>>319
つづき

https://math.stackexchange.com/questions/428295/for-g-group-and-h-subgroup-of-finite-index-prove-that-n-subset-h-normal
For G group and H subgroup of finite index, prove that N⊂H normal subgroup of G of finite index exists

Let G be a group and H be a subgroup of G with finite index. I want to show that there exists a normal subgroup N of G with finite index and N⊂H. The hint for this exercise is to find a homomorphism G→Sn for n:=[G:H] with kernel contained in H.

The standard solution suggests to choose φ as the homomorphism induced by left-multiplication φ:G→S(G/H)?Sn. I'm not 100% sure if I understand this correctly. What exactly does φ do? We take g∈G and send it to a bijection φg:G/H→G/H,xH?gxH? If so, how can I see that its kernel is contained in H? Also, the standard solution claims its image is isomorphic to G/N and thus N has a finite index in G, how can I see that the image is isomorphic to G/N?
Thanks in advance for any help.

asked Jun 24 '13 at 13:53
Huy

2 Answers
1
I don't know it's true or false but I try this as this
H is a subgroup of G. (G:H)=n, we can get atleast one normal subgroup N⊆H.
Let ［G:H］={g1,g2,...,gn}
Now we define a mapping f:G→Sn such that f(a)=gi where a∈gi,N⊆giH Clearly mapping is well defined.
Let f(b)=gj where gj∈gj,N⊆gjH.
Now a∈giN and b∈gjN therefore ab∈gi.gjN⊆gigjH Therefore f(ab)=gi.gj=f(a)f(b), f is homomorphism.
Let x∈kerf .Then x∈N⊆H i.e kerf=N⊆H

answered Nov 3 '17 at 17:19
Manldipa Sarkar

https://ja.wikipedia.org/wiki/%E5%8D%98%E7%B4%94%E7%BE%A4
1.2 無限単純群
無限交代群A_∞、つまり整数全体の偶置換の群は単純群である。この群は有限群A_nの(標準埋め込み A_n → A_n+1 に関する)単調増加列の合併として定義できる。
(引用終り)
以上