# Finitely generated group has only finitely many subgroups of given index

In my previous question i have got a comment that :

If a group is finitely generated, then there are finitely many subgroups of a given finite index.

I do not yet see the beauty of this problem but, i wanted to prove this atleast.

So,what i have tried is :

I fix $n\in \mathbb{N}$ and assume $H\leq G$ with $|G/H|=n$

As we have $H\leq G$ of finite index, we have the action of $G$ on left cosets of $H$ and

we get $\eta: G\rightarrow S_n$

I assume $\{g_1,g_2,\dots, g_r\}$ generate $G$

what i am thinking is once $H$ is fixed, the homomorphism is fixed $\eta$ (after all it is left multiplication of cosets of H). (otherway may not be true.. i dont see it immediately)

and any homomorphism of $G$ is fixed by all these $g_i :1\leq i\leq r$

each $g_i$ have $n!$ possibilities as its image.

So all $g_i : 1\leq i\leq r$ would have $n!n!\dots n!$ repeating $r$ times i.e., $(n!)^r$ times..

and as these homomorphism (i am not sure but somethings may not correspond to H) are finite.

Thus , I see that No.of $H$ should be atmost $(n!)^r$ (after excludind some useless cases)

So, I would like to say that

If a group is finitely generated, then there are finitely many subgroups of a given finite index.

I am very much sure that there are some gaps which i am unable to see..

please help e to fill this in detail (if this way is partially true)by giving some hints or please suggest me another approach (if this is a blunder).

Thank you

-
I don't see any problem with your reasoning. You prove that there are only finitely many actions of $G$ on a finite set, and a subgroup is equivalent to a transitive action with a chosen element of the set ($H$ is then the stabilizer of the element) – user8268 Sep 15 '13 at 14:19
@user8268 : i am not able to understand what you have meant by "You prove that there are only finitely many actions of G on a finite set, and a subgroup is equivalent to a transitive action with a chosen element of the set (H is then the stabilizer of the element)" .......finitely many actions on a finite set?? as my set is fixed, it will e unique.. i am calculating "how many sets".... Subgroup equivalent to transitive action???.... I would really like to know what exactly do you mean by that... AM i missing something?? – Praphulla Koushik Sep 16 '13 at 1:42
Yes, the claim which you are proving is a theorem by M. Hall and its proof from “Group theory” by A. G. Kurosh is the same as yours. Since the number of different homomorphisms from $G$ to $S_n$ is finite and any such homomorhism $\eta$ uniquely determines the respective subgroup $H$ (it seems because $g\in H$ iff $\eta(g)(1)=(1)$). So you can write your reasoning as an answer, accept it, and do not disturb MSE users. :-D – Alex Ravsky Sep 26 '13 at 5:29