TITLE: Groups having at most one subgroup of any given finite index
QUESTION [14 upvotes]: Cyclic groups have at most one subgroup of any given finite index. Can we describe the class of all groups having such property?
Thank you!

REPLY [3 votes]: Let $G$ be a group.  The canonical residually finite quotient of $G$ is $R(G)=G/K$ where $K$ is the intersection of all the finite-index subgroups of $G$.
Lemma: If $G$ is finitely generated (update) then $G$ has at most one subgroup of each index if and only if $R(G)$ is cyclic.
Proof: First, note that $R(G)$ is residually finite.  If every finite quotient of $R(G)$ is cyclic then $R(G)$ is residually cyclic, and it follows that $R(G)$ is abelian.  So $R(G)$ has a non-cyclic finite quotient unless $R(G)$ is cyclic.  Therefore, if $R(G)$ is not cyclic then $R(G)$, and hence $G$, has a finite non-cyclic quotient, and hence, by Artuto's answer, has a two distinct finite-index subgroups of the same index.
Conversely, suppose that $R(G)$ is cyclic.  Every finite-index subgroup of $G$ contains $K$, so the quotient map $G\to R(G)$ maps finite-index subgroups to finite-index subgroups bijectively and preserves the index.  Therefore, if $R(G)$ is cyclic then $G$ has at most one subgroup of each index.  QED
I believe that it is an open question whether or not there is an algorithm to determine whether a fp group has a proper finite-index subgroup, ie whether or not $R(G)$ is non-trivial.  So it may be open whether or not it is possible to determine if $R(G)$ is cyclic, too.
Note: Earlier, I forgot to mention that I had implicitly assumed that $G$ is finitely generated.  This assumption is clearly necessary; otherwise the additive group of the rationals is a counterexample.  If $G$ is not finitely generated, then the same argument shows that if $G$ has at most one subgroup of each finite index then $R(G)$ is residually cyclic.  But it's not clear to me that the converse of this statement is true.   So I'll finish with a question:

If $G$ is residually cyclic, does $G$ have at most one subgroup of each finite index?