Let $G$ be an abelian group (say countable). I know it is possible for $G$ and the subgroup $2G=\{g+g: g\in G\}$ to be isomorphic, e.g. $\mathbb{Z}$. In this case the map $s:G \to G$ that maps $g\mapsto g+g$ is injective. My question is as follows: Is it ever possible that $s: G \to G$ is a group isomorphism?
Note that I am not asking about $s:G \mapsto 2G$, for which it suffices to have injectivity. Equivalently, this would mean that $G=2G$ (or $[G:2G]=1$) and that $|\ker(s)|=1$.
Sure. Consider the integers modulo $3$ under addition.
Or, if you want an infinite example, you can just take an infinite direct sum of that group with itself.
Suppose that $G$ is a finite group. Then the map $f:G\rightarrow G$; $g\mapsto g^2$ is a homomorphism if and only if $G$ is abelian (so we can think of $g^2$ as $g+g$). Then $f$ is an isomorphism if and only if $|G|$ is odd:
For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd
Hence the "doubling map" is an isomorphism for every abelian group of odd order.
