Totally right preorderable groups

Are there any known non-trivial sufficient conditions, or full characterizations, of a totally right-preorderable group?

More precisely:

• totally right-preorderable: has a non-trivial total right-preorder
• non-trivial total right-preorder: transitive and symmetric relation ("preorder") $\le$ with $a\le b$ or $b\le a$ for all $a,b$ ("total"), with $a\le b$ iff $ac\le bc$ ("right"), and with $a\not\le b$ for some $a,b$ ("non-trivial")

I couldn't find anything in the literature on total right preorderability (but perhaps I didn't know the terminology to look under).

• having a right-orderable quotient is sufficient
• in particular, indicability (non-trivial homomorphism to $\mathbb Z$) is sufficient
• not being generated by elements of finite order is necessary
• equivalent to total left-preorderability (but perhaps with a different preorder--same proof as for the equivalence of right- and left-orderability)

A more specific question then is whether the first or the third fact has a true converse? (The second doesn't, since there are right-orderable groups that have a non-indicable subgroup.) Or at least a converse given some nice assumptions?

A related question: Are there non-trivial sufficient conditions for having a right-orderable quotient?

• What you call "complete" pre-orders are also commonly called linear pre-orders, and also sometimes they are called total pre-orders. Nov 6 '13 at 19:23
• to admit a faithful order-preserving action on a totally ordered set and to be left-orderable are equivalent. So it follows from my answer that your condition (being totally left-preorderable) is indeed equivalent to the existence of a nontrivial left-orderable quotient.
– YCor
Nov 6 '13 at 19:31
• a torsion-free finite index subgroup in $\mathrm{SL}_3(\mathbf{Z})$ is not left-orderable (I guess this is due to Witte), and all its proper quotients are finite (Bass-Milnor-Serre), hence is not totally left-preorderable. This shows that the converse of your third fact is far from true.
– YCor
Nov 6 '13 at 19:33
• I don't know why I used "complete" instead of "total" in the question. I think it's because one of the last things I had read on preorders used "complete". I think "total" or "linear" is indeed more standard, so I changed it to "total". Nov 6 '13 at 20:01

A group $$G$$ admits a nontrivial preorder iff it admits a nontrivial order-preserving action on a totally ordered set (which can be chosen to be $$\mathbf{Q}$$ or its completion $$\mathbf{R}$$ if $$G$$ is countable).

Proof: [I like left actions rather than right actions so I'll go ahead with left-invariant instead of right-invariant: you can pass from one to another by inversion.]

Indeed suppose that $$G$$ admits such an action. Let $$x$$ be not fixed by all of $$G$$. Define $$g\le h$$ if $$gx\le hx$$. Then this is a non-trivial left-invariant total preorder on $$G$$.

Conversely, suppose that $$G$$ has a left-invariant total preorder $$\le$$. Let $$H$$ be the set of elements $$h$$ such that $$1\le h\le 1$$. Then $$H$$ is a subgroup: left-multiplying the latter by $$h^{-1}$$ shows it's stable by inversion, and if $$1\le g\le 1$$ as well, multiplying the former on the left by $$g$$ yields $$g\le gh\le g$$, whence $$1\le gh\le 1$$.

The preorder is actually right-$$H^2$$-invariant, in the sense that $$g\le g'$$ and $$h,h'\in H$$ implies $$gh\le g'h'$$. Indeed, we have $$gh\le g\le g'\le g'h'$$. The coset space $$G/H$$ inherits a $$G$$-invariant total order by $$gH\le g'H$$ iff $$g\le g'$$, this does not depend on the choices of $$g,g'$$ because of the above right $$H^2$$-invariance.

If moreover $$G/H$$ is countable, you can just take the lexicographic product $$(G/H)\times\mathbf{Q}$$ (with trivial action on $$\mathbf{Q}$$) to get an action on $$\mathbf{Q}$$. Taking completion also yields an action on $$\mathbf{R}$$.

In some cases you have an even simpler characterization. Indeed, a theorem of Witte (Alg. Geom. Topol. 2006, arXiv link; ProjectEuclid) can be restated as: a finitely generated amenable group $$G$$ admits a nontrivial order-preserving action on some totally ordered set (or on the real line, it's the same) if and only if it admits $$\mathbf{Z}$$ as a quotient.

Edit: Here's a direct proof (not using actions) that if $$G$$ admits a nontrivial left-invariant total preorder $$\le$$, then it admits a nontrivial left-orderable quotient: fix a strict well-ordering $$\prec$$ on the set $$G$$ (unrelated to the group structure and $$\le$$: if $$G$$ is countable just take an enumeration of $$G$$), and say that $$g\not\le' g'$$ if there exists $$h\in G$$ such that $$gh\not\le g'h$$ and $$gh'\le g'h'\le gh'$$ for every $$h'\prec h$$. Then $$\le'$$ is another left-invariant total preorder, but has the additional feature that the set $$N$$ of $$g$$ such that $$1\le' g\le' 1$$ is a normal subgroup ($$N$$ is indeed, using that $$\prec$$ is a well-ordering, the set of $$g$$ such that $$1\le h^{-1}gh\le 1$$ for all $$h\in G$$). Moreover $$N$$ is contained in the subgroup $$\{g:1\le g\le 1\}$$ and hence is not all of $$G$$. So the quotient $$G/N$$ is a nontrivial left-orderable group.

• Your answer here helped me to get a pretty much complete answer to another question of mine: mathoverflow.net/questions/146440/… . So thank you again! Nov 7 '13 at 14:38
• I am having some trouble following your last direct proof. By your definition, we have $g\le' g'$ iff for all $h\in H$ we have $gh\le g'h$ or there is an $h'\prec h$ such that not: $gh'\le g'h'\le gh'$. But as you showed, $\le$ is right $H$-invariant, so the first disjunct is just equivalent to $g\le g'$. Thus $\le'$ is weaker than $\le$ and so $\{ g : 1 \le g \le 1 \}\subseteq N$. But you claim the opposite inclusion (and surely in general they aren't equal). What am I missing? Dec 29 '13 at 21:40
• $\le'$ is stronger than $\le$, in the sense that the relation $\le'$ is contained in the relation $\le$.
– YCor
Dec 29 '13 at 21:49
• Let's restate the definition: say that $g\equiv g'$ if $g\le g'\le g$, and $g\equiv\!\!\!\!\!/\;g'$ otherwise. If $g\neq g'$, then there exists $h$ such that $gh\equiv\!\!\!\!\!/\;g'h$; we pick such $h$ minimal for $\prec$ and define $g\le'g'$ or $g\ge'g'$ according to whether $gh\le gh'$ or $gh\ge gh'$.
– YCor
Dec 29 '13 at 21:59
• But since $\le$ is right $H$-invariant, if $g\ne g'$ and $h\in H$, then $gh \equiv g'h$ iff $g \equiv g'$. Thus the minimal $h$ will always be the same (i.e., the $\prec$-minimal member of $H$). Perhaps you have a typo, though, and meant $h\in G$ instead of $h\in H$? I was assuming $H=\{ g : 1 \le g \le 1 \}$, as earlier in your post. Dec 30 '13 at 1:17