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.

  • $\begingroup$ 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? $\endgroup$
    – Derek Holt
    Commented May 10 at 10:23
  • 3
    $\begingroup$ @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". $\endgroup$ Commented May 10 at 10:49

1 Answer 1


This question is answered here.

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .