Groups having at most one subgroup of any given finite index 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!
 A: 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?

A: Just thought, I would make this obvious remark, extending Arturo's answer: since, even in the infinite case, any subgroup of finite index in $G$ must be normal for $G$ to satisfy the requirement, it follows that for any $H$ of finite index in $G$, any subgroup of the quotient $G/H$ will also correspond to a subgroup of $G$ of finite index. In particular, for any $H$ of finite index, $H$ must be normal and the quotient $G/H$ must be cyclic by Arturo's argument.
A class of such groups considerably extending that of cyclic groups are the pro-cyclic group, i.e. inverse limits of cyclic ones. Examples include $\mathbb{Z}_p$ or any product $\prod_p \mathbb{Z}_p$ over distinct primes $p$. In particular, $\hat{\mathbb{Z}}$ is another example. In fact, Arturo's argument shows that any pro-finite group satisfying the above condition must be pro-cyclic.
A: Since $G$ has exactly one subgroup of each finite index, and the index of a conjugate of $H$ equals the index of $H$, then every subgroup of finite index is normal. 
If $G$ is finite, then every subgroup is normal, so the group must be a Dedekind group (also known as Hamiltonian groups). 
All such groups that are nonabelian are of the form $G = Q_8 \times B \times D$, where $Q_8$ is the quaternion group of $8$ elements, $B$ is a direct sum of copies of the cyclic group of order $2$, and $D$ is an abelian group of odd order. Any of the factors may be missing.
Since $Q_8$ contains several subgroups of index $2$ (exactly three, in fact), if a factor of $Q_8$ appears then $G$ would have several subgroups of the same index, hence $G$ must in fact be an abelian group. 
Since $G$ is finite and abelian, it is isomorphic to a direct sum of cyclic groups, $G = C_{a_1}\oplus\cdots\oplus C_{a_k}$, where $1\lt a_1|a_2|\cdots|a_k$. If $k\gt 1$, then $G$ contains at least two subgroups of order $a_{k-1}$; thus $k=1$ so $G$ is in fact cyclic. So the only finite groups with the desired property are the cyclic groups.
If $G$ is infinite, you can have other possibilities. One example is the Prüfer group, Added: but only by vacuity: it has no proper subgroups of finite index.
In general, if $H$ if a subgroup of finite index in $G$ then $H$ is normal, as above, and $G/H$ also has the desired property and is finite; thus, $G/H$ is cyclic for every subgroup of finite index by the argument above. I'm sure there's more to be said, but I'll think about it a bit first...
A: This is a very old question by now, but since OP was interested in the infinite case, it is nice with some concrete examples: it is not too difficult to show that Baumslag's one-relator group $G = \operatorname{Gp}\langle a, b \mid [a, b^{-1}ab] = a\rangle$ has exactly one finite index (normal) subgroup for every index $n$. This group was introduced in the very short paper [1] (available here). As expected by HJRW's answer, one can find that $R(G) \cong \mathbb{Z}$ for this group.
$\\$
[1] Baumslag, G., A non-cyclic one-relator group all of whose finite quotients are cyclic, J. Aust. Math. Soc. 10, 497-498 (1969). ZBL0214.27402.
