Groups containing each other as finite index subgroups

Question

Suppose that $G, H$ are finitely generated groups such that $H$ is isomorphic to a finite index subgroup of $G$ and vice versa. Does it follow that $G$ is isomorphic to $H$?

I am sure that the answer is negative but cannot find an example. I am mostly interested in the case of finitely presented groups. The assumption of finite index is, of course, necessary, otherwise one can take any two nonabelian free groups of finite rank. Here is what I know: Given a pair of groups $G, H$ as above, there is, of course, a sequence of isomorphic proper subgroups of finite index
$$ ... G_n\lneq G_{n-1}\lneq ...\lneq G_1\lneq G $$ One can rule out the existence of such a sequence when $G$ is nonelementary hyperbolic, but this does not say much.

PS. Noam's example makes me feel rather silly since I have seen such groups in vivo: The affine Coxeter groups $\tilde{B}_n, \tilde{C}_n$, $n\ge 3$.

Answer 1

Simple counterexample: $G$ is the square of an infinite dihedral group, consisting of symmetries of the ${\bf Z}^2$ lattice of the form $(x,y) \mapsto (\pm x + a, \pm y + b)$ with $a,b \in \bf Z$; and $H$ is the index-$2$ subgroup where $a \equiv b \bmod 2$. Then $H$ has index-$2$ subgroup consisting of the symmetries $(x,y) \mapsto (\pm x + 2a, \pm y + 2b)$, and this subgroup is isomorphic with $G$, but $G \not \cong H$.

Answer 2

The answer is "no". It is proved in https://arxiv.org/abs/0711.1014 for R. Thompson group $F$. That group contains a finite index subgroup $H$ which is not isomorphic to $F$ but contains a copy of $F$ as a subgroup of finite index.