When does an abelian "linearly ordered group " has the property that any non-empty subset of the set of "non-negative" elements has a "least" element

Question

By a theorem of F.W. Levi , we know that an abelian group can be equipped with a linear order (https://en.wikipedia.org/wiki/Linearly_ordered_group) iff the group is torsion free . So let $(G,\le )$ be an abelian linearly ordered group , then what condition(s) on $G$ ensure that any non-empty subset of $G$ that is bounded below ($S\subseteq G$ is called bounded below if $\exists g \in G$ such that $g \le s , \forall s \in S$ ) , has a least element (i.e. $\exists s_o\in S$ such that $s_o\le s , \forall s \in S$ ) ? Due to compatibility of the order with group operation , equivalently I am asking that when can we say that any non-empty subset of $\{g \in G : 0\le g\}$ ($0$ being the identity of $G$) has a least element ? Has there been any work done on this ?

Answer 1

The requirement is that the positive cone $\{x\in G:x\ge0\}$ is well ordered (if not empty, but in this case the group would be trivial).

In particular, $G$ has a minimum positive element $x_0$. I contend that $G=\langle x_0\rangle$.

Suppose $x\in G$, $x>0$. There exists $n\in\mathbb{N}$ such that $nx_0>x$. Otherwise the set $\{nx_0:n>0\}$ has no maximum and consequently the set of nonnegative elements $$ \{x-nx_0:n>0\} $$ has no minimum.

Let $n_x$ be the minimum nonnegative integer such that $(n_x+1)x_0>x$. Then $n_xx_0\le x$.

If $n_xx_0<x$, then $x-n_xx_0<x_0$, a contradiction to the minimality of $x_0$. Therefore $n_xx_0=x$.

If $x<0$, then $-x>0$ and we can repeat the argument.