Let $H,G$ be abelian groups with $H \leq G$. We say that $H$ is a pure subgroup of $G$ if for every $n \in \mathbb N$ and $h \in H$ the following holds: If $h$ is $n$-divisible in $G$, then $h$ is $n$-divisible in $H$. In formulas: $$ \forall n \in \mathbb N : nG \cap H = nH. $$ There is a whole theory around pure subgroups, however I could not find anything in the context of ordered groups.

We say that $(G,G_+)$ is an ordered abelian group (with $G_+ \subseteq G$) if $G_+ + G_+ \subseteq G_+$, $G_+ - G_+ = G$ and $G_+ \cap - G_+ = \{0\}$.

So my question is the following: If $(H,H_+)$ and $(G,G_+)$ are ordered abelian groups with $H \leq G$ pure and $H_+ \subseteq G_+$, can one characterize this differently ? (As in the case of non-ordered pure subgroups).

It would also make sense to define pure subgroups in the ordered case by asking for positive divisibility.