近世代数答案(一、二章)

Chapter 11、proof Let A,B,C be sets .Suppose that x ∈B,we get x ∈A ∩B orx A B A ∈- ,andx A C∈ or x A C A ∈- sinceA B A C= andA B A C= .so x ∈C andB C⊆.Similarly ,we haveC B⊆and soB=C .2、proof ① First,consider()x A B A ∈- .Then x A ∈ or x B ∈,butx A ∉.This implies if x is not an element of A ,then x B ∈.Hence x A B∈ and ()A B A - ⊆A B .Conversely, ifx A B∈ ,then by definition , x A ∈ or x B ∈.This generates two cases: (a1) If x A ∈,clearly()x A B A ∈- ;(b2) Ifx B ∈,then either x A ∈ or not . i.e.,either x B ∈andx A ∈or x B ∈but x A ∉, in either case, we have()x A B A ∈- .Hence A B ⊆()A B A - .Therefore()A B A - =A B . ② Suppose that()x A B C ∈- .Then x A ∈but x B ∉ and x C∉.Sox A ∈-B and x A C ∈- and ()()x A B A C ∈--bydefinition .Hence()A B C - ⊆()()A B A C -- .Converssely, Assume that()()x A B A C ∈-- ,then x A ∈-Bandx A C∈-,and we have ,x A ∈but x B ∉ andx C ∉.Hence x B C ∉ , x A ∈, i.e., ()x A B C ∈- .Therefore ()()A B A C -- ⊆()A B C - and,so ()A B C - =()()A B A C --3.(a) surjective (b) bijective (c) bijective4.proof if f: X →Y and g: Y →Z are functions,then their composite denoted by g ︒f, is the function X →Z given by g ︒f: X →g(f(x))(i) suppose that (g ︒f)(a)= (g ︒f)(b), where a,b ∈X. we haveg(f(a))=g(f(b)) by definition, and f(a)=f(b) since g is injective, similarly, a=b for f is injective. Therefore, g ︒f is injective. (ii) For each Z ∈Z, there is y ∈Y with g(y)=z since g is surjective,and for each y ∈ Y , there exists a ∈ x with f(a)=y since f is surjective. So for∀z ∈Z, there is a ∈ x with(g ︒f)(a)=g(f(a))=g(y)=z. which implies g ︒f is surgective.5. proof clearly,α:R →R is a function. Suppose thatα(a)=α(b)where a, b ∈R are distinct. Then 332211aba b =++, cross multiplyingyields332323a ab b a b+=+, which simplifies to33a b= and hence a b =,so α is injective. for ∀given y ∈R,321xyx =+from,we getequation 320x yx y --=, which can be solved for x, i.e .for each y ∈R,there is at least x ∈x such that 321xyx =+.whic impliesαissurjective. Thereforeαis bijective.6、(a) R is reflexive, symmetric, transitive. (b) R is reflexive, not symmetric, transitive. (c ) R is reflexive, symmetric, transitive. (d) R is reflexive, symmetric, transitive.7、proof (1) For every a ∈R-{0}，we have 20aa a =>, and so,aR a(2) If aRb , where ,{0}a b R ∈-, i.e.,0,ab > then0ba >, i.e., bRa ,(3) If ,aR b bR c , where ,,a b c ∈{0}R -, i.e.0,ab >0bc >, then 0ac >.i.e.,aRc .Therefore, therelation ～ is an equivalence relation .8、 There are 