純粋・応用数学（含むガロア理論）6

1 ：１３２人目の素数さん：2020/12/12(土) 11:50:07.88 ID:+J6pglya.net

テンプレは後で

319 ：現代数学の系譜 雑談 ：2021/01/02(土) 00:14:00.39 ID:k00K5jWz.net

>>318
つづき

https://mathoverflow.net/questions/2630/infinite-subgroups-with-finite-index
mathoverflow
Infinite subgroups with finite index
Is there a general method to prove that an infinite subgroup of a group has finite index? Or, in other words, to prove that the quotient group is finite? I am particularly interested in classical groups, such as GL(n), SL(n), etc, over a nonarchimedean local field but I am looking for a general method, if there exists one.
asked Oct 26 '09 at 15:53
Sergio Mendes

7 Answers
14
This is a somewhat tautological answer, but: if you can show that the subgroup contains the kernel of a finite representation (i.e. a homomorphism to a finite group), you're done. Intuitively: "I only need a finite number of things to go my way in order to belong to this subgroup."
If the group (or some representation of that group) is compact in some topology, and the subgroup contains the connected component of the identity (or an open neighbourhood of the identity), you're also done.
answered Oct 26 '09 at 16:08
Terry Tao

For this type of problem, "general methods" tend to be quite general indeed, but here are two ideas:
1) Find an action of G on a finite set X and an element x∈X such that H is the stabilizer of x. Then, by the Orbit-Stabilizer Theorem, G/H is isomorphic to the orbit space Gx, so is finite.
2) Find a finite group X and a homomorphism f:G→X such that H contains the kernel of f. Then f:G/ker(f)?X, so ker(f) has finite index, so H, which contains ker(f), has finite index.
Note that both of these will, in principle, always work. In Case 1, take X=G/H. In Case 2, let H′=?g∈GgHg?1 be the normal core of H. It is easy to show that (since H has finite index), H′ is a finite index normal subgroup of G which is contained in H. Take X=G/H′ and f to be the quotient map.
answered Oct 26 '09 at 16:15
Pete L. Clark

つづく

320 ：現代数学の系譜 雑談 ：2021/01/02(土) 00:14:23.80 ID:k00K5jWz.net

>>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