TITLE: If there are injective homomorphisms between two groups in both directions, are they isomorphic?
QUESTION [29 upvotes]: 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?

REPLY [8 votes]: 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.