Solution Manual for First Course in Abstract Algebra, A, 8th Edition

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’SSOLUTIONSMANUALJOHNB.FRALEIGH ANDNEALBRANDAFIRSTCOURSE INABSTRACTALGEBRAEIGHTHEDITIONJohn B. FraleighUniversity of Rhode IslandNeal BrandUniversity of North Texas

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CONTENTS0. Sets and Relations01I. Groups and Subgroups1. Binary Operations052. Groups083. Abelian Examples144. Nonabelian Examples195. Subgroups226. Cyclic Groups277. Generators and Cayley Digraphs32II. Structure of Groups8. Groups of Permutations349. Finitely Generated Abelian Groups4010. Cosets and the Theorem of Lagrange4511. Plane Isometries50III. Homomorphisms and Factor Groups12. Factor Groups5313. Factor Group Computations and Simple Groups5814. Group Action on a Set6515. Applications of G-Sets to Counting70VI. Advanced Group Theory16. Isomorphism Theorems7317. Sylow Theorems7518. Series of Groups8019. Free Abelian Groups8520. Free Groups8821. Group Presentations91

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V. Rings and Fields22. Rings and Fields9523. Integral Domains10224. Fermat’s and Euler’s Theorems10625. RSA Encryption109VI. Constructing Rings and Fields26. The Field of Quotients of an Integral Domain11027. Rings of Polynomials11228. Factorization of Polynomials over a Field11629. Algebraic Coding Theory12330. Homomorphisms and Factor Rings12531. Prime and Maximal Ideals13132. Noncommutative Examples137VII. Commutative Algebra33. Vector Spaces14034. Unique Factorization Domains14535. Euclidean Domains14936. Number Theory15437. Algebraic Geometry16038. Gröbner Bases for Ideals163VIII. Extension Fields39. Introduction to Extension Fields16840. Algebraic Extensions17441. Geometric Constructions17942. Finite Fields182IX. Galois Theory43. Automorphisms of Fields18544. Splitting Fields19145. Separable Extensions19546. Galois Theory199

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47. Illustrations of Galois Theory20348. Cyclotomic Extensions21149. Insolvability of the Quintic214APPENDIX:Matrix Algebra216

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0. Sets and Relations10. Sets and Relations1.{3,3}−2.{2, –3}.3.{1, −1, 2, −2, 3, −3, 4, −4, 5, −5, 6, −6, 10, −10, 12, −12, 15, −15, 20, −20, 30, −30,60, −60}4.{2, 3, 4, 5, 6, 7, 8}5.It is not a well-defined set. (Some may argue that no element of+is large,because every element exceeds only a finite number of other elements but isexceeded by an infinite number of other elements. Such people might claim theanswer should be∅.)6.∅7.The set is∅because 33= 27 and 43= 64.8.{2narr∈=for some aa+∈and some integer n≥0}.9.It is not a well-defined set.10.Thesetcontainingallnumbersthatare(positive,negative,orzero)integermultiples of 1, 1/2, or 1/3.11.{(a, 1), (a, 2), (a,c), (b, 1), (b, 2), (b,c), (c, 1), (c, 2), (c,c)}12. a.This is a function which is both one-to-one and onto B.b.This not a subset ofA×B, and therefore not a function.c.It is not a function because there are two pairs with first member 1.d.Thisisafunctionwhichisneitherone-to-one(6appearstwiceinthesecondcoordinate) nor onto B ( 4 is not in the second coordinate).e.It is a function. It is not one-to-one because there are two pairs with second member 6.It is not ontoBbecause there is no pair with second member 2.f.This is not a function mapping A into B since 3 is not in the first coordinate of anyordered pair.13.Draw the line throughPandx, and letybe its point of intersection with the linesegmentCD.14. a.[][]():0,10, 2where2xxφφ→=b.[][]():1, 35, 25wher23exxφφ=+→c.[][]()():,,wheredcabcdxcxabaφφ−→=+−−15.Let()12:be defined bytan(()).Sxxφφπ→=−16. a.; cardinality 1∅b.,; cardinali{ }ty 2a∅c.{ } { } {},,,,; cardinality 4abab∅d.{ } { } { } {} {},,,,,,,,,,,,; cardinalit{}y{}8abcabacbcabc∅

