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 .


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.

  • $\begingroup$ 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 $\endgroup$ – user Aug 6 '17 at 19:59
  • $\begingroup$ Yes, and in fact $\mathbb{Q}$ itself is not free. $\endgroup$ – D_S Aug 6 '17 at 19:59
  • $\begingroup$ 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 $\endgroup$ – user Aug 6 '17 at 20:00
  • $\begingroup$ That's fine actually. We finish the proof by showing that $G$ contains a nonfree subgroup (even if that subgroup is $G$ itself). $\endgroup$ – D_S Aug 6 '17 at 20:04
  • $\begingroup$ Um but that won't give any contradiction . Our hypothesis is all "proper non-trivial " subgroups of $G$ is free ... $\endgroup$ – user Aug 6 '17 at 20:07

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