflection across the horizontal axis throughthe middle of theH’s is a symmetry; reflection across any vertical axismidway between twoH’s or bisecting anyHis a symmetry. All othersymmetries are compositions of finitely many of those already described.The group is non-Abelian.19. In each case the group isD6.20.D2821. First observe thatX26=R0. SinceR0andR180are the only elements inD4that are squares we haveX2=R180. SolvingX2Y=R90forYgivesY=R270.22.X2=Fhas no solutions; the only solution toX3=FisF.23. 180◦rotational symmetry.24.Z4,D5,D4,Z2D4,Z3,D3,D16D7,D4,D5,Z1025. Their only symmetry is the identity.

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9CHAPTER 2Groups1.c,d2.c,d3. none4.a,c5. 7; 13;n−1;13−2i=13−2i3+2i3+2i=313+213i6.a.−31−ib.5c.112[2−3−86]d.[2446].7. The set does not contain the identity; closure fails.8. 1, 3, 7, 9, 11, 13, 17, 19.9. Under multiplication modulo 4, 2 does not have an inverse. Undermultiplication modulo 5,{1,2,3,4}is closed, 1 is the identity, 1 and 4 aretheir own inverses, and 2 and 3 are inverses of each other. Modulomultiplication is associative.10.[1101] [1011]6=[1011] [1101].11.a11, a6, a4, a112. 5,4,813. (a) 2a+ 3b; (b)−2a+ 2(−b+c); (c)−3(a+ 2b) + 2c= 014. (ab)3=abababand(ab−2c)−2= ((ab−2c)−1)2= (c−1b2a−1)2=c−1b2a−1c−1b2a−1.15. Observe thata5=eimplies thata−2=a3andb7=eimplies thatb14=eand thereforeb−11=b3. Thus,a−2b−11=a3b3. Moreover,(a2b4)−2= ((a2b4)−1)2= (b−4a−2)2= (b3a3)2.16. The identity is 25.17. Since the inverse of an element inGis inG,H⊆G. Letgbelong toG.Theng−1belongs toGand therefore (g−1)−1=gbelong toG. So,G⊆H.18.K={R0, R180};L={R0, R180, H, V, D, D′}.

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2/Groups1019. The set is closed because det (AB) = (detA)(detB). Matrixmultiplication is associative.[1001]is the identity.Since[abcd]−1=[d−b−ca]its determinant isad−bc= 1.20. 12= (n−1)2= 1.21. Using closure and trial and error, we discover that 9·74 = 29 and 29 is noton the list.22. Considerxyx=xyx.23. Forn≥0, we use induction. The case thatn= 0 is trivial. Then notethat (ab)n+1= (ab)nab=anbnab=an+1bn+1. Forn <0, note thate= (ab)0= (ab)n(ab)−n= (ab)na−nb−nso thatanbn= (ab)n. In anon-Abelian group (ab)nneed not equalanbn.24. The “inverse” of putting on your socks and then putting on your shoes istaking off your shoes then taking off your socks. UseD4for the examples.(An appropriate name for the property (abc)−1=c−1b−1a−1is“Socks-Shoes-Boots Property.”)25. Suppose thatGis Abelian. Then by Exercise 24,(ab)−1=b−1a−1=a−1b−1. If (ab)−1=a−1b−1then by Exercise 24e=aba−1b−1. Multiplying both sides on the right bybayieldsba=ab.26. By definition,a−1(a−1)−1=e. Now multiply on the left bya.27. The case wheren= 0 is trivial. Forn >0, note that(a−1ba)n= (a−1ba)(a−1ba)· · ·(a−1ba) (nterms). So, cancelling theconsecutiveaanda−1terms givesa−1bna. Forn <0, note thate= (a−1ba)n(a−1ba)−n= (a−1ba)n(a−1b−na) and solve for (a−1ba)n.28. (a1a2· · ·an)(a−1na−1n−1· · ·a−12a−11) =e29. By closure we have{1,3,5,9,13,15,19,23,25,27,39,45}.30.Z105;Z44andD22.31. Supposexappears in a row labeled withatwice. Sayx=abandx=ac.Then cancellation givesb=c. But we use distinct elements to label thecolumns.32.157111157115511177711151111751