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20. Sets and Relations17. Conjecture:()||.|2|2sAPA==ProofThe number of subsets of a setAdepends only on the cardinality ofA, not onwhat the elements ofAactually are. SupposeB= {1,2,3, · · · , s−1}andA= {1,2,3, · · · , s}. ThenAhas all the elements ofBplus the one additional elements. Allsubsets ofBare also subsets ofA; these are precisely the subsets ofAthat do notcontains, so the number of subsets ofAnot containingsis|P(B)|. Any other subsetofAmust contains, and removal of theswould produce a subset ofB. Thus thenumber of subsets ofAcontainingsis also|P(B)|. Because every subset ofAeithercontainssor does not contains(but not both), we see that the number of subsets ofAis 2|P(B)|.We have shown that ifAhas one more element thatB, then|P(A)|= 2|P(B)|.Now|P(∅)|= 1, so if|A|=s, then|P(A)|= 2s.18.Wedefineaone-to-onemapφofBAontoP(A).Letf∈BA,andlet()(){|1 .}fxAfxφ=∈=Supposeφ(f) =φ(g). Thenf(x) = 1 if and only ifg(x)= 1. Because the only possible values forf(x) andg(x) are 0 and 1, we see thatf(x)= 0 if and only ifg(x) = 0. Consequentlyf(x) =g(x) for allx∈Asof=gandφisone to one. To show thatφis ontoP(A), letS⊆A, and leth:A→{0,1}bedefined byh(x) = 1 ifx∈Sandh(x) = 0 otherwise. Clearlyφ(h) =S, showing thatφis indeed ontoP(A).19.Pickingupfromthehint,let(){|}.ZxA xxφ=∈∈Weclaimthatforany(),.aAaZφ∈=Either(),aaφ∈in which case(), or,aZaaφ∈∈in whichcase.aZ∈ThusZandφ(a) are certainly different subsets ofA; one of themcontainsaand the other one does not.Based on what we just showed, we feel that the power set ofAhas cardinalitygreater than|A|. Proceeding naively, we can start with the infinite set,form itspower set, then form the power set of that, and continue this process indefinitely. Ifthere were only a finite number of infinite cardinal numbers, this process wouldhave to terminate after a fixed finite number of steps. Since it doesn’t, it appearsthat there must be an infinite number of different infinite cardinal numbers.The set of everything is not logically acceptable, because the set of all subsets ofthe set of everything would be larger than the set of everything, which is a fallacy.20. a.The set containing precisely the two elements ofAand the three (different)elements ofBisC= {1,2,3,4,5}which has 5 elements.i)LetA= {−2,−1,0}andB= {1,2,3, · · ·}=.+Then|A|= 3 and|B|=ℵ0, andAandBhave no elements in common. The setCcontaining all elements in eitherAorBisC= {−2,−1,0,1,2,3, · · ·}. The mapφ:C→Bdefined byφ(x) =x+ 3 is one toone and ontoB, so|C|=|B|=ℵ0. Thus we consider 3 +ℵ0=ℵ0.ii)LetA= {1,2,3, · · ·}andB= {1/2,3/2,5/2, · · ·}. Then|A|=|B|=ℵ0andAandBhave no elements in common.The setCcontaining all elements in eitherAofBisC={1/2,1,3/2,2,5/2,3, · · ·}. The mapφ:C→Adefined byφ(x) = 2xis one to oneand ontoA, so|C|=|A|=ℵ0. Thus we considerℵ0+ℵ0=ℵ0b.We leave the plotting of the points inA × Bto you. Figure 0.15 in the text, wherethere areℵ0rows each havingℵ0entries, illustrates that we would consider thatℵ0·ℵ0=ℵ0.

