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

Solution 1:

There is a missing step in your proof, at the line:

"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)"

This question asks how to plug this gap.