A "residually finite group" is group for which the intersection of all finite index subgroups is trivial. Suppose $G$ and $G'$ are two quasiisometric finitely generated groups. Does the residual finiteness of $G$ implies the same property for $G'$?

2$\begingroup$ Actually this was already answered inside the related question of Misha here: mathoverflow.net/questions/136431/… $\endgroup$ – YCor Mar 2 at 11:57
No: let $Q$ be a nonabelian group of order 8. Then the standard lamplighter groups $(\mathbf{Z}/2\mathbf{Z})\wr\mathbf{Z}$ (which is RF) and the wreath product $Q\wr\mathbf{Z}$ (which is not RF: exercise; initially due to Gruenberg 1957) are QI.
Indeed, $(\mathbf{Z}/2\mathbf{Z})\wr\mathbf{Z}$ has a unique normal subgroup of index 3, isomorphic to $(\mathbf{Z}/2\mathbf{Z})^3\wr\mathbf{Z}$, and the latter shares a (nonlabeled) Cayley graph with $Q\wr\mathbf{Z}$.
Also, BurgerMozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.
Also, various finitebyRF f.g. groups are known not to be RF: examples of Deligne and then Raghunathan were mentioned many times here too; Erschler (J. Algebra