1,3,5,15 equivalence relations on a set S with 1,2,3 or 4elements, separately.9、 We can list the elements of the residue classes of modulo 3: [0]={…,-9,-6,-3,0,6,9,…} [1]={…,-8,-5,-2,1,4,7,10,…} [2]={…，-7,-4,-1,2,5,8,11,…}Chapter 21、proof i)ii ）For every x,y,z ∈G ,(x*y)*z=(xy-x-y+2)*z=(xy-x-y+2)z-z-(xy-x-y+2)+2=xyz-xz-yz+z-xy+x+yx*(y*z)=x*(yz-y-z+2)=x(yz-y-z+2)-x-(yz-y-z+2)+2=xyz-xy-xz+x-yz+y +zwe have (x*y)*z=x*(y*z). And so the associative law holds.3、Solution Straightforward calculation shows that 46A IB ==.()nAB I ≠, since 1()01nn AB I -⎛⎫=≠⎪⎝⎭(0)n ≠.4、proof Suppose222()ab a b = for all ,a b G ∈，then 2()ab =()()ab ab =22a b =()()aa bb ,i.e., abab aabb =. Applying left cancellation , this becomesbab abb =,and by right cancellation, this reduces to ba ab =.5、proof For every a G ∈, there is a ,1a G -∈ such that 1a a e -=(identity)So 11()naba aba--=1aba- (1)aba-1aba-=1n ab a-。

6、proof Firstly ,the identity element belongs to N since it is equal tone e =,(e is an identity of G ). The product of any two elements of N is an element of N , since G is abelianAnd ()n n na b ab =.Also the inverse of an element of N is an element of N .Since 11()()nna a --=For some positive integer. Thus N is a subgroup of G .7、Proof Clearly ，10cos 0sin 001sin 0cos 0e H -⎡⎤⎡⎤==∈⎢⎥⎢⎥⎣⎦⎣⎦. For every 1111cos sin sin cos θθθθ-⎡⎤⎢⎥⎣⎦, 2222cos sin sin cos θθθθ-⎡⎤⎢⎥⎣⎦H∈,we have1111cos sin sin cos θθθθ-⎡⎤⎢⎥⎣⎦2222cos sin sin cos θθθθ-⎡⎤⎢⎥⎣⎦= 12121212cos()sin()sin()cos()θθθθθθθθ+-+⎡⎤⎢⎥++⎣⎦H ∈ ,and the inverse ofcos sin sin cos θθθθ-⎡⎤⎢⎥⎣⎦is cos()sin()sin()cos()θθθθ---⎡⎤⎢⎥--⎣⎦H ∈Thus H is a subgroup of G .8、Proof Firstly ,the identity element1001e ⎡⎤=⎢⎥⎣⎦∈K For every 2ab ba ⎡⎤⎢⎥-⎣⎦,2cd d c ⎡⎤⎢⎥-⎣⎦∈k,2222()2ab c d ac bd ad bc k b a d c ad bc ac bd -+⎡⎤⎡⎤⎡⎤=∈⎢⎥⎢⎥⎢⎥---+-⎣⎦⎣⎦⎣⎦, And the inverse of 2a b b a ⎡⎤⎢⎥-⎣⎦ is 22122a b k ba a b⎡⎤∈⎢⎥+⎣⎦, so k is a subgroup of 2()G L R .9. Hint: Theorem 2.4.6.10 .Proof For every a ∈G , we have ax=xa since G is abelian. For a subgroup H, naturally, Ha=aH and H is normal.11.o(0)=1, (1)6,(2)3,(3)2,(4)3,(5) 6.o o o o o =====12. Solution AU positive divisors of 6 are 1,2,3,6, so the subgroup of6Z are :<1>=6Z ={0,1,2,3,4.5}, <2>=26Z ={0,2,4}, <3>=36Z ={0,3}, <6>=66Z ={0}.13. Yes. 14. need not. 15. H={mz/z ∈Z},where m is integer . 16.Answers may vary.17. proof suppose that o(ab)=r, then ()rnrnrnrna ab ab e === since o(b)=n, and we have m/r n. From gcd(m,n)=1, so m/r. Similary n/r, and mn/r since gcd(m,n)=1. On the other hand,()mnmn mnab a be e e ==⋅= and so r/mn ,Hence r=mn.19、(14)(265),odd. 20、5！ 