Is the property of left invariant orderability for finitely generated groups preserved by quasiisometrics? More precisely, if $G$ is a left orderable (finitely generated) group and $H$ is a torsionfree group quasiisometric (in the sense of Gromov) to $G$, can we conclude $H$ is left orderable? If the answer is no in the general case, what about 3manifold groups?

3$\begingroup$ $\mathbb Z$ and $\mathbb Z\times (\mathbb Z/2)$ are quasiisometric groups, both are fundamental groups of 3manifolds, $\mathbb Z$ is left orderable, but $\mathbb Z\times (\mathbb Z/2)$ is not. $\endgroup$ – Anton Petrunin Sep 7 '14 at 4:11

$\begingroup$ Thanks! I edited the question to exclude trivial counterexamples. $\endgroup$ – Mahdi Teymuri Garakani Sep 7 '14 at 4:32

9$\begingroup$ All closed hyperbolic 3manifold groups are quasiisometric; however there are both orderable and nonorderable such groups. $\endgroup$ – Ian Agol Sep 7 '14 at 4:39

4$\begingroup$ Note that all the previous examples are virtually leftorderable (VLO). There are candidates to show that for torsionfree groups, VLO is not QIinvariant. For instance, BurgerMozes groups and irreducible cocompact lattices in $SL_2(\mathbf{Q}_p)^2$ are QI to products of two free groups; still I don't know if they are all known to be not leftorderable. Same question for irreducible cocompact lattices in $SL_2(\mathbf{R})^2$, which are QI to products of 2 surface groups. $\endgroup$ – YCor Sep 7 '14 at 10:54

3$\begingroup$ finally I found examples not VLO: let $H,H'$ be two torsionfree, bilipschitz groups, with $H$ LO and $H'$ not LO and perfect (e.g. virtually abelian, or cocompact Kleinian). Then the wreath products $G=H\wr\mathbf{Z}$ and $G'=H'\wr\mathbf{Z}$ are QI, but $G$ is LO while $G'$ is not VLO. $\endgroup$ – YCor Sep 7 '14 at 21:03
This answer adds some details, and some references, following Ian's comment.
First the references:
https://arxiv.org/abs/math/0211110, https://arxiv.org/abs/math/0203192v2
Now, for the answer: In the Boyer, Rolfsen, Wiest paper (first reference) we find:
Theorem 1.8 For each of the eight 3dimensional geometries, there exist closed, connected, orientable 3manifolds with the given geometric structure whose fundamental groups are leftorderable. There are also closed, connected, orientable 3manifolds with the given geometric structure whose groups are not leftorderable.
As Ian mentions, for any pair of closed threemanifolds $M$ and $N$ with the same Thurston geometry, the fundamental groups $\pi_1(M)$ and $\pi_1(N)$ are quasiisometric.
In the second reference, by Calegari and Dunfield, there is a table of closed hyperbolic rational homology threespheres, of low volume, with the orderability of the fundamental group given as orderable, nonorderable, or unknown. It looks like this is a difficult property to determine. Also, in this situation, orderability appears to be rare (but possible!).
There are also lots of amenable examples: by a theorem of P. Linnell and D. WitteMorris (in this paper), an amenable group is leftorderable if and only if it is locally indicable (i.e. any nontrivial subgroup surjects onto $\mathbb Z$). It is not hard to see that this implies that any crystallographic group with nonsolvable holonomy is not leftorderable; on the other hand any such group contains a free abelian group of finite index, and the latter is certainly orderable. One can also get examples more complicated olycyclic examples.
A simpler crystallographic example is given by the subgroup of ${\rm Isom}(\mathbb{R}^3)$ generated by three halfturns with disjoint axes of orthogonal directions (its abelianization is finite; in fact it has a presentation given by $$ \langle a,b  ab^2a^{1}=b^{2},\, ba^2b^{1} = a^{2}\rangle $$ (see for example the last section in this paper of Bowditch).