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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 $\tilde{B}_n, \tilde{C}_n$, $n\ge 3$.

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    $\begingroup$ Easy abelian (and not finitely generated) counterexample: Let $G$ be the direct sum of an infinite sequence of cyclic groups of order $4$. Let $H$ be the product of $G$ with a group of order $2$. $\endgroup$ – Tom Goodwillie Dec 21 '19 at 0:45
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    $\begingroup$ @TomGoodwillie: Oh, of course, I should have thought about this. I will add the finite generation assumption. $\endgroup$ – Moishe Kohan Dec 21 '19 at 0:47
  • $\begingroup$ Another example can be found with $G$ the integral Heisenberg group (exercise). $\endgroup$ – YCor Dec 21 '19 at 10:16
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    $\begingroup$ @YCor: Sela proved that torsion-free one-ended hyperbolic groups are co-hopfian. The assumption of being torsion-free is removed in Moioli's thesis, Graphe de groupes et groupes co-hopfiens. Removing the assumption of being one-ended is done in Quasiregular self-mappings of manifolds and word hyperbolic groups but only in the torsion-free case. I didn't read Lemma 4.2 carefully yet, but it seems reasonable to think that a similar argument can be followed for groups splitting over finite subgroups.The article Complexity volumes of splittable groups may also be relevant. $\endgroup$ – AGenevois Dec 22 '19 at 9:22
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    $\begingroup$ See arxiv.org/abs/1012.1785. Page 1. $\endgroup$ – user6976 Dec 22 '19 at 15:25
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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 $G \not \cong H$.

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    $\begingroup$ Great, thank you! $\endgroup$ – Moishe Kohan Dec 21 '19 at 1:02
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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.

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