抽象代数 孟道骥版 习题解答 第四章

Chapter4

4.1 Ӯ

1. G 4. G 4 Klein K4 .

đ Ϝ4 S4 . .

(i)G 4 đ G 4 .

(ii)G 4 đ ∀a∈G,a2=e.Ֆ ∀a,b∈G,(ab)2= e,, ab=(ab)−1=b−1a−1=ba, G Abel đ G∼=K4.

2. G 6. G 6 S3 .

G с 3 đ Ԣ 2 đ с

Abel đ a=b∈G, a=e,b=e, a,b 4 đ .

G с 2 đ Ԣ 3 đ |G| đ .

G 2 a, 3 b.

1):a,b đ ab 6 đՖ G= ab 6 .

2):a,b҂ đ G 6 . G k 3 đj 2 đ

2k+j+1=6, (k,j)=(2,1) (1,3). k=2, G

3 {x,x−1,y,y−1}. xy҂ 3 đ xy

2 đ yx đ xy=yx, x,y 9 đ

. (k,j)=(1,3). թ G S6 ֆ ϕ,҂ ϕ(b)= (1,2,3), ϕ(a)=σ. G 3 đ σ(1,2,3)σ−1= (σ(1),σ(2),σ(3)), {σ(1),σ(2),σ(3)}={1,2,3}. σ (1,2,3)҂ đ ҂ σ(1)=1,σ(2)=2,σ(3)=3,҂ σ(1)=1,σ(2)=

3,σ(3)=2.

α=

456

σ(4)σ(5)σ(6)

σ=(2,3)α, σ2=e, α2=e.

σ,(1,2,3) ={(1,2,3),(1,3,2),e,(2,3)α,(1,2)α,(3,1)α} S3

64

65

G ∼=S 3.

3. G r =st đH G t .

H ={g s |g ∈G }={h ∈G |h

=e }.

G = g 0 , {g s |g ∈G }={g s 0,g 2s 0,···,g ts 0},{h ∈G |h t =e }={g s 0,g 2s 0,···,g ts 0}, {g s 0,g 2s 0,···,g ts 0} G t đ G t . G ={g s |g ∈G }={h ∈G |h t =e }

4. G đa,b ∈G.ӫ[a,b ]=aba −1b −1 a,b .

{aba −1b −1|a,b ∈G } Ӯ G (1)ӫ G

. :1) α∈Aut G , α(G (1))=G (1);2) H G. G/H Abel ԉ H ⊇G (1). 1)α(G (1))=α( {aba −1b −1|a,b ∈G } )= {σ(a )σ(b )σ(a )−1σ(b )−1|a,b ∈G } =G (1).2)G/H Abel ⇔(G/H )(1)={e }⇔G (1)⊆H .

5. S G Ӯ đ ϕ,ψ G H đ ϕ(x )=ψ(x ),∀x ∈S. ϕ=ψ. ∀a ∈G , G = S , a =y 1y 2···y n , y i ∈S y −1i ∈S . ϕ(x )=ψ(x ),∀x ∈S , ϕ(x −1)=ψ(x −1),∀x ∈S ,Ֆ ϕ(y i )=ψ(y i ),∀1≤i ≤n , ϕ(a )=ψ(a ), ϕ=ψ.

6. H G đ H =G . G = G −H . H =G ∃a ∈G , GH , aH ∩H =∅, aH ⊇H , G −H ⊃H ∪(G −H )=G , G = G −H .

7. G đ G с 2 . G k m čm >1Ďđ m kϕ(m ) đ ϕ . m đ ϕ(m ) đ Ԣ . |G | đ с đՖ թ 2 .

8. α∈S 3 ҂ . α= 1234567836548271

α= 1234567836548271

=(1358)(26).

66

9. S n= {(12),(23),···,(n−1n)} .

. n=2 đ Ӯ . n=k Ӯ .

n=k+1 đ

{(12),···,(k−1k)} =S k

(1k)(k k+1)(1k)=(1k+1)

{(12),···,(k−1k),(k k+1)} = {(12),···,(1k),(1k+1)} =S k+1

{(12),···,(n−1n)} =S n

n=k+1 Ӯ .Ֆ n∈N Ӯ .

10. (i1i2···i r)−1=(i r i r−1···i1).

(i1···i r)(i r i r−1···i1)=e

11. ∀σ∈S n,

σ(i1i2···i r)σ−1=(σ(i1)σ(i2)···σ(i r)).

j=1,2,···,r−1,(σ(i1i2···i r)σ−1)(σ(i j))=

σ(i1i2···i r)(i j)=σ(i j+1),

j=r,(σ(i1i2···i r)σ−1)(σ(i r))=σ(i1i2···i r)(i r)=σ(i1),

j=r+1,···,n,(σ(i1i2···i r)σ−1)(σ(i j))=σ(i1i2···i r)(i j)=

σ(i j),

∴σ(i1i2···i r)σ−1=(σ(i1)σ(i2)···σ(i r)).

12. G 2k, k k>1. G с 2 .

Cayley ğ

G L G

|G|=2k թ 2 a∈G, ∀x∈G,L a(x)2=x,

L a k đ k đՖ L a đ L G թ

2 đ G .