If there are injective homomorphisms between two groups in both directions, are they isomorphic? If I have two groups, $G$ and $H$ and two injective homomorphisms $\phi:G \to H$ and $\psi: H \to G$, then by the first isomorphism theorem applied to $\phi$, we have that $G \cong \mathrm{Im} (\phi)$, a subgroup of $H$. Similarly, $H$ is isomorphic to a subgroup of $G$. For finite groups, this guarantees that $G \cong H$ but does this hold for infinite groups?
Weird things can happen for infinite groups, e.g $n\mathbb Z \subsetneq \mathbb Z$ but $n \mathbb Z \cong \mathbb Z$. I'm wondering if this kind of thing stops an isomorphism from occurring.
I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?
 A: 
I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?

The answer is no for rings and fields. It should be no for modules but I'm having trouble coming up with a snappy counterexample off the top of my head. 
Rings: Let $k$ be a field, for simplicity. The ring $k[t^2, t^3]$ injects, by construction, into the ring $k[t]$. The ring $k[t]$ also injects into $k[t^2, t^3]$ via the map $t \mapsto t^2$. So there are injections
$$k[t] \hookrightarrow k[t^2, t^3] \hookrightarrow k[t]$$
but these two rings are not isomorphic because, for example, $k[t]$ is integrally closed and $k[t^2, t^3]$ is not. 
Fields: The field $\mathbb{C}$ injects into the field $\mathbb{C}(t)$, and hence injects into its algebraic closure $\overline{ \mathbb{C}(t) }$. Now, fun theorem: (Edit: uncountable) algebraically closed fields of characteristic $0$ are completely determined by their cardinality, and $\overline{ \mathbb{C}(t) }$ has the same cardinality as $\mathbb{C}$. So they are isomorphic. So there are injections 
$$\mathbb{C} \hookrightarrow \mathbb{C}(t) \hookrightarrow \mathbb{C}$$
but these two fields are not isomorphic because $\mathbb{C}$ is algebraically closed and $\mathbb{C}(t)$ is not. 
A: A nonconstructive example: there are injective homomorphisms $$(\mathbb R,+)\to(\mathbb R\times\mathbb Z,+)\to(\mathbb R\times\mathbb R,+)\cong(\mathbb R,+)$$ but $(\mathbb R,+)\not\cong(\mathbb R\times\mathbb Z,+).$
A: For groups this property - sometimes called the Cantor-Schröder-Bernstein property after the corresponding theorem for plain sets - is wrong. Let $G = F_2$ the free group on two generators $\{a,b\}$ and $H = F_3$ the free group on three generators $x,y,z$. Then there are monomorphisms $f \colon G \to H$, given by $f(a) = x$, $f(b) = y$ and $g \colon H \to G$ given by $g(x) = a^2$, $g(y) = b^2$, $g(z) = ab$, there is no isomorphism.
A: No. Any counterexample must be infinite, and one way to build such an example  is to construct an analogue of the well-known injective homomorphism $S_n \hookrightarrow A_{n + 2}$ for the infinite analogues of those groups.
Consider the group $S_{\infty}$ of (finite) permutations of a set $\{1, 2, \ldots\}$ of countably many elements. Then, let $A_{\infty}$ denote the subgroup of $S_{\infty}$ comprised of even elements (products of even numbers of permutations), or equivalently, the subgroup generated by $3$-cycles. Then, inclusion $A_{\infty} \to S_{\infty}$ is a homomorphism.
On the other hand, let $S_{\infty}'$ denote the subgroup of $S_{\infty}$ of permutations of the set $\{3, 4, \ldots\}$, and consider the map $\Phi: S_{\infty} \to S_{\infty}'$ that maps a permutation $(a_1 \cdots a_n)$ to $((a_1 + 2) \cdots (a_n + 2))$ (this amounts to relabeling and so is manifestly an isomorphism). Then, we can define a map $\Psi: S_{\infty}' \to A_{\infty}$ by
$$\Psi(\sigma) :=
\left\{
\begin{array}{cc}
\sigma     , & \sigma \text{ even} \\
\sigma (12), & \sigma \text{ odd}
\end{array}
\right.
$$
and readily see that it is an injective homomorphism. So, $\Psi \circ \Phi$ is an injective homomorphism $S_{\infty} \to A_{\infty}$, and hence we have injective homomorphisms in both directions, but the two groups are not isomorphic.