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0. Sets and Relations321.There are 102= 100 numbers (.00 through .99) of the form .##, and 105= 100, 000numbers (.00000 through .99999) of the form .#####. Thus for .#####· · ·, weexpect 100ℵsequences representing all numbersx∈such that 0 ≤x≤ 1, but asequence trailing off in 0’s may represent the samex∈as a sequence trailing ofin 9’s. At any rate, we should have100ℵ≥ |[0, 1]| =;see Exercise 15. On theother hand, we can represent numbers inusing any integer basen >1, and thesesame100ℵsequences using digits from 0 to 9 in basen= 12 would not represent allx∈[0,1], so we have10.0ℵ≤Thus we consider the value of 100ℵto be.Wecould make the same argument using any other integer basen >1, and thusconsidern0ℵ=forn∈,+n >1. In particular,1212.00ℵℵ==22.( 2)||2)(2)0(,, 2, 2|2|,ℵ23.1. There is only one partition {{a}} of a one-element set {a}.24.There are two partitions of {a,b}, namely {{a,b}} and {{a}, {b}}.25.There are five partitions of {a,b,c}, namely {{a,b,c}}, {{a}, {b,c}}, {{b},{a,c}}, {{c}, {a,b}}, and {{a}, {b}, {c}}.26.15. The set {a, b, c, d}has 1 partition into one cell, 7 partitions into two cells (fourwith a 1,3 split and three with a 2,2 split), 6 partitions into three cells, and 1partition into four cells for a total of 15 partitions.27.52. The set {a, b, c, d, e}has 1 partition into one cell, 15 into two cells, 25 intothree cells, 10 into four cells, and 1 into five cells for a total of 52. (Do acombinatorics count for each possible case, such as a 1,2,2 split where there are 15possible partitions.)28.Reflexive:In order forx R xto be true,xmust be in the same cell of the partition asthe cell that containsx. This is certainly true.Transitive:Suppose thatx R yandy R z. Thenxis in the same cell asyso,xy=and y is inthe same cell aszso that.yz=By the transitivity of the set equalityrelation on the collection of cells in the partition, we see thatxz=so thatxis inthe same cell asz. Consequently,x R z.29.Not an equivalence relation; 0 is not related to 0, so it is not reflexive.30.Not an equivalence relation; 3≥2 but 23,≥so it is not symmetric.31.Not an equivalence relation since transitivity fails: 3R15 and 15R5, but 3R5.Also not reflexive: 1R1.32.0(0, 0)=and(),xyis the circle centered at the origin with radius22.xy+33.(See the answer in the text.)34.It is an equivalence relation;11, 11, 21, 31,· · · , 22, 12, 22, 32,{}{}{· · · , · · ·, 1010, 20, 30, 40,··}·.===35.a.{ . . . , –3, 0, 3, . . . }, { . . . , –2, 1, 4, . . .}, { . . . , –1, 2, 5, . . .}b.{ . . . , –4, 0, 4, . . . }, { . . . , –3, 1, 4, . . .}, { . . . , –6, –2, 2, . . .}, { . . . , –5, –1, 3, . . .}c.{ . . . , –5, 0, 5, . . . }, {. . . , –4, 1, 6, . . . }, { . . . , –3, 2, 7, . . .}, { . . . , –2, 3, 8, . . . }.{ . . . , –1, 4, 9, . . . }

