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Let $G$ be a finitely generated group. $n>0$ is a fixed integer. $\{K_\alpha\}$ is the set of all normal subgroups of $G$ with index $n$, that is $[G:K_\alpha]=n$. Consider $\bigcap_\alpha K_\alpha$. It is a normal group. But why is $[G: \bigcap_\alpha K_\alpha]$ finite?

The integer ring $\mathbb Z$ is an example. Its normal subgroup with index $n$ is $n\mathbb Z$.

Shaun
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Functor
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  • @Alessandro $n$ is the index of the $K_\alpha$ in $G$, not the number of normal subgroups of index $n$. The main point to observe is that the fact that $G$ is finitely generated implies that the number of normal subgroups of index $n$ is finite. – Derek Holt Apr 04 '22 at 07:33
  • The proof of the fact mentioned by Derek Holt can be found here https://math.stackexchange.com/q/88719/464908 – Erica Apr 04 '22 at 09:41
  • @DerekHolt, I noticed, that's why I deleted my comment – Alessandro Apr 04 '22 at 09:42
  • The idea of the proof is that with each subgroup of index $n$ we can associate a homomorphism $\phi:G \to S_n$ arising from the action of $G$ on the cosets of the subgroup, and $G$ finitely generated implies that there are only finitely many homomorphisms to $S_n$. – Derek Holt Apr 04 '22 at 10:20
  • Thank you very much. Now I see that there are only finite normal subgourps with index n. So the index of their intersections is finite. – Functor Apr 04 '22 at 10:52

1 Answers1

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Let G be a group generated by m generators. $n>0$ is an integer. Then G has at most $(n!)^m$ subgroups of index n.

Proof : Every subgroup K of index n corresponds to a group homomorphism

$\phi_K : G \longrightarrow S(G/K)\cong S_n$

$g \mapsto (gxK \mapsto xK )$

We always let $K$ be the first element in $\{1,2,...n\}$. If H and K are different subgroups of index n, then choose $g \in K-H$. $\phi_K(g)(K) = K, \phi_H(g)(H) = gH \not= H$. Therefore, $\phi_K \not= \phi_H$. Therefore, $K \mapsto \phi_K$ is bijective.

Such a map is defined on $m$ generators of G. For each generators, there are $n!$ choices of images. There are at most $(n!)^m$ such maps, so are subgroups of index n.

Functor
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