21、3023、Proof firstly, we will prove H a =H iffa ∈H. If H a =H, thena =e a ∈Ha=H .i.e. a ∈H. Conversely , If a ∈H,H is a subgroup of thegroup G , then1a-∈H and H a ⊆H ⋅H=H. Also H=H a 1a -=(H a )1a-⊆H a .Suppose .So H a =H. Hence H a =H iff a ∈H.Suppose that H a =H a , then H a 1b-=Hb 1b -=H. anda 1b-∈H. from theabove proof .Ifa 1b-∈H, then H a 1b-=H and so H a =Hb.24、Proof First, a 1a -=e ∈H, then a ～a .If a ～b ,then a 1b-∈H and H a = Hb by 23.So Hb 1a -=H andb 1a-∈H.we get b ～a .If a ～b, b ～c, thena 1b-∈H, b 1c-∈H, anda 1c-=(a1b-)b 1c-∈H, soa ～c. therefore a ～biffa 1b-∈H defines an equivalence relation.26、proof Firstly Let 1:aH Haϕ-→.IfaH bH=,then1a b H-∈ by exeraise23,and11111()()HaHa bb H a b b-----==, this shows that ϕ is a mapping fromaHto1H a-.Secondly , LetaH ,/bH G H∈,if ()()aH bH ϕϕ=, i.e.,1H a-=1H b -, then1b a -H∈, and sobH=aH by exercivse 23. ϕ is injective.For∀1H a -∈\H G, we haveaH/G H∈ and ()aH ϕ=1H a -. So ϕ is surjective from/G Hto \H G. Where {}/|G H gH g G =∈,}{\|H G Hg g G =∈. 28. proof For a H∀∈, ifa ∈H, Then aH H Ha ==. Ifa H∉, thenaHandHare distinct cosets., and G H aH= since]:2G H ⎡=⎣, similarly , G H Ha= , sinceaH H =∅, we haveaH G H Ha⊆= andaH Ha⊆. Similarly ,H a⊆aH. So aH=H a , and H is a normal subgroup of G .29. Hint : computation by the definition of left coset and right coset . 32. proof Let :a f G G→1x axa-→Clearly , af is a mapping from G to G .For x G∀∈. There exist 1a xa -G∈ such that11()()a f a xa a a xa x--== andsoaf is surjective .(3)If x1a-=ay1a-,where x、y∈G, then x=y by the cancellation law, and fa is injective.(4)For∀x、y∈G, we have a f (xy)=axy1a-=ax(1a-a)y1a-=(ax1a-)(ay1a-)=a f(x)a f(y) so a f is a homomorphism from G to G.Therefore fa is an automorphism of G.33、Proof (1)For∀x∈G, we have ex= xe, and e∈Z(G).So Z(G)≠φ. (2)If a,b∈Z(G),then for∀x∈G,(ab0x=a(bx)=(ax)b=(xa)b=x(ab).and so ab∈Z(G).(3)If a∈Z(G),then ax=xa for ∀x∈G and 1a-(ax) 1a-,i.e. , x1a-=1a-x, so1a-∈Z(G).Hence Z(G) is an abelian subgroup of group G., and Z(G) is a normal subgroup of G Since a Z(G)={x∈Z(G)}={xa | x∈Z(G)} Z(G)a, ∀a∈G.36、Proof (1)For ∀1010001a b⎛⎫⎪⎪⎪⎝⎭,1010001c d⎛⎫⎪⎪⎪⎝⎭∈G,then1010001a b⎛⎫⎪⎪⎪⎝⎭1010001c d⎛⎫⎪⎪⎪⎝⎭=1010001a cb d++⎛⎫⎪⎪⎪⎝⎭∈G . G is closed.(2)Clearly,the associative law holds.(3)The identity element is100 010 001⎛⎫ ⎪ ⎪ ⎪⎝⎭.(4)For ∀1010001a bG⎛⎫⎪∈⎪⎪⎝⎭, we get11010001a b-⎛⎫⎪⎪⎪⎝⎭1010001a b--⎛⎫⎪=⎪⎪⎝⎭, so G is a subgroup ofG L R(3,)。