A finitely generated group has only finitely many subgroups of a given index

Let $G$ be a finitely generated group and let $n$ be a positiveinteger. Let $H$ be a subgroup of $G$ of index $n$ and consider theaction of $G$ on the coset space $(G:H)$ by right multiplication.Label the cosets $1,\mathrm{\dots },n$, with the coset $H$ labelled by $1$.This gives a homomorphism $\varphi :G\to S_n$. Now, $x\in H$ if andonly if $Hx=H$, that is, $G$ fixes the coset $H$. Therefore, $H={\mathrm{Stab}}_G(1)=\{g\in G\mid 1(g\varphi )=1\}$, and this is completelydetermined by $\varphi$. Now let $X$ be a finite generating set for $G$.Then $\varphi$ is determined by the images $x\varphi$ of the generators $x\in X$. There are $|S_n|=n!$ choices for the image of each $x\in X$, so there are at most ${n!}^{|X|}$ homomorphisms $G\to S_n$. Hence,there are only finitely many possibilities for $H$.