Solution Manual For First Course In Abstract Algebra, A, 7th Edition

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CONTENTS0. Sets and Relations1I. Groups and Subgroups1. Introduction and Examples42. Binary Operations73. Isomorphic Binary Structures94. Groups135. Subgroups176. Cyclic Groups217. Generators and Cayley Digraphs24II. Permutations, Cosets, and Direct Products8. Groups of Permutations269. Orbits, Cycles, and the Alternating Groups3010. Cosets and the Theorem of Lagrange3411. Direct Products and Finitely Generated Abelian Groups3712. Plane Isometries42III. Homomorphisms and Factor Groups13. Homomorphisms4414. Factor Groups4915. Factor-Group Computations and Simple Groups5316. Group Action on a Set5817. Applications of G-Sets to Counting61IV. Rings and Fields18. Rings and Fields6319. Integral Domains6820. Fermat’s and Euler’s Theorems7221. The Field of Quotients of an Integral Domain7422. Rings of Polynomials7623. Factorization of Polynomials over a Field7924. Noncommutative Examples8525. Ordered Rings and Fields87V. Ideals and Factor Rings26. Homomorphisms and Factor Rings8927. Prime and Maximal Ideals9428. Gr¨obner Bases for Ideals99iii

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VI. Extension Fields29. Introduction to Extension Fields10330. Vector Spaces10731. Algebraic Extensions11132. Geometric Constructions11533. Finite Fields116VII. Advanced Group Theory34. Isomorphism Theorems11735. Series of Groups11936. Sylow Theorems12237. Applications of the Sylow Theory12438. Free Abelian Groups12839. Free Groups13040. Group Presentations133VIII. Groups in Topology41. Simplicial Complexes and Homology Groups13642. Computations of Homology Groups13843. More Homology Computations and Applications14044. Homological Algebra144IX. Factorization45. Unique Factorization Domains14846. Euclidean Domains15147. Gaussian Integers and Multiplicative Norms154X. Automorphisms and Galois Theory48. Automorphisms of Fields15949. The Isomorphism Extension Theorem16450. Splitting Fields16551. Separable Extensions16752. Totally Inseparable Extensions17153. Galois Theory17354. Illustrations of Galois Theory17655. Cyclotomic Extensions18356. Insolvability of the Quintic185APPENDIXMatrix Algebra187iv

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0. Sets and Relations10. Sets and Relations1.{√3,−√3}2.The set is empty.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.{−10,−9,−8,−7,−6,−5,−4,−3,−2,−1,0,1,2,3,4,5,6,7,8,9,10,11}5.It is not a well-defined set. (Some may argue that no element ofZ+is large, because every elementexceeds only a finite number of other elements but is exceeded by an infinite number of other elements.Such people might claim the answer should be∅.)6.∅7.The set is∅because 33= 27 and 43= 64.8.It is not a well-defined set.9.Q10.The set containing all numbers that are (positive, negative, or zero) integer multiples of 1, 1/2, or1/3.11.{(a,1),(a,2),(a, c),(b,1),(b,2),(b, c),(c,1),(c,2),(c, c)}12. a.It is a function. It is not one-to-one since there are two pairs with second member 4. It is not ontoBbecause there is no pair with second member 2.b.(Same answer as Part(a).)c.It is not a function because there are two pairs with first member 1.d.It is a function.It is one-to-one.It is ontoBbecause every element ofBappears as secondmember of some pair.e.It is a function. It is not one-to-one because there are two pairs with second member 6. It is notontoBbecause there is no pair with second member 2.f.It is not a function because there are two pairs with first member 2.13.Draw the line throughPandx, and letybe its point of intersection with the line segmentCD.14. a.φ: [0,1]→[0,2] whereφ(x) = 2xb.φ: [1,3]→[5,25] whereφ(x) = 5 + 10(x−1)c.φ: [a, b]→[c, d] whereφ(x) =c+d−cb−a(x−a)15.Letφ:S→Rbe defined byφ(x) = tan(π(x−12)).16. a.∅; cardinality 1b.∅,{a}; cardinality 2c.∅,{a},{b},{a, b}; cardinality 4d.∅,{a},{b},{c},{a, b},{a, c},{b, c},{a, b, c}; cardinality 817. Conjecture:|P(A)|= 2s= 2|A|.ProofThe number of subsets of a setAdepends only on the cardinality ofA, not on what theelements ofAactually are. SupposeB={1,2,3,· · ·, s−1}andA={1,2,3,· · ·, s}. ThenAhas allthe elements ofBplus the one additional elements.All subsets ofBare also subsets ofA; theseare precisely the subsets ofAthat do not contains, so the number of subsets ofAnot containingsis|P(B)|. Any other subset ofAmust contains, and removal of theswould produce a subset ofB.Thus the number 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)|.

