Let $G$ and $H$ be groups such that there exist group homomorphisms $\iota_{HG}: G \hookrightarrow H$, $\iota_{GH}: H \hookrightarrow G$, $\pi_{HG}: G \twoheadrightarrow H$, $\pi_{GH}: H \twoheadrightarrow G$ where $\iota_{HG}$, $\iota_{GH}$ are injective and $\pi_{HG}$, $\pi_{GH}$ are surjective.
Does it necessarily follow that $G \cong H$?
Ideas for counterexamples are taken from the following threads:
- If there are injective homomorphisms between two groups in both directions, are they isomorphic?
- Does existence of surjective and injective homomorphisms imply isomorphism?
$G = F_2 = \langle a, b \rangle$ and $H = F_3 = \langle c, d, e \rangle$. It is well-known that they are not isomorphic. There is obviously the inclusion $\iota_{HG}(a) = c$, $\iota_{HG}(b) = d$. There is also the reverse injection $\iota_{GH}(c) = a^2$, $\iota_{GH}(d) = ab$, $\iota_{GH}(e) = b^2$. We have a projection $\pi_{GH}(c) = a$, $\pi_{GH}(d) = b$, $\pi_{GH}(e) = \operatorname{id}$. However, there cannot be a surjection $\pi_{HG}: G \twoheadrightarrow H$, otherwise, $F_3$ would be generated by two elements, a contradiction. So this is not a counterexample.
$G = A_{\infty}$ the group of even permutations of $\mathbb{N}$ with finite support and $H = S_{\infty}$ the group of all permutations of $\mathbb{N}$ with finite support. It is well-known that these groups are not isomorphic. There is an obvious inclusion $\iota_{HG}$ and it is simple to construct an injection $\iota_{GH}$ that shifts the permutation to the right and then repairs the sign. However, there cannot be a surjection $\pi_{HG}$. Since $A_{\infty}$ is simple, such a surjection would either have kernel $\{ \operatorname{id} \}$, in which case it would be injective, meaning $\pi_{HG}$ would be an isomorphism, a contradiction, or it would have kernel $A_{\infty}$, in which case it would be trivial and not surjective, a contradiction. So this is not a counterexample.
$G = \mathbb{R}$ and $H = \mathbb{R} \times \mathbb{Z}$. These groups are not isomorphic because $G$ is divisible but $H$ is not divisible. There is an obvious inclusion $\iota_{HG}$. Because $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$ have the same dimension as $\mathbb{Q}$-vector spaces, they are isomorphic as additive groups. Composing the inclusion of $H$ into $\mathbb{R} \times \mathbb{R}$ with the isomorphism of $\mathbb{R} \times \mathbb{R}$ with $G$, we get an inclusion $\iota_{GH}$. There is an obvious projection $\pi_{GH}$. However, there can be no surjection $\pi_{HG}$, because otherwise $H$ would be a quotient of the divisible group $G$, meaning that $H$ itself would be disivible, a contradiction. So this is not a counterexample.
$G = \prod_{i = 1}^{\infty} \mathbb{Z}$ and $H = \mathbb{Z}_2 \times \prod_{i = 1}^{\infty} \mathbb{Z}$. These groups are not isomorphic since there is an element of order 2 in $H$, but not in $G$. There is an obvious inclusion $\iota_{HG}$, an obvious projection $\pi_{GH}$ and an obvious projection $\pi_{HG}$. However, there can be no injection $\iota_{GH}$, since otherwise $\iota_{GH}(\overline{1}, 0)$ would be of order 2 in $G$, but all nontrivial elements of $G$ are of infinite order. So this is not a counterexample.
Best Answer
Possibly the standard example, which is a variation on the last one you mention, is given by $G = \prod_{i=1}^{\infty} F_{i}$ and $H = T \times G$, where the $F_{i}$ are cyclic of order four, and $T$ is of order two.
Injections and surjections should be immediate. The two groups are not isomorphic, because in $G$ all elements of order two are squares, whereas in $H$ a generator of $T$ is not a square.