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Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$.

Here is a sketch of a proof. It consists of assuming such a subgroup $H$ exists, and allowing $G$ to act on $G/H$ by left multiplication. Such action defines a homomorphism $$\phi :G \rightarrow \text{Sym}(G/H)\cong \mathbb{S}_n$$

If $G=\langle g_1, \ldots ,g_r\rangle$, then the homomorphism is uniquely determined by where it sends the generators $g_i$. In particular, there are at most $(n!)^r$ such homomorphisms.

The last step of the proof requires one to notice that a different subgroup $H'$ of index $n$ in $G$ would produce a different homomorphism, and here is where I'm struggling. I've been able to show that $\text{ker}(\phi )=\text{core}(H)$, yet haven't been able to rule out the possibility of $\text{core}(H)=\text{core}(H')$.

Shaun
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Sam
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  • What's the stabilizer of a point? – Steve D Feb 07 '20 at 23:45
  • @SteveD, I believe $\text{Stab}(gH)={ x\in G : xgH=gH} = { x\in G : x\in gHg^{-1}} = gHg^{-1}$. – Sam Feb 07 '20 at 23:52
  • There are a few ways to proceed. (1) you need to make a choice in defining the homomorphism to $S_n$. If you always choose the number '1' for the coset containing the identity, you can recover $H$ as a point stabilizer. (2) you can make your choices arbitrarily, noting only finitely many groups yield a given homomorphism [the point stabilizers]. (3) you can show only finitely many groups have a given core, since $G/\text{ker}$ is finite. – Steve D Feb 08 '20 at 17:26

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