# Is a subgroup in $\mathbb Q^n$ satisfying some condition free abelian?

Consider $$\mathbb Q^n$$ as an abelian group under addtion, $$G\subset \mathbb Q^n$$ is a subgroup, such that:

For any $$v\in\mathbb Q^n$$, the subgroup $$(\mathbb Q\cdot v)\cap G$$ is either isomorphic to $$0$$ or $$\mathbb Z$$.

In other words, looking at any "direction" of some element in $$G$$, one always find a generator.

Could we prove that $$G$$ is a finitely generated free Abelian group? Of course if we can prove that $$G$$ is always finitely generated as an abelian group, since it is torsion free, $$G$$ would be a free group automatically.

• I haven't thought this through, but is something like $\langle (1/2^n,1/2^{2n}) : n \in {\mathbb Z}_{\ge 0} \rangle < {\mathbb Q}^2$ a counterexample? Commented May 10 at 10:23
• @DerekHolt, I was also thinking of things like that. I don't think this one works, because we then also have $2^n (1/2^n, 1/2^{2n}) - 2^{n + 1}(1/2^{n + 1}, 1 / 2^{2n + 2}) = (0, 1/2^n - 1/2^{n + 1})$ for each $n$, so there is a dense "vertical line subgroup". Commented May 10 at 10:49

The example constructed in the paper even has the property that every quotient by a pure rank 1 subgroup (which are exactly the subgroups you are considering) is isomorphic to $$\mathbb{Q}$$ (so that the group is, in some sense, "maximally" far away from beeing free abelian).
In fact, the example is not too difficult to write down: Fix any automorphism $$\phi$$ of $$\mathbb{Q}/\mathbb{Z}$$ that acts as multiplication by a transcendental element of $$\mathbb{Z}_p$$ on the $$p$$-component for each prime $$p$$. The subgroup $$A=\{(a,b)\in \mathbb{Q}^2: \phi(\overline{a})=\overline{b}\}$$ (where $$\overline{x}=x+\mathbb{Z}$$) has the desired properties.