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).

Obvious facts about total right-preorderability:

  • 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?

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    $\begingroup$ What you call "complete" pre-orders are also commonly called linear pre-orders, and also sometimes they are called total pre-orders. $\endgroup$ Nov 6 '13 at 19:23
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Nov 6 '13 at 19:31
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Nov 6 '13 at 19:33
  • $\begingroup$ 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". $\endgroup$ 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.

  • $\begingroup$ 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! $\endgroup$ Nov 7 '13 at 14:38
  • $\begingroup$ 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? $\endgroup$ Dec 29 '13 at 21:40
  • $\begingroup$ $\le'$ is stronger than $\le$, in the sense that the relation $\le'$ is contained in the relation $\le$. $\endgroup$
    – YCor
    Dec 29 '13 at 21:49
  • $\begingroup$ 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 $gh\ge gh'$. $\endgroup$
    – YCor
    Dec 29 '13 at 21:59
  • $\begingroup$ 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. $\endgroup$ Dec 30 '13 at 1:17

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