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40. Sets and Relations36.a.{}0, 1, 2b.{}0, 1, 2, 3c.{}0, 1, 2, 3, 437.{}1 =haremainders1xxn∈÷depends on the value ofn.38. a.Leth, k,andmbe positive integers. We check the three criteria.Reflexive: h–h = n0 soh∼h.Symmetric:Ifh∼kso thath–k=nsfor somes∈,thenk–h=n(–s) sok∼h.Transitive:Ifh∼kandk∼m, then for somes, t∈,we haveh–k=nsandk–m=nt. Thenh–m= (h−k) + (k–m) =ns+nt=n(s+t), soh∼m.b.Leth, k∈.In the sense of this exercise,h∼kif and only ifh – k = nqfor someq∈.In the sense of Example 0.19,h≡k(modn) if and only ifhandkhave the sameremainder when divided byn. Writeh = nq1+ r1andk = nq2+ r2where 0 ≤r1<nand 0 ≤r2<n. Thenh–k=n(q1–q2) + (r1–r2)and we see thath–kis a multiple ofnif and only ifr1=r2. Thus the conditions arethe same.39.()()()()11221212ababaabb+−+=−+−which is the sum of two multiples of n.Since the sum of two multiples ofnis also a multiple of n,()()1122.abab++40.()()()()()()()()112211121222112122a ba ba ba ba ba babbaab−=−+−=−+−whichisthe sum of two multiples of n. Since the sum of two multiples of n is also a multipleofn,()()1122.a ba b41.The nametwo-to-two functionsuggests that such a functionfshould carry everypair of distinct points into two distinct points. Such a function is one-to-one in theconventional sense. (If the domain has only one element, the function cannot fail tobe two-to-two, because the only way it can fail to be two-to-two is to carry twopoints into one point, and the set does not have two points.) Conversely, everyfunction that is one-to-one in the conventional sense carries each pair of distinctpoints into two distinct points. Thus the functions conventionally called one-to-oneare precisely those that carry two points into two points, which is a much moreintuitive unidirectional way of regarding them. Also, the standard way of trying toshow that a function is one-to-one is precisely to show that it does not fail to betwo-to-two. That is, proving that a function is one-to-one becomes more natural inthe two-to-two terminology.

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1. Binary Operations51. Binary Operations1.b∗d=e,c∗c=b,[(a∗c)∗e]∗a= [c∗e]∗a=a∗a=a2.(a∗b)∗c=b∗c=aanda∗(b∗c) =a∗a=a, so the operation might beassociative, but we can’t tell without checking all other triple products.3.(b∗d)∗c=e∗c=aandb∗(d∗c) =b∗b=c, so the operation is notassociative.4.It is not commutative becauseb∗e=cbute∗b=b.5.Nowd∗a=dso fill indfora∗d. Also,c∗b=aso fill inaforb∗c. Nowb∗d=cso fill incford∗b. Finally,c∗d=bso fill inbford∗c.6.d∗a= (c∗b)∗a=c∗(b∗a) =c∗b=d. In a similar fashion, substitutingc∗bfordand using the associative property, we find thatd∗b=c, d∗c=c,andd∗d=d. ais an identity.7.It is not commutative because 1 – 2≠2 – 1. It is not associative because 2 = 1 –(2 – 3)≠(1 – 2) – 3 = –4. No identity.8.Commutative since 2ab+ 3 = 2ba+ 3. Not associative since (1∗2)∗3 = 45 and1∗(2∗3) = 33. No identity since 0∗e= 3≠0.9.Commutative sincea∗b=ab+a+b=b∗a. Associative sincea∗b= (a+ 1)(b+ 1) – 1 making it easy to see that (a*b)∗c= (a+ 1) (b+ 1) (c+ 1) – 1 =a∗(b∗c). The identity is 0.10.It is commutative because 2ab= 2bafor all,.a b+∈It is not associative because(2)22()ababcabcc∗∗=∗=, but()222()bcbcabaca=∗∗=∗. No identity.11.It is not commutative because 2∗3 = 23= 8≠9 = 32= 3∗2. It is not associativebecausea∗(b∗c) =a∗bc=(),cbabut (a∗b)∗c=ab∗c= (ab)c=abc, andbc≠bcfor some,.b c+∈No identity.12.IfShas just one element, there is only one possible binary operation onS; the tablemust be filled in with that single element. IfShas two elements, there are 16possible operations, for there are four places to fill in a table, and each may befilled in two ways, and 2⋅2⋅2⋅2 = 16. There are 19,683 operations on a setSwith three elements, for there are nine places to fill in a table, and 39= 19,683.Withnelements, there aren2places to fill in a table, each of which can be done innways, so there are()2nnpossible tables.13.A commutative binary operation on a set withnelements is completely determinedby the elements on or above themain diagonalin its table, which runs from theupper left corner to the lower right corner. The number of such places to fill in is22.22nnnnn−++=Thus there are()2/2nnn+possible commutative binary operations on ann-elementset. Forn= 2, we obtain 23= 8, and forn= 3 we obtain 36= 729.14.nn(n– 1))since there aren2–n=n(n– 1) spots to be filled once the diagonal isfilled.

