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I'm thinking of the following problem: If we have two groups $G$ and $H$, with surjective homomorphisms $\psi:G\to H$ and $\phi:H \to G$, is it possible to prove that $G$ and $H$ are isomorphic? Or could we construct a counterexample?

$G$ and $H$ must have the same cardinality (see this question), but I don't know that whether they are necessarily isomorphic.

I know that the Cantor-Schroder-Bernstein theorem doesn't hold for groups (see here), but I'm wondering that whether the situation would be different for surjective homomorphisms.

Shaun
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AnyonicFugue
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    No. For example, it fails for the Baumslag-Solitar BS(4, 6) group; see https://arxiv.org/pdf/1308.5122.pdf . (The Baumslag-Solitar groups are generally great examples for this sort of bad behavior.) – anomaly Feb 09 '23 at 04:04
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    @anomaly But what are the non-isomorphic groups $G$ and $H$ in this example? – Derek Holt Feb 09 '23 at 08:21
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    Take $G = BS(4, 6)$. For $H$, there's a construction outlined in the paper that gives an finite family of groups; see Cor. 6.8. – anomaly Feb 09 '23 at 13:26

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No, it is not true. We can, for example, take $G = \prod_{\mathbb{N}} S_3$ to be the product of countably many copies of $S_3$, and take $H = G \times C_2$. $G$ and $H$ are not isomorphic because $G$ has trivial center but $H$ has nontrivial center. But there is a surjection $G \to H$ given by mapping the first copy of $S_3$ to $C_2$ via the sign homomorphism and then mapping the $n^{th}$ copy of $S_3$ to the $(n-1)^{th}$ copy of $S_3$, and there is a surjection $H \to G$ given by quotienting by $C_2$.

This is related to the existence of non-Hopfian groups.

Qiaochu Yuan
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  • Very nice! Seems like this construction should work for any non-abelian group $G$ with trivial center which admits a surjection onto some group $A$ with nontrivial center (playing the roles of $S_{3}$ and $\mathbb{Z}/2\mathbb{Z}$ respectively). – Alex Wertheim Feb 09 '23 at 04:09
  • Right. Probably various generalizations are possible too. – Qiaochu Yuan Feb 09 '23 at 04:31
  • Another (probably standard) example, with abelian groups: use $C_4$ instead of $S_3$. In these examples we also have $G$ embedding into $H$ and $H$ embedding into $G$. Any example where this doesn't happen? – spin Feb 09 '23 at 05:17
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    @spin Something like $F$ and $F\times \mathbb Z$, where $F$ is a countable-rank free group? Maybe $F\times F$ and $F\times \mathbb Z$ means neither is a subgroup of the other. – David A. Craven Feb 09 '23 at 08:28