# Is a group with only finitely many subgroups of index n (for all n) finitely generated

Let $G$ be a group with the following property. For all integers $n$, there are only finitely many subgroups of index $n$.

Question. Is $G$ finitely generated?

The converse is true. That is, if $G$ is finitely generated, then $G$ has only finitely many subgroups of index $n$ (for all $n$). Here's an argument for normal subgroups that you can also make work in the general case.

To give a normal subgroup of index $n$ of $G$ is to give a finite group $H$ of cardinality $n$ and a surjection $G\to H$. There are only finitely many groups of cardinality $n$ and for each finite group $H$ there are only finitely many $G\to H$ (because it suffices to designate the image of each generator of $G$ in $H$). QED

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It is not true because $\Bbb Q / \Bbb Z$ is not finitely generated whereas it has no subgroup of index n for any n.

You can also make examples that have subgroups of finite index : let $G = \bigoplus \Bbb Z / p \Bbb Z$ where $p$'s range is over the prime numbers.

It is not finitely generated because any finite subset of $G$ is included in a finite direct sum of those cyclic groups, and is not $G$.

Let $n \ge 1$. Since $G$ is commutative, every subgroup is normal and so a subgroup of index $n$ corresponds to a map into a (commutative) group of order $n$. Let $1_p$ be the generator of $\Bbb Z/ p\Bbb Z$, such that the set $\{1_p ;$p$\text{ prime}\}$ generates $G$. Now any such map is defined by the image of those generators. But if $H$ is of order $n$, and $p$ is prime with $n$, there is no element of order $p$ in $H$, and so $1_p$ must be sent to $0_H$. Since there are finitely many primes that are not coprime with $n$, there can only be finitely many such maps, and thus finitely many subgroups of $G$ of index $n$.

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@Harry: Theorem: An abelian group $G$ is finitely generated if and only if it is a quotient of a free abelian group of finite rank. This generalizes mercio's example. –  B. S. Jan 24 '13 at 10:30
Great! I guess one can also use this to obtain a non-abelian example by taking the product of $\mathbf Q/\mathbf Z$ with a finite non-abelian group $G$. Then $\mathbf Q/\mathbf Z \times G$ is not finitely generated, but it also has only finitely many subgroups of index $n$, right? –  Harry Jan 24 '13 at 11:11
@mercio : can you give an example of subgroup of index 2 ? It seems to me that $Q/Z$ is 2-divisible, hence all its quotients are $2$-divisible. –  user10676 Jan 24 '13 at 12:00
@user10676 : yes, I think I misunderstood with "subgroup of order $n$". Now I can't delete this. –  mercio Jan 24 '13 at 12:29
Anyway this Is a counter-exemple, because it has no subgroup of finite index (and $Q$ is a more trivial counter-exemple). Congrats for the upvotes ! –  user10676 Jan 24 '13 at 12:38
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Here there is an interesting family:

A profinite group is a compact topological group which is the inverse limite of finite groups. Examples of these are the profinite completion of a group which is defined as follows: For a group $G$ take the inverse limite of the finite quotients $G/N$ and call this group the profinite completion of $G$ which is denoted by $\widehat{G}$. It can be proved that the image of $G$ in $\widehat{G}$ is dense.

A profinite group is said to be finitely generated if it has a finitely generated subgroup in the abstract sense which is dense. Examples are the profinite completion of finitely generated groups. Infinite profinite groups are not finitely generated. You can prove that every finitely generated profinite have for each n just a finite number of open subgroups of index $n$. Then it is a theorem by Nicolov-Segal that every finite index subgrouop of a finitely generated profinite group is open. Then every infinte finitely generated profinite group is a counterexample. Example: The $p$-adic numbers $\mathbb{Z}_p$.

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