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20. Sets and RelationsWe have shown that ifAhas one more element thatB, then|P(A)|= 2|P(B)|.Now|P(∅)|= 1, soif|A|=s, then|P(A)|= 2s.18.We define a one-to-one mapφofBAontoP(A). Letf∈BA, and letφ(f) ={x∈A|f(x) = 1}.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) forallx∈Asof=gandφis one to one. To show thatφis ontoP(A), letS⊆A, and leth:A→ {0,1}be defined byh(x) = 1 ifx∈Sandh(x) = 0 otherwise. Clearlyφ(h) =S, showing thatφis indeedontoP(A).19.Picking up from the hint, letZ={x∈A|x /∈φ(x)}. We claim that for anya∈A, φ(a)6=Z. Eithera∈φ(a), in which casea /∈Z, ora /∈φ(a), in which casea∈Z.ThusZandφ(a) are certainlydifferent subsets ofA; one of them containsaand the other one does not.Based on what we just showed, we feel that the power set ofAhas cardinality greater than|A|.Proceeding naively, we can start with the infinite setZ, form its power set, then form the power setof that, and continue this process indefinitely. If there were only a finite number of infinite cardinalnumbers, this process would have to terminate after a fixed finite number of steps. Since it doesn’t,it appears that 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 of the set ofeverything 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,· · ·}=Z+.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 to one 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 one and ontoA,so|C|=|A|=ℵ0. Thus we considerℵ0+ℵ0=ℵ0.b.We leave the plotting of the points inA×Bto you. Figure 0.14 in the text, where there areℵ0rows each havingℵ0entries, illustrates that we would consider thatℵ0· ℵ0=ℵ0.21.There are 102= 100 numbers (.00 through .99) of the form.##, and 105= 100,000 numbers (.00000through .99999) of the form.#####. Thus for.#####· · ·, we expect 10ℵ0sequences representingall numbersx∈Rsuch that 0≤x≤1, but a sequence trailing off in 0’s may represent the samex∈Ras a sequence trailing of in 9’s. At any rate, we should have 10ℵ0≥ |[0,1]|=|R|; see Exercise15.On the other hand, we can represent numbers inRusing any integer basen >1, and thesesame 10ℵ0sequences using digits from 0 to 9 in basen= 12 would not represent allx∈[0,1], so wehave 10ℵ0≤ |R|. Thus we consider the value of 10ℵ0to be|R|. We could make the same argumentusing any other integer basen >1, and thus considernℵ0=|R|forn∈Z+, n >1.In particular,12ℵ0= 2ℵ0=|R|.22.ℵ0,|R|,2|R|,2(2|R|),2(2(2|R|))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}}.

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0. Sets and Relations325.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 (four with a 1,3 splitand three with a 2,2 split), 6 partitions into three cells, and 1 partition into four cells for a total of15 partitions.27.52. The set{a, b, c, d, e}has 1 partition into one cell, 15 into two cells, 25 into three cells, 10 into fourcells, and 1 into five cells for a total of 52. (Do a combinatorics count for each possible case, such asa 1,2,2 split where there are 15 possible partitions.)28.Reflexive:In order forxRxto be true,xmust be in the same cell of the partition as the cell thatcontainsx. This is certainly true.Transitive:Suppose thatxRyandyRz. Thenxis in the same cell asysox=y, andyis in thesame cell aszso thaty=z. By the transitivity of the set equality relation on the collection of cellsin the partition, we see thatx=zso thatxis in the same cell asz. Consequently,xRz.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.It is an equivalence relation; 0 ={0}anda={a,−a}fora∈R, a6= 0.32.It is not an equivalence relation; 1R3 and 3R5 but we do not have 1R5 because|1−5|= 4>3.33.(See the answer in the text.)34.It is an equivalence relation;1 ={1,11,21,31,· · ·},2 ={2,12,22,32,· · ·},· · ·,10 ={10,20,30,40,· · ·}.35.(See the answer in the text.)36. a.Leth, k,andmbe positive integers. We check the three criteria.Reflexive:h−h=n0 soh∼h.Symmetric:Ifh∼kso thath−k=nsfor somes∈Z, thenk−h=n(−s) sok∼h.Transitive:Ifh∼kandk∼m, then for somes, t∈Z, we haveh−k=nsandk−m=nt. Thenh−m= (h−k) + (k−m) =ns+nt=n(s+t), soh∼m.b.Leth, k∈Z+. In the sense of this exercise,h∼kif and only ifh−k=nqfor someq∈Z. In thesense of Example 0.19,h≡k(modn) if and only ifhandkhave the same remainder when dividedbyn. 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 are the same.c. a.0 ={· · ·,−2,0,2,· · ·},1 ={· · ·,−3,−1,1,3,· · ·}b.0 ={· · ·,−3,0,3,· · ·},1 ={· · ·,−5,−2,1,4,· · ·},2 ={· · ·,−1,2,5,· · ·}c.0 ={· · ·,−5,0,5,· · ·},1 ={· · ·,−9,−4,1,6,· · ·},2 ={· · ·,−3,2,7,· · ·},3 ={· · ·,−7,−2,3,8,· · ·},4 ={· · ·,−1,4,9,· · ·}