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61. Binary Operations15.2–(()1)nnsince after the first row and column are determined there are (n– 1)2spotsto be filled.16.It is incorrect. Mention should be made of the underlying set for∗and theuniversal quantifier,for all, should appear.A binary operation∗on a setSiscommutativeif and only ifa∗b=b∗aforall,.a bS∈17.The definition is correct.18.It is incorrect. Replace the finalSbyH.19.An identity in the setSwith operation∗is elementeS∈such that for all,aS∈a∗e=e∗a=a.20.No, becausee1∗e2=e1ande1∗e2=e2.21.This is an operation.22.No. Condition 2 is violated. 1∗2 should be 0, but0.+∉23.No. Condition 2 is violated. 2∗1 should be 0, but0.+∉24.No. Condition 1 is violated since the value of 1∗2 is not well defined as it couldeither be 1 or –1. Also, Condition 2 is violated since –1∗2 is undefined.25.It is not a binary operation. Condition 1 is violated, for 2∗3 might be any integergreater than 9.26.It is not a binary operation. Condition 2 is violated, for 1∗1 = 0 and0.+∉27.a.Yes.() .abcdacbdbadcbdac−−+−++=++b.Yes.() .abcdacbdadbcbadcadbcacbd−−−−+=+−28.F T F F F T T T T F F F T F29.(See the answer in the text.)30.We have (a∗b)∗(c∗d) = (c∗d)∗(a∗b) = (d∗c)∗(a∗b) = [(d∗c)∗a]∗b,where we used commutativity for the first two steps and associativity for the last.31.The statement is true. Commutativity and associativity assert the equality ofcertain computations. For a binary operation on a set with just one element, thatelement is the result of every computation involving the operation, so the operationmust be commutative and associative.∗ab32.abaThe statement is false. Consider the operation on {a,b} defined bybaathe table. Then (a∗a)∗b=b∗b=abuta∗(a∗b) =a∗a=b.33.It is associative.Proof:[(f+g) +h](x) = (f+g)(x) +h(x) = [f(x) +g(x)] +h(x) =f(x) + [g(x) +h(x)] =f(x) + [(g+h)(x)] = [f+ (g+h)](x) because addition inis associative.

