Pure subgroup of torsion-free abelian group not implies direct factor

Statement for pure subgroups
It is possible to have the following:


 * $$G$$ is a torsion-free abelian group
 * $$H$$ is a pure subgroup of $$G$$ (this means that for any $$h \in H$$ and $$n \in \mathbb{N}$$ such that $$nx = h$$ has a solution for $$x \in G$$, there is a solution for $$x \in H$$).
 * $$H$$ is not a direct factor of $$G$$.

Statement for local powering-invariant subgroups
It is possible to have the following:


 * $$G$$ is a torsion-free abelian group
 * $$H$$ is a local powering-invariant subgroup of $$G$$ (this means that for any $$h \in H$$ and $$n \in \mathbb{N}$$ such that $$nx = h$$ has a unique solution for $$x \in G$$, that unique solution is in $$H$$).
 * $$H$$ is not a direct factor of $$G$$.

Related facts

 * Pure subgroup implies direct factor in torsion-free abelian group that is finitely generated as a module over the ring of integers localized at a set of primes

Proof
Example where $$H = \mathbb{Z}$$, $$G/H = \mathbb{Z}[p^{-1}]$$ for some prime number $$p$$.