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41. Introduction and Examples37.The nametwo-to-two functionsuggests that such a functionfshould carry every pair of distinct pointsinto two distinct points. Such a function is one-to-one in the conventional sense. (If the domain hasonly one element, the function cannot fail to be two-to-two, because the only way it can fail to betwo-to-two is to carry two points into one point, and the set does not have two points.) Conversely,every function that is one-to-one in the conventional sense carries each pair of distinct points into twodistinct points. Thus the functions conventionally called one-to-one are precisely those that carry twopoints into two points, which is a much more intuitive unidirectional way of regarding them.Also,the standard way of trying to show that a function is one-to-one is precisely to show that it doesnot fail to be two-to-two. That is, proving that a function is one-to-one becomes more natural in thetwo-to-two terminology.1. Introduction and Examples1.i3=i2·i=−1·i=−i2.i4= (i2)2= (−1)2= 13.i23= (i2)11·i= (−1)11·i= (−1)i=−i4.(−i)35= (i2)17(−i) = (−1)17(−i) = (−1)(−i) =i5.(4−i)(5 + 3i) = 20 + 12i−5i−3i2= 20 + 7i+ 3 = 23 + 7i6.(8 + 2i)(3−i) = 24−8i+ 6i−2i2= 24−2i−2(−1) = 26−2i7.(2−3i)(4 +i) + (6−5i) = 8 + 2i−12i−3i2+ 6−5i= 14−15i−3(−1) = 17−15i8.(1 +i)3= (1 +i)2(1 +i) = (1 + 2i−1)(1 +i) = 2i(1 +i) = 2i2+ 2i=−2 + 2i9.(1−i)5= 15+5114(−i) +5·42·113(−i)2+5·42·112(−i)3+5111(−i)4+ (−i)5= 1−5i+ 10i2−10i3+ 5i4−i5=1−5i−10 + 10i+ 5−i=−4 + 4i10.|3−4i|=√32+ (−4)2=√9 + 16 =√25 = 511.|6+4i|=√62+ 42=√36 + 16 =√52 = 2√1312.|3−4i|=√32+ (−4)2=√25 = 5and 3−4i= 5(35−45i)13.| −1 +i|=√(−1)2+ 12=√2and−1 +i=√2(−1√2+1√2i)14.|12 + 5i|=√122+ 52=√169and 12 + 5i= 13(1213+513i)15.| −3 + 5i|=√(−3)2+ 52=√34and−3 + 5i=√34(−3√34+5√34i)16.|z|4(cos 4θ+isin 4θ) = 1(1 + 0i) so|z|= 1 and cos 4θ= 1 and sin 4θ= 0. Thus 4θ= 0 +n(2π) soθ=nπ2which yields values 0,π2, π,and3π2less than 2π. The solutions arez1= cos 0 +isin 0 = 1,z2= cosπ2 +isinπ2 =i,z3= cosπ+isinπ=−1,andz4= cos 3π2+isin 3π2=−i.17.|z|4(cos 4θ+isin 4θ) = 1(−1 + 0i) so|z|= 1 and cos 4θ=−1 and sin 4θ= 0. Thus 4θ=π+n(2π) soθ=π4+nπ2which yields valuesπ4,3π4,5π4,and7π4less than 2π. The solutions arez1= cosπ4 +isinπ4 =1√2 +1√2i,z2= cos 3π4+isin 3π4=−1√2 +1√2i,z3= cos 5π4+isin 5π4=−1√2−1√2i,andz4= cos 7π4+isin 7π4=1√2−1√2i.

