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.