# Infinite abelian group with every proper nontrivial subgroup being free abelian

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 .

## 1 Answer

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.

• Ah yes . If $G$ is divisible then $G$ contains a copy of $\mathbb Q$ , but $\mathbb Q$ contains a proper non-trivial subgroup which is not free , contradicting our hypothesis. Hence $G$ is not divisible, so $nG$ is a proper nontrivial subgroup for some $n \in \mathbb N$ ; so for that $n$ , $nG$ is free abelian ; but $G$ being torsion free ; $G \cong nG$ ; thus $G$ is free abelian – user Aug 6 '17 at 19:59
• Yes, and in fact $\mathbb{Q}$ itself is not free. – D_S Aug 6 '17 at 19:59
• Yes , but that won't do . Because it might be that $\mathbb Q=G$ , so we in fact have to say that $\mathbb Q$ has a non-free proper nontrivial subgroup – user Aug 6 '17 at 20:00
• That's fine actually. We finish the proof by showing that $G$ contains a nonfree subgroup (even if that subgroup is $G$ itself). – D_S Aug 6 '17 at 20:04
• Um but that won't give any contradiction . Our hypothesis is all "proper non-trivial " subgroups of $G$ is free ... – user Aug 6 '17 at 20:07