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?

Has anything been proven about this? Is there an easy counterexample? Thank you all.

  • $\begingroup$ Can you say what you mean by "ordered group". A group with a total ordering that is both left and right invariant? $\endgroup$ – YCor Jul 4 '18 at 14:37
  • $\begingroup$ 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". $\endgroup$ – 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.

  • $\begingroup$ Could you provide some literature that might reference this? As in any paper that uses the Szpilrajn extension theorem with a nontrivial group? $\endgroup$ – lunchmeat Aug 20 '18 at 11:38
  • $\begingroup$ 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. $\endgroup$ – shane.orourke Aug 20 '18 at 20:25

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