Let $A$ be a countable torsion-free abelian group. The following conditions are well known to be equivalent:
- $A$ is free abelian,
- every finite rank pure subgroup of $A$ is free abelian.
Consider the following condition:
- every rank one pure subgroup of $A$ is free abelian.
Is this condition equivalent to the previous two? This is surely known but I was not able to (dis)prove it or find it anywhere.