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.



Freeness of torsion-free abelian groups
No, there is a torsion-free abelian group of rank 2 such that all subgroups of rank 1 are free abelian but the whole group isn't. In fact, all its rank-1 quotients are divisible. An example is constructed in "On the cancellation of modules in direct sums over Dedekind domains" by L. Fuchs and F. Loonstra [Nederl. Akad. Wetensch. Proc. Ser. A 74 ( = Indag. Math. 33) (1971) pp. 163-169].