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1. Binary Operations734.It is not commutative. Letf(x) = 2xandg(x) = 5x. Then (f–g)(x) =f(x) –g(x) =2x– 5x= –3xwhile (g–f)(x) =g(x) –f(x) = 5x– 2x= 3x.35.It is not associative. Letf(x) = 2x,g(x) = 5x, andh(x) = 8x. Then [f– (g–h)](x) =f(x) – (g–h)(x) =f(x) – [g(x) –h(x)] =f(x) –g(x) +h(x) = 2x– 5x+ 8x= 5x, but[(f–g) –h](x) = (f–g)(x) –h(x) =f(x) –g(x) –h(x) = 2x– 5x– 8x= –11x.36.No identity.37.The constant functionf(x) = 1 is an identity element inF.38.It is commutative.Proof:(f⋅g)(x) =f(x)⋅g(x) =g(x)⋅f(x) = (g⋅f)(x) because multiplication inis commutative.39.It is associative.Proof:[(f⋅g)⋅h](x) = (f⋅g)(x)⋅h(x) = [f(x)⋅g(x)]⋅h(x) =f(x)⋅[g(x)⋅h(x)] =[f⋅(g⋅h)](x) because multiplication inis associative.40.It is not commutative. Letf(x) =x2andg(x) =x+ 1. Then (fog)(3) =f(g(3)) =f(4) = 16 but (gof)(3) =g(f(3)) =g(9) = 10.41.It is not true. Let∗be + and let∗′be⋅and letS=. Then 2 + (3⋅5) = 17 but (2 +3)⋅(2 + 5) = 35.42.Let,.a bH∈By definition ofH, we havea∗x=x∗aandb∗x=x∗bfor all.xS∈Using the fact that∗is associative, we then obtain, for all,xS∈()()()()()(.)abxabxaxbaxbxabxab∗∗=∗∗=∗∗=∗∗=∗∗=∗∗This shows thata∗bsatisfies the defining criterion for an element ofH, so().abH∗∈43.Let,.a bH∈By definition ofH, we havea∗a=aandb∗b=b. Using, one stepat a time, the fact that∗is associative and commutative, we obtain()()[()][()][()][()]()().ababababababaabbaabbabbabbab∗∗∗=∗∗∗=∗∗∗=∗∗∗=∗∗∗=∗∗=∗∗=∗This show thata∗bsatisfies the defining criterion for an element ofH, so()abH∗∈.44.For any,,xyS∈x∗y= (x∗y)∗(x∗y) = ((x∗y)∗x)∗y= ((y∗x)∗x)∗y= ((x∗x)∗y)∗y= (x∗y)∗y= (y∗y)∗x=y∗x. So∗is commutative. Since∗iscommutative, (x∗y)∗z= (y∗z)∗x=x∗(y∗z) for and,.xyS∈So∗isassociative.

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82. Groups2. Groups1.No.G3fails.2.Yes3.No.G1fails.4.No.G3fails.5.No.G1fails.6.No.G2fails.7.No.G3fails.8.No.G2fails.9.No.G1fails:()()abbabb∗∗=∗∗/10.a.Closure:Letnrandnsbe two elements ofn. Now()nrnsn rsn+=+∈sonis closed under addition.Associative:We know that addition of integers is associative.Identity:00,nn=∈and 0 is the additive identity element.Inverses:Foreachnmn∈,wealsohave()nmn−∈and()()00nmnmn mmn+−=−==.b.Let:n→fbe defined by() =mnmfform∈. Clearlyfis one to oneandmapsonton.Forr,s∈,wehave()()( )( )rsn rsnrnsrs+=+=+=+fff. Thusfis an isomorphism of,+with,.n+11.Yes, it is a group. Addition of diagonal matrices amounts to adding inentries incorrespondingpositions on the diagonals, and that addition is associative. Thematrix with all entries 0 is theadditive identity, and changing the sign of theentries in a matrix yields the additive inverse of thematrix.12.No, it is not a group. Multiplication of diagonal matrices amounts to multiplying inentries incorresponding positions on the diagonals. The matrix with 1 at allplaces on the diagonal is theidentity element, but a matrix having a diagonal entry0 has no inverse.13.Yes, it is a group. See the answer to Exercise 12.14.Yes, it is a group. See the answer to Exercise 12.15.No. The matrix with all entries 0 is upper triangular, but has no inverse.16.Yes, it is a group. The sum of upper-triangular matrices is again upper triangular,and addition amounts to just adding entries inin corresponding positions.17.Yes, it is a group.Closure:LetAandBbe upper triangular with determinant 1. Then entrycijin rowiand columnjinC=ABis 0 ifi>j, because for each productaikbkjwherei > jappearing in the computation ofcij, eitherk < iso thataik= 0 ork≥i>jso thatbkj= 0. Thus the product of two upper-triangular matrices is again upper triangular.The equation det(AB) = det(A)·det(B), shows that the product of two matrices ofdeterminant 1 again has determinant 1.Associative:We know that matrix multiplication is associative.Identity:Then × nidentity matrixnIhas determinant 1 and is upper triangular.Inverse:The product property11()(1detdetdet) det)(( )nIA AAA−−===⋅showsthat if det(A) = 1, then det(A−1) = 1 also.