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1. Introduction and Examples518.|z|3(cos 3θ+isin 3θ) = 8(−1 + 0i) so|z|= 2 and cos 3θ=−1 and sin 3θ= 0. Thus 3θ=π+n(2π) soθ=π3+n2π3which yields valuesπ3, π,and5π3less than 2π. The solutions arez1= 2(cosπ3 +isinπ3 ) = 2( 12 +√32i) = 1 +√3i,z2= 2(cosπ+isinπ) = 2(−1 + 0i) =−2,andz3= 2(cos 5π3+isin 5π3 ) = 2( 12−√32i) = 1− √3i.19.|z|3(cos 3θ+isin 3θ) = 27(0−i) so|z|= 3 and cos 3θ= 0 and sin 3θ=−1. Thus 3θ= 3π/2 +n(2π)soθ=π2+n2π3which yields valuesπ2,7π6,and11π6less than 2π. The solutions arez1= 3(cosπ2 +isinπ2 ) = 3(0 +i) = 3i,z2= 3(cos 7π6+isin 7π6 ) = 3(−√32−12i) =−3√32−32iandz3= 3(cos 11π6+isin 11π6) = 3(√32−12i) = 3√32−32i.20.|z|6(cos 6θ+isin 6θ) = 1 + 0iso|z|= 1 and cos 6θ= 1 and sin 6θ= 0.Thus 6θ= 0 +n(2π) soθ= 0 +n2π6which yields values 0,π3,2π3, π,4π3,and5π3less than 2π. The solutions arez1= 1(cos 0 +isin 0) = 1 + 0i= 1,z2= 1(cosπ3 +isinπ3 ) = 12 +√32i,z3= 1(cos 2π3+isin 2π3 ) =−12 +√32i,z4= 1(cosπ+isinπ) =−1 + 0i=−1,z5= 1(cos 4π3+isin 4π3 ) =−12−√32i,z6= 1(cos 5π3+isin 5π3 ) = 12−√32i.21.|z|6(cos 6θ+isin 6θ) = 64(−1 + 0i) so|z|= 2 and cos 6θ=−1 and sin 6θ= 0. Thus 6θ=π+n(2π)soθ=π6+n2π6which yields valuesπ6,π2,5π6,7π6,3π2and11π6less than 2π. The solutions arez1=2(cosπ6 +isinπ6 ) = 2(√32+ 12i) =√3 +i,z2=2(cosπ2 +isinπ2 ) = 2(0 +i) = 2i,z3=2(cos 5π6+isin 5π6 ) = 2(−√32+ 12i) =−√3 +i,z4=2(cos 7π6+isin 7π6 ) = 2(−√32−12i) =−√3−i,z5=2(cos 3π2+isin 3π2 ) = 2(0−i) =−2i,z6=2(cos 11π6+isin 11π6) = 2(√32−12i) =√3−i.22.10 + 16 = 26>17,so 10 +1716 = 26−17 = 9.23.8 + 6 = 14>10,so 8 +106 = 14−10 = 4.24.20.5 + 19.3 = 39.8>25,so 20.5 +2519.3 = 39.8−25 = 14.8.25.12+78=118>1,so12+1 78=118−1 =38.26.3π4+3π2=9π4>2π,so3π4+2π3π2=9π4−2π=π4.

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61. Introduction and Examples27.2√2 + 3√2 = 5√2>√32 = 4√2,so 2√2 +√323√2 = 5√2−4√2 =√2.28.8 is not inR6because 8>6,and we have only defineda+6bfora, b∈R6.29.We need to havex+ 7 = 15 + 3, sox= 11 will work.It is easily checked that there is no othersolution.30.We need to havex+3π2= 2π+3π4=11π4,sox=5π4will work.It is easy to see there is no othersolution.31.We need to havex+x= 7 + 3 = 10,sox= 5 will work.It is easy to see that there is no othersolution.32.We need to havex+x+x= 7 + 5, sox= 4 will work. Checking the other possibilities 0, 1, 2, 3, 5,and 6, we see that this is the only solution.33.An obvious solution isx= 1. Otherwise, we need to havex+x= 12 + 2,sox= 7 will work also.Checking the other ten elements, inZ12, we see that these are the only solutions.34.Checking the elements 0,1,2,3∈Z4, we find that they are all solutions. For example, 3+43+43+43 =(3 +43) +4(3 +43) = 2 +42 = 0.35.ζ0↔0,ζ3=ζ2ζ↔2 +85 = 7,ζ4=ζ2ζ2↔2 +82 = 4,ζ5=ζ4ζ↔4 +85 = 1,ζ6=ζ3ζ3↔7 +87 = 6,ζ7=ζ3ζ4↔7 +84 = 336.ζ0↔0,ζ2=ζζ↔4 +74 = 1,ζ3=ζ2ζ↔1 +74 = 5,ζ4=ζ2ζ2↔1 +71 = 2,ζ5=ζ3ζ2↔5 +71 = 6,ζ6=ζ3ζ3↔5 +75 = 337.If there were an isomorphism such thatζ↔4,then we would haveζ2↔4 +64 = 2 andζ4=ζ2ζ2↔2 +62 = 4 again, contradicting the fact that an isomorphism↔must give aone-to-one correpondence.38.By Euler’s fomula,eiaeib=ei(a+b)= cos(a+b) +isin(a+b). Also by Euler’s formula,eiaeib=(cosa+isina)(cosb+isinb)=(cosacosb−sinasinb) +i(sinacosb+ cosasinb).The desired formulas follow at once.39.(See the text answer.)40. a.We havee3θ= cos 3θ+isin 3θ. On the other hand,e3θ=(eθ)3= (cosθ+isinθ)3=cos3θ+ 3icos2θsinθ−3 cosθsin2θ−isin3θ=(cos3θ−3 cosθsin2θ) +i(3 cos2θsinθ−sin3θ).Comparing these two expressions, we see thatcos 3θ= cos3θ−3 cosθsin2θ.b.From Part(a), we obtaincos 3θ= cos3θ−3(cosθ)(1−cos2θ) = 4 cos3θ−3 cosθ.

