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
Best Answer
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 $n$.