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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.

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  • $\begingroup$ Instead of asking that $G/H$ be torsion-free, require the following in the partially ordered case: (i) that $H$ be convex (that is, if $h_1 \leq g \leq h_2$ with $h_i \in H$ and $g \in G$, then $g \in H$); (ii) that $G/H$ with the obvious quotient pre-ordering (a coset is positive if it contains a positive element), is unperforated (that is, if a positive integer multiple of something is positive (or zero), then the something is already positive or zero). This has been somewhat studied. $\endgroup$
    – David Handelman
    Commented Feb 14, 2019 at 18:43

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