The height of an element $g$ in an abelian group $G$ is the largest $n\in \mathbb{N}$ such that there exist an element $h\in G$ such that $n*h=g$. If $g$ has no such largest integer than $g$ is of infinite height.
Suppose that a group $G$ is countable, torsion free,and has no elements of infinite height, then must it be a free group?
If we restrict to finitely generated groups this follows immediately from the classification of finitely generated abelian groups. If, on the other hand, we allow uncountable groups (of any size) then the Baer-Specker Group ($\mathbb{Z}^\mathbb{N})$ serves as a counterexample. I can't figure out a way to adapt the proof of the Baer-Specker Group to any countable group though.
No, and in fact there's a counterexample that's a subgroup of $\mathbb{Q}^2$.
See https://mathoverflow.net/questions/90586/are-these-abelian-groups-free for more details.