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2. Binary Operations72. 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 be associative, but we can’ttell without checking all other triple products.3.(b∗d)∗c=e∗c=aandb∗(d∗c) =b∗b=c, so the operation is not associative.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 theassociative property, we find thatd∗b=c, d∗c=c,andd∗d=d.7.It is not commutative because 1−26= 2−1. It is not associative because 2 = 1−(2−3)6= (1−2)−3 =−4.8.It is commutative becauseab+ 1 =ba+ 1 for alla, b∈Q. It is not associative because (a∗b)∗c=(ab+ 1)∗c=abc+c+ 1 buta∗(b∗c) =a∗(bc+ 1) =abc+a+ 1, and we need not havea=c.9.It is commutative becauseab/2 =ba/2 for alla, b∈Q. It is associative becausea∗(b∗c) =a∗(bc/2) =[a(bc/2)]/2 =abc/4, and (a∗b)∗c= (ab/2)∗c= [(ab/2)c]/2 =abc/4 also.10.It is commutative because 2ab= 2bafor alla, b∈Z+. It is not associative because (a∗b)∗c= 2ab∗c=2(2ab)c, buta∗(b∗c) =a∗2bc= 2a(2bc).11.It is not commutative because 2∗3 = 23= 86= 9 = 32= 3∗2.It is not associative becausea∗(b∗c) =a∗bc=a(bc), but (a∗b)∗c=ab∗c= (ab)c=abc, andbc6=bcfor someb, c∈Z+.12.IfShas just one element, there is only one possible binary operation onS; the table must be filled inwith that single element. IfShas two elements, there are 16 possible operations, for there are fourplaces to fill in a table, and each may be filled in two ways, and 2·2·2·2 = 16. There are 19,683operations 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 therearen(n2)possible tables.13.A commutative binary operation on a set withnelements is completely determined by the elementson or above themain diagonalin its table, which runs from the upper left corner to the lower rightcorner. The number of such places to fill in isn+n2−n2=n2+n2.Thus there aren(n2+n)/2possible commutative binary operations on ann-element set. Forn= 2, weobtain 23= 8, and forn= 3 we obtain 36= 729.14.It is incorrect. Mention should be made of the underlying set for∗and the universal quantifier,forall, should appear.A binary operation∗on a setSiscommutativeif and only ifa∗b=b∗afor alla, b∈S.

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82. Binary Operations15.The definition is correct.16.It is incorrect. Replace the finalSbyH.17.It is not a binary operation. Condition 2 is violated, for 1∗1 = 0 and 0/∈Z+.18.This does define a binary operation.19.This does define a binary operation.20.This does define a binary operation.21.It is not a binary operation. Condition 1 is violated, for 2∗3 might be any integer greater than 9.22.It is not a binary operation. Condition 2 is violated, for 1∗1 = 0 and 0/∈Z+.23. a.Yes.[a−bba]+[c−ddc]=[a+c−(b+d)b+da+c].b.Yes.[a−bba] [c−ddc]=[ac−bd−(ad+bc)ad+bcac−bd].24.F T F F F T T T T F25.(See the answer in the text.)26.We have (a∗b)∗(c∗d) = (c∗d)∗(a∗b) = (d∗c)∗(a∗b) = [(d∗c)∗a]∗b, where we used commutativityfor the first two steps and associativity for the last.27.The statement is true. Commutativity and associativity assert the equality of certain computations.For a binary operation on a set with just one element, that element is the result of every computationinvolving the operation, so the operation must be commutative and associative.28.∗abababaaThe statement is false.Consider the operation on{a, b}defined by the table.Then(a∗a)∗b=b∗b=abuta∗(a∗b) =a∗a=b.29.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 inRis associative.30.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.31.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.32.It is commutative.Proof:(f·g)(x) =f(x)·g(x) =g(x)·f(x) = (g·f)(x) because multiplication inRis commutative.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·(g·h)](x)because multiplication inRis associative.

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3. Isomorphic Binary Structures934.It is not commutative. Letf(x) =x2andg(x) =x+ 1. Then (f◦g)(3) =f(g(3)) =f(4) = 16 but(g◦f)(3) =g(f(3)) =g(9) = 10.35.It is not true. Let∗be + and let∗′be·and letS=Z. Then 2 + (3·5) = 17 but (2 + 3)·(2 + 5) = 35.36.Leta, b∈H. By definition ofH, we havea∗x=x∗aandb∗x=x∗bfor allx∈S. Using the factthat∗is associative, we then obtain, for allx∈S,(a∗b)∗x=a∗(b∗x) =a∗(x∗b) = (a∗x)∗b= (x∗a)∗b=x∗(a∗b).This shows thata∗bsatisfies the defining criterion for an element ofH, so (a∗b)∈H.37.Leta, b∈H. By definition ofH, we havea∗a=aandb∗b=b. Using, one step at a time, the factthat∗is associative and commutative, we obtain(a∗b)∗(a∗b)=[(a∗b)∗a]∗b= [a∗(b∗a)]∗b= [a∗(a∗b)]∗b=[(a∗a)∗b]∗b= (a∗b)∗b=a∗(b∗b) =a∗b.This show thata∗bsatisfies the defining criterion for an element ofH, so (a∗b)∈H.3. Isomorphic Binary Structures1.i)φmust be one to one.ii)φ[S] must be all ofS′.iii)φ(a∗b) =φ(a)∗′φ(b) for alla, b∈S.2.It is an isomorphism;φis one to one, onto, andφ(n+m) =−(n+m) = (−n) + (−m) =φ(n) +φ(m)for allm, n∈Z.3.It is not an isomorphism;φdoes not mapZontoZ. For example,φ(n)6= 1 for alln∈Z.4.It is not an isomorphism becauseφ(m+n) =m+n+ 1 whileφ(m) +φ(n) =m+ 1 +n+ 1 =m+n+ 2.5.It is an isomorphism;φis one to one, onto, andφ(a+b) =a+b2=a2+b2=φ(a) +φ(b).6.It is not an isomorphism becauseφdoes not mapQontoQ.φ(a)6=−1 for alla∈Q.7.It is an isomorphism becauseφis one to one, onto, andφ(xy) = (xy)3=x3y3=φ(x)φ(y).8.It is not an isomorphism becauseφis not one to one. All the 2×2 matrices where the entries in thesecond row are double the entries above them in the first row are mapped into 0 byφ.9.It is an isomorphism because for 1×1 matrices, [a][b] = [ab], andφ([a]) =asoφjust removes thebrackets.10.It is an isomorphism. For any basea6= 1, the exponential functionf(x) =axmapsRone to one ontoR+, andφis the exponential map witha= 0.5. We haveφ(r+s) = 0.5(r+s)= (0.5r)(0.5s) =φ(r)φ(s).11.It is not an isomorphism becauseφis not one to one;φ(x2) = 2xandφ(x2+ 1) = 2x.12.It is not an isomorphism becauseφis not one to one:φ(sinx) = cos 0 = 1 andφ(x) = 1.13.No, becauseφdoes not mapFontoF. For allf∈F, we see thatφ(f)(0) = 0 so, for example, nofunction is mapped byφintox+ 1.

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103. Isomorphic Binary Structures14.It is an isomorphism. By calculus,φ(f) =f, soφis the identity map which is always an isomorphismof a binary structure with itself.15.It is not an isomorphism becauseφdoes not mapFontoF. Note thatφ(f)(0) = 0·f(0) = 0. Thusthere is no element ofFthat is mapped byφinto the constant function 1.16. a.Forφto be an isomorphism, we must havem∗n=φ(m−1)∗φ(n−1) =φ((m−1) + (n−1)) =φ(m+n−2) =m+n−1.The identity element isφ(0) = 1.b.Using the fact thatφ−1must also be an isomorphism, we must havem∗n=φ−1(m+ 1)∗φ−1(n+ 1) =φ−1((m+ 1) + (n+ 1)) =φ−1(m+n+ 2) =m+n+ 1.The identity element isφ−1(0) =−1.17. a.Forφto be an isomorphism, we must havem∗n=φ(m−1)∗φ(n−1) =φ((m−1)·(n−1)) =φ(mn−m−n+ 1) =mn−m−n+ 2.The identity element isφ(1) = 2.b.Using the fact thatφ−1must also be an isomorphism, we must havem∗n=φ−1(m+ 1)∗φ−1(n+ 1) =φ−1((m+ 1)·(n+ 1)) =φ−1(mn+m+n+ 1) =mn+m+n.The identity element isφ−1(1) = 0.18. a.Forφto be an isomorphism, we must havea∗b=φ(a+ 13)∗φ(b+ 13)=φ(a+ 13+b+ 13)=φ(a+b+ 23)=a+b+ 1.The identity element isφ(0) =−1.b.Using the fact thatφ−1must also be an isomorphism, we must havea∗b=φ−1(3a−1)∗φ−1(3b−1) =φ−1((3a−1) + (3b−1)) =φ−1(3a+ 3b−2) =a+b−13.The identity element isφ−1(0) = 1/3.19. a.Forφto be an isomorphism, we must havea∗b=φ(a+ 13)∗φ(b+ 13)=φ(a+ 13·b+ 13)=φ(ab+a+b+ 19)=ab+a+b−23.The identity element isφ(1) = 2.b.Using the fact thatφ−1must also be an isomorphism, we must havea∗b=φ−1(3a−1)·φ−1(3b−1) =φ−1((3a−1)·(3b−1)) =φ−1(9ab−3a−3b+ 1) = 3ab−a−b+ 23.The identity element isφ−1(1) = 2/3.

