Ok, let me start by saying that I have seen this post and the one linked within, but they are quite old and didn't really answer my problem... I have read before that, in such a case, the standard guideline is to create a new question, which is what prompted me to write this inquiry.
The problem in question is as follows:
Given a finitely generated group $G$, prove that, for any $m \in \mathbb{N}$, there are finitely many subgroups of index $m$ in $G$.
My attempt
Let $H \leq G$ such that $[G:H] = m \in \mathbb{N}$, meaning $\mathcal{P}(G/H) \simeq S_m$, where $\mathcal{P}(G/H)$ denotes the permutations of the set $G/H$.
Let's define a homomorphism $$\begin{align}\varphi_H: G \to S_m \\ a \mapsto p_{a, H}\end{align}$$ where $$\begin{align}p_{a,H}: G/H \to G/H \\ gH \mapsto (ag)H\end{align}$$
Now, let's consider the sets $A = \{H \leq G \mid [G:H] = m\}$ and $\{\varphi_H \mid H \in A\}$. We can then define a mapping $f$ in between them, such that $f(H) = \varphi_H$. If $\varphi_H = \varphi_L$, then their kernels are the same. But it is quite simple to check that the kernel of $\varphi_H$ is $H$, meaning $f$ is an injection.
This, in turn, implies that the number of subgroups of index $m$ in $G$ is bounded by the number of $\varphi_H$, which is bounded by the number of homomorphisms from $G$ to $S_m$. Since $G$ is finitely generated, this is at most $(m!)^r$, where $r$ denotes the number of generators of $G$. Hence, the number of subgroups of index $m$ is finite $\square$
Is the above correct? I ask because the aforementioned posts contained proofs which I thought were fine, but which seemed to contain problems, and the sheer amount of mappings interacting in my proof made me a little confused... In case of a mistake, could you please point out where and why?
Thanks in advance!
EDIT: Attempt 2
After reading the comments and giving more thought, I decided to try a different approach. Let's first recall that every subgroup of index $m$ in $G$ induces a homomorphism $G \to S_m$, such that its kernel is contained in said subgroup, using the ideas of my previous attempt. (I don't think I have to fix the isomorphism in this case, because even if $H$ induces plenty of homomorphisms, it does not change the general argument anymore).
However, we can now use that, if we call the homomorphism induced by $H \leq G$ by $\alpha_H$, then $H$ is isomorphic to a subgroup of $G/\ker(\alpha_H)$, due to the Correspondence Theorem.
Ergo, the number of such subgroups is bounded by $\sum s(h)$, where $h$ runs through all homomorphisms from $G$ to $S_m$ (finite) and $s(\phi)$ is the number of subgroups of $G/\ker(\phi)$ (finite, because it's bounded by the number of subsets of the finite set $G/\ker(\phi)$). Thus, the sum is finite and we have the result $\square$
Is this attempt better? Or is there some other problem still lingering around?
