Infinite abelian group with every proper nontrivial subgroup being free abelian

Question

Let $G$ be an infinite abelian group such that all proper non-trivial subgroups of $G$ are free abelian groups . Then is it true that $G$ is a free abelian group ? I think the statement is true , but I cannot come up with a proof . I can see that $G$ is torsion free ; so if $G$ is finitely generated , then $G$ is free abelian . But I don't know what happens if $G$ is not finitely generated . Please help .

Answer 1

If $n$ is an integer, and $nG$ is properly contained in $G$, then by your hypothesis, $nG$ must be free abelian with some basis $S$. Then $\frac{1}{n}S$ is a basis for $G$, and you are done.

Otherwise, $nG = G$ for all $n \in \mathbb{N}$. This means that it makes sense to divide things in $G$ by integers. Fixing a given $g \neq 1_G$ in $G$, the homomorphism $\mathbb{Q} \rightarrow G$ given by

$$\frac{a}{b} \mapsto \frac{a}{b} \cdot g$$

is well defined and injective. Then $G$ contains a nonfree subgroup.