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2/Groups1133. Proceed as follows. By definition of the identity, we may complete the firstrow and column. Then complete row 3 and column 5 by using Exercise 31.In row 2 onlycanddremain to be used. We cannot usedin position 3 inrow 2 because there would then be twod’s in column 3. This observationallows us to complete row 2. Then rows 3 and 4 may be completed byinserting the unused two elements. Finally, we complete the bottom rowby inserting the unused column elements.34. (ab)2=a2b2⇔abab=aabb⇔ba=ab.(ab)−2=b−2a−2⇔b−1a−1b−1a−1=b−1b−1a−1a−1⇔a−1b−1=b−1a−1⇔ba=ab.35.axb=cimplies thatx=a−1(axb)b−1=a−1cb−1;a−1xa=cimplies thatx=a(a−1xa)a−1=aca−1.36. Observe thatxabx−1=bais equivalent toxab=baxand this is true forx=b.37. Sinceeis one solution it suffices to show that nonidentity solutions comein distinct pairs. To this end note that ifx3=eandx6=e, then(x−1)3=eandx6=x−1. So if we can find one nonidentity solution we canfind a second one. Now suppose thataanda−1are nonidentity elementsthat satisfyx3=eandbis a nonidentity element such thatb6=aandb6=a−1andb3=e. Then, as before, (b−1)3=eandb6=b−1. Moreover,b−16=aandb−16=a−1. Thus, finding a third nonidentity solution gives afourth one. Continuing in this fashion we see that we always have an evennumber of nonidentity solutions to the equationx3=e.To prove the second statement note that ifx26=e, thenx−16=xand(x−1)26=e. So, arguing as in the preceding case we see that solutions tox26=ecome in distinct pairs.38. InD4, HR90V=DR90HbutHV6=DH.39. Observe thataa−1b=ba−1a. Cancelling the middle terma−1on bothsides we obtainab=ba.40.X=V R270D′H.41. IfF1F2=R0thenF1F2=F1F1and by cancellationF1=F2.42. Observe thatF1F2=F2F1implies that (F1F2)(F1F2) =R0. SinceF1andF2are distinct andF1F2is a rotation it must beR180.43. SinceF Rkis a reflection we have (F Rk)(F Rk) =R0. Multiplying on theleft byFgivesRkF Rk=F.44. SinceF Rkis a reflection we have (F Rk)(F Rk) =R0. Multiplying on theright byR−kgivesF RkF=R−k. IfDnwere Abelian, thenF R360◦/nF=R360◦/n. But (R360◦/n)−1=R360◦(n−1)/n6=R360◦/nwhenn≥3.

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2/Groups1245.a.R3b.Rc.R5F46. Closure and associativity follow from the definition of multiplication;a=b=c= 0 gives the identity; we may find inverses by solving theequationsa+a′= 0,b′+ac′+b= 0,c′+c= 0 fora′, b′, c′.47. Sincea2=b2= (ab)2=e, we haveaabb=abab. Now cancel on left andright.48. Ifasatisfiesx5=eanda6=e, then so doesa2, a3, a4. Now, usingcancellation we have thata2, a3, a4are not the identity and are distinctfrom each other and distinct froma. If these are all of the nonidentitysolutions ofx5=ewe are done. Ifbis another solution that is not a powerofa, then by the same argumentb, b2, b3andb4are four distinctnonidentity solutions. We must further show thatb2, b3andb4are distinctfroma, a2, a3, a4. Ifb2=aifor somei, then cubing both sides we haveb=b6=a3i,which is a contradiction. A similar argument applies tob3andb4. Continuing in this fashion we have that the number of nonidentitysolutions tox5=eis a multiple of 4. In the general case, the number ofsolutions is a multiple of 4 or is infinite.49. The matrix[abcd]is in GL(2, Z2) if and only ifad6=bc. This happenswhenaanddare 1 and at least 1 ofbandcis 0 and whenbandcare 1and at least 1 ofaanddis 0. So, the elements are[1001] [1101] [1011] [1110] [0111] [0110].[1101]and[1011]do not commute.50. Ifnis not prime, we can writen=ab, where 1< a < nand 1< b < n.Thenaandbbelong to the set{1,2, . . . , n−1}but 0 =abmodndoes not.51. Letabe any element inGand writex=ea. Thena−1x=a−1(ea) = (a−1e)a=a−1a=e. Then solving forxwe obtainx=ae=a.52. Suppose thatab=eand letb′be the element inGwith the property thatbb′=e. Then observe thatba= (ba)e=ba(bb′) =b(ab)b′=beb′= (be)b′=bb′=e.

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