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3. Isomorphic Binary Structures1120.Computingφ(x∗y) is done by first executing the binary operation∗, and then performing the mapφ.Computingφ(x)∗′φ(y) is done by first performing the mapφ, and then executing the binaryoperation∗′. Thus, reading in left to right order of peformance, the isomorphism property is(binary operation)(map) = (map)(binary operation)which has the formal appearance of commutativity.21.The definition is incorrect. It should be stated that〈S,∗〉and〈S′,∗′〉are binary structures,φmust beone to one and ontoS′, and the universal quantifier “for alla, b∈S” should appear in an appropriateplace.Let〈S,∗〉and〈S′,∗′〉be binary structures. A mapφ:S→S′is anisomorphismif and onlyifφis one to one and ontoS′, andφ(a∗b) =φ(a)∗′φ(b) for alla, b∈S.22.It is badly worded. The “for alls∈S” applies to the equation and not to the “is an identity for∗”.Let∗be a binary operation on a setS. An elementeofSis anidentity elementfor∗if andonly ifs∗e=e∗s=sfor alls∈S.23.Suppose thateandeare two identity elements and, viewing each in turn as an identity element,computee∗ein two ways.24. a.Let∗be a binary operation on a setS. An elementeLofSis aleft identity elementfor∗ifand only ifeL∗s=sfor alls∈S.b.Let∗be a binary operation on a setS. An elementeRofSis aright identity elementfor∗ifand only ifs∗eR=sfor alls∈S.A one-sided identity element is not unique. Let∗be defined onSbya∗b=afor alla, b∈S.Then everyb∈Sis a right identity. Similarly, a left identity is not unique. If in the proof of Theorem3.13, we replaceebyeLandebyeLeverywhere, and replace the word “identity” by “left identity”,the first incorrect statement would be, “However, regardingeLas left identity element, we must haveeL∗eL=eL.”25.No, if〈S∗〉has a left identity elementeLand a right identity elementeR, theneL=eR.ProofBecauseeLis a left identity element we haveeL∗eR=eR, but viewingeRas right identityelement,eL∗eR=eL. ThuseL=eR.26.One-to-one:Suppose thatφ−1(a′) =φ−1(b′) fora′, b′∈S′. Thena′=φ(φ−1(a′)) =φ(φ−1(b′)) =b′,soφ−1is one to one.Onto:Leta∈S. Thenφ−1(φ(a)) =a, soφ−1mapsS′ontoS.Homomorphism property:Leta′, b′∈S′. Nowφ(φ−1(a′∗′b′)) =a′∗′b′.Becauseφis an isomorphism,φ(φ−1(a′)∗φ−1(b ,′)) =φ(φ−1(a′))∗′φ(φ−1(b′)) =a′∗′b′

