Is there a special name for a linearly ordered group $G$ such that for every positive element $g\in G$ there exists an element $h\in G$ such that $e<h<g$?
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2$\begingroup$ I don't know if there is a standard name for this, but, assuming $e$ in the OP is the identity of the group, a possibility would be "non-atomic l.o. groups" (cf. en.wikipedia.org/wiki/Atom_%28order_theory%29). $\endgroup$– Salvo TringaliCommented Sep 22, 2015 at 6:53
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$\begingroup$ Yes, $e$ is the identity of the group. $\endgroup$– Alex RavskyCommented Sep 22, 2015 at 6:55
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2$\begingroup$ I assume the order is translation invariant. Then I think people usually express this condition by saying that the order is "dense" (meaning that if $a < b$, then there exists $x$, such that $a < x < b$). So the group is "densely ordered". However, an alternative would be to say that the order has no isolated points (if $a < x < b$, then there exists $y \neq x$ with $a < y < b$). $\endgroup$– Dave Witte MorrisCommented Sep 22, 2015 at 7:25
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I assume that the order is left-invariant. All orders here are total.
First, by homogeneity, the order is either dense or discrete (in the sense that the order topology is discrete). Beware that being discrete doesn't make it simpler: indeed for every left-ordered group $G$, the direct product with lexicographic order $G\times\mathbf{Z}$ is also left-ordered and the order topology is discrete. A better "far opposite" to being dense is being scattered, as defined by Ville Salo in this later question. Back to the OP's question:
Then a left-ordered group has a discrete order topology (i.e., does not satisfy the OP's condition) iff it is trivial or a (necessarily unique) convex subgroup isomorphic to $\mathbf{Z}$.
Uniqueness is clear. For existence, assuming $G$ nontrivial then implies that each element has a successor and a predecessor, so there is a unique map $f:\mathbf{Z}\to G$ such that $f(0)=1_G$ and $f(n+1)$ is the successor of $f(n)$ for every $n$. If $s=f(1)$, then $1<s$ and the open interval $]1,s[$ is empty, hence by left invariance, $s^n<s^{n+1}$ and the open interval $]s^n,s^{n+1}[$ is empty for every $n\in\mathbf{Z}$, which implies that $f(n)=s^n$ for every $n\in\mathbf{Z}$.
Note that if the order is bi-invariant, then this unique convex infinite cyclic subgroup is central (indeed, the positive cone is invariant by conjugation, so its minimum $s$ is invariant). As corollary:
For every center-free bi-ordered group, the underlying order is dense.
This applies to many polycyclic groups of exponential growth, as well as in Thompson's group $F$.
Note: This more generally applies if the center $Z$ has no infinite cyclic subgroup $C$ such that $Z/C$ is torsion-free, e.g., if $Z$ is isomorphic to a non-cyclic subgroup of $\mathbf{Q}$.
In the opposite direction, we have, by a straightforward argument:
Let $G$ be a group with a central subgroup $Z$, such that $G/Z$ is bi-orderable. Then $G$ is bi-orderable in a way that makes $Z$ convex. In particular, if $Z$ is infinite cyclic then the resulting order topology on $G$ is discrete.
The latter can be applied to every nontrivial, torsion-free finitely generated nilpotent group.