Let $G$ be a group and $H$ a finite index subgroup, one could study the cardinality of the set of subgroups $K$ of $H$ of fixed finite index $[H:K]=n$.

At first I thought this was always a finite set, but I soon hit on an example where it is not so, e.g. the direct product of infinite copies of $\mathbb{Z}$, where we can take $G=H$ and we have infinite different subgroups $K$ of index 2, by choosing $K$ as the product of $2 \mathbb Z \times \prod \mathbb Z$ where $2 \mathbb Z$ is each time the subgroup of a different copy of $\mathbb Z$.

So my question is: How do we characterize groups $G$ that has the following property

"For every finite index subgroup $H$, and every fixed $n>0$ there is only a finite number of subgroups $K$ of $H$ such that $[H:K]=n$?

Note for example that every finitely generated abelian group $G$ has this property, thanks to the classification theorem: every subgroup $H$ of $G$ is of the form $H_{\text {free}} \bigoplus H_{\text{ torsion}}$, so asking it is of finite index means that the ranks of $G$, $H$ and $K$ are the same. But then $H_{\text{torsion}}$ is finite, thus it has finitely many subgroups.

I don't know if this is a trivial question or an uninteresting one, so any reference or comment is welcome. Thanks!

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    $\begingroup$ I think your subgroup $H$ is an unnecessary distraction! Why not just say, how do we characterize groups such that for all $n \ge 1$ there are only finitely many subgroups of index $n$? I don't expect there to be any sensible answer. For example, all simple groups have this property, since an infinite simple group has no proper subgroup of finite index, and hence any direct product of a simple group with a finitely generated group will have it. $\endgroup$ – Derek Holt Sep 3 '12 at 11:38
  • $\begingroup$ You're right, $H$ is just a distraction. And thanks for the insight. $\endgroup$ – Niccolò Sep 4 '12 at 8:37

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