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2. Groups918.Matrix multiplication is associative, so it remains to show thatGis closed undermatrix multiplication,Ghas an identity and each element ofGhas an inverse. The∗eabtable for G iseeabfrom which all of these properties are easily spotted.aabebbea19.a.We must show thatSis closed under∗, that is, that1abab++= −/fora,.bS∈Nowa+b+ab=−1 if and only if(011)().1ababab=+++=++This isthe case if and only if eithera=−1 orb=−1, which is not the case fora,.bS∈b.Associative:We have()()()()abcabcbcabcbca bcbcabcabacbcabc∗∗=∗++=++++++=++++++and()()()().abcababcababcabab cabcabacbcabc∗∗=++∗=++++++=++++++Identity:0 acts as identity element for∗, for00.aaa∗=∗=Inverses:1aa−+acts as inverse ofa, for2(1)00.11111aaaa aaaaaaaaaaa−−−+−−∗=++===+++++c.Because the operation is commutative,232311.xxx∗∗=∗∗=∗Now theinverse of 11 is –11/12 by Part(b). From 11∗x= 7, we obtain11111111847741777.12121212123x−−−−+−−=∗=++=== −20.eabceabceabceeabceeabceeabcaaecbaaecbaabcebbceabbcaebbceaccbaeccbeacceabTable ITable IITable IIITable I is structurally different from the others because every element is its owninverse. Table II can be made to look just like Table III by interchanging thenamesaandbeverywhere to obtain.

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102. Groupsebaceebacbbecaaacbcccaeband rewriting this table in the ordere,a,b,c.a.The symmetry of each table in its main diagonal shows that all groups of order4 are commutative.b.Relabel1001ase,0110−asa,1001−−asb,0110−ascto obtainTable III.c.Taken= 2. There are four 2×2 diagonal matrices with entries ±1, namely10101010,,, and.01010101EABC−−====−−If we write the table for this group using the lettersE,A,B,Cin that order, weobtain Table I with the letters capitalized.21.A binary operation on a set {x,y} of two elements that produces a group iscompletely determined by the choice ofxoryto serve as identity element, so just2 of the 16 possible tables give groups. For a set {x,y,z} of three elements, agroup binary operation is again determined by the choicex,y, orzto serve asidentity element, so there are just 3 of the 19,683 binary operations that givegroups. (Recall that there is only one way to fill out a group table for {e,a} and for{e,a,b} if you requireeto be the identity element.)22.The ordersG1G3G2,G3G1G2, andG3G2G1are not acceptable. The identity elementeoccurs in the statement ofG3, which must not come beforeeis defined inG2.23.Ignoring spelling, punctuation and grammar, here are some of the mathematicalerrors.a.The statement “x= identity” is wrong.b.The identity element should bee, not (e). It would also be nice to give theproperties satisfied by the identity element and by inverse elements.c.Associativity is missing. Logically, the identity element should be mentionedbefore inverses. The statement “an inverse exists” is not quantified correctly: foreach element of the set, an inverse exists. Again, it would be nice to give theproperties satisfied by the identity element and by inverse elements.d.Replace “such that for alla,bG∈” by “if for allaG∈”. Delete “underaddition” in line 2. The element should bee, not {e}. Replace “=e” by “=a” inline 3.

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