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.
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.