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123. Isomorphic Binary Structuresalso. Becauseφis one to one, we conclude thatφ−1(a′∗b′) =φ−1(a′)∗′φ−1(b′).27.One-to-one:Leta, b∈Sand suppose (ψ◦φ)(a) = (ψ◦φ)(b). Thenψ(φ(a)) =ψ(φ(b)). Becauseψisone to one, we conclude thatφ(a) =φ(b). Becauseφis one to one, we must havea=b.Onto:Leta′′∈S′′. BecauseψmapsS′ontoS′′, there existsa′∈S′such thatψ(a′) =a′′. BecauseφmapsSontoS′, there existsa∈Ssuch thatφ(a) =a′. Then (ψ◦φ)(a) =ψ(φ(a)) =ψ(a′) =a′′,soψ◦φmapsSontoS′′.Homomorphism property:Leta, b∈S. Sinceφandψare isomorphisms, (ψ◦φ)(a∗b) =ψ(φ(a∗b)) =ψ(φ(a)∗′φ(b)) =ψ(φ(a))∗′′ψ(φ(b)) = (ψ◦φ)(a)∗′′(ψ◦φ)(b).28.Let〈S,∗〉,〈S′,∗′〉and〈S′′,∗′′〉be binary structures.Reflexive:Letι:S→Sbe the identity map. ThenιmapsSone to one ontoSand fora, b∈S, wehaveι(a∗b) =a∗b=ι(a)∗ι(b), soιis an isomorphism ofSwith itself, that isS'S.Symmetric:IfS'S′andφ:S→S′is an isomorphism, then by Exercise 26,φ−1:S′→Sis anisomorphism, soS′'S.Transitive:Suppose thatS'S′andS′'S′′, and thatφ:S→S′andψ:S′→S′′are isomorphisms.By Exercise 27, we know thatψ◦φ:S→S′′is an isomorphism, soS'S′′.29.Let〈S,∗〉and〈S′,∗′〉be isomorphic binary structures and letφ:S→S′be an isomorphism. Supposethat∗is commutative. Leta′, b′∈S′and leta, b∈Sbe such thatφ(a) =a′andφ(b) =b′. Thena′∗′b′=φ(a)∗′φ(b) =φ(a∗b) =φ(b∗a) =φ(b)∗′φ(a) =b′∗a′, showing that∗′is commutative.30.Let〈S,∗〉and〈S′,∗′〉be isomorphic binary structures and letφ:S→S′be an isomorphism. Supposethat∗is associative.Leta′, b′, c′∈S′and leta, b, c∈Sbe such thatφ(a) =a′, φ(b) =b′andφ(c) =c′. Then(a′∗′b′)∗′c′=(φ(a)∗′φ(b))∗′φ(c) =φ(a∗b)∗′φ(c) =φ((a∗b)∗c))=φ(a∗(b∗c)) =φ(a)∗′φ(b∗c) =φ(a)∗′(φ(b)∗′φ(c)) =a′∗′(b′∗′c′),showing that∗′is associative.31.Let〈S,∗〉and〈S′,∗′〉be isomorphic binary structures and letφ:S→S′be an isomorphism. SupposethatShas the property that for eachc∈Sthere existsx∈Ssuch thatx∗x=c.Letc′∈S′,and letc∈Ssuch thatφ(c) =c′. Findx∈Ssuch thatx∗x=c. Thenφ(x∗x) =φ(c) =c′, soφ(x)∗′φ(x) =c′.If we denoteφ(x) byx′, then we see thatx′∗x′=c′, soS′has the analagousproperty.32.Let〈S,∗〉and〈S′,∗′〉be isomorphic binary structures and letφ:S→S′be an isomorphism. SupposethatShas the property that there existsb∈Ssuch thatb∗b=b. Letb′=φ(b). Thenb′∗′b′=φ(b)∗′φ(b) =φ(b∗b) =φ(b) =b′, soS′has the analogous property.33.Letφ:C→Hbe defined byφ(a+bi) =[a−bba]fora, b∈R. Clearlyφis one to one and ontoH.

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4. Groups13a.We haveφ((a+bi) + (c+di)) =φ((a+c) + (b+d)i) =[a+c−(b+d)b+da+c]=[a−bba]+[c−ddc]=φ(a+bi) +φ(c+di).b.We haveφ((a+bi)·(c+di)) =φ((ac−bd) + (ad+bc)i) =[ac−bd−(ad+bc)ad+bcac−bd]=[a−bba]·[c−ddc]=φ(a+bi)·φ(c+di).34.Let the set be{a, b}. We need to decide whether interchanging the names of the letters everywherein the table and then writing the table again in the orderafirst andbsecond gives the same tableor a different table.The same table is obtained if and only if in the body of the table, diagonallyopposite entries are different. Four such tables exist, since there are four possible choices for the firstrow; Namely, the tables∗abaaabbb∗abaabbab∗abababbaand∗ababbbaa.The other 12 tables can be paired off into tables giving the same algebraic structure.One table ofeach pair is listed below. The number of different algebraic structures is therefore 4 + 12/2 = 10.∗abaaabaa∗abaaabab∗abaaabba∗abaabbaa∗abababaa∗abaabbba4. Groups1.No.G3fails.2.Yes3.No.G1fails.4.No.G3fails.5.No.G1fails.6.No.G2fails.7.The group〈U1000,·〉of solutions ofz1000= 1 inCunder multiplication has 1000 elements and isabelian.8.·81357113573317555713775319.Denoting the operation in each of the three groups by∗and the identity element byefor the moment,the equationx∗x∗x∗x=ehas four solutions in〈U,·〉, one solution in〈R,+〉, and two solutions in〈R∗,·〉.10. a.Closure:Letnrandnsbe two elements ofnZ. Nownr+ns=n(r+s)∈nZsonZis closed underaddition.Associative:We know that addition of integers is associative.

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