Ordered group acting freely on partially ordered set

Let $(G, <)$ be a totally ordered group, and let $<$ be left-invariant. Let $G$ act (freely?) on a partially ordered set $(S, <)$, such that this group action preserves the ordering:

$$s_1 < s_2 \Rightarrow gs_1 < gs_2;$$ $$g_1 < g_2 \Rightarrow g_1s < g_2s$$

Can the partial ordering $(S, <)$ be extended to a total ordering such that the above is preserved?

Thinking the action must be free $\Rightarrow$ no element has nontrivial stabilizer, so group must be infinite? (for it could act on itself?) The partial ordered set must also be infinite?

• Can you say what you mean by "ordered group". A group with a total ordering that is both left and right invariant? – YCor Jul 4 '18 at 14:37
• Obviously such an action is free. Indeed if $g$ fixes $x$, we can suppose $g\ge 1$ (changing to $g^{-1}$ if necessary) and then by the second axiom $1<g$ implies $x=1x<gx=x$, contradiction. Note that what's usually called "preserve the ordering" is just the first axiom. I'd rather say "preserves both orderings" or "is compatible with both orderings". – YCor Jan 30 '20 at 21:34

Yes, such a partial order can be extended to a total order: we can amend the proof of the Szpilrajn extension theorem which essentially establishes the result in the case $$G=1$$.

Starting with the given partial order $$\leq_S$$ on $$S$$, consider partial orders $$\leq$$ on $$S$$ satisfying

(1) $$s\leq_S t$$ implies $$s\leq t$$; and

(2) $$s\leq t$$ implies $$gs\leq gt$$.

The set of such partial orders is itself partially ordered by inclusion and contains $$\leq_S$$. As usual, every chain in this poset is bounded above by its union, and therefore by Zorn's Lemma there is a maximal partial order $$\leq'$$. I claim that this is a total order: for otherwise if $$s$$ and $$t$$ are incomparable with respect to $$\leq'$$, we can extend it to an order in which, say, $$s\leq'' t$$ and more generally $$\gamma s\leq'' \gamma t$$ for all $$\gamma\in G$$. (Note that since $$s$$ and $$t$$ are $$\leq'$$ incomparable, so are $$\gamma s$$ and $$\gamma t$$, thanks to condition (2); therefore decreeing $$\gamma s\leq'' \gamma t$$ does not conflict with $$\leq'$$.) This contradicts the maximality of $$\leq'$$; therefore $$\leq'$$ is a total order on $$S$$. Moreover it satisfies the conditions in the OP.

To answer your other questions: yes the action of $$G$$ on $$S$$ is necessarily free, since if $$g\neq 1$$ then we have $$gs>1s$$ or $$gs<1s$$ for all $$s$$. Any non-trivial (left-)ordered group is torsion-free and therefore infinite. And finally each orbit is in 1-1 correspondence with $$G$$ since the action is free, so $$G$$ and $$S$$ are infinite assuming $$G$$ is non-trivial.

• Could you provide some literature that might reference this? As in any paper that uses the Szpilrajn extension theorem with a nontrivial group? – lunchmeat Aug 20 '18 at 11:38
• I'm not aware of any published version of the argument I've given, though I wouldn't be surprised to learn that there is one. – shane.orourke Aug 20 '18 at 20:25