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


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*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:


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*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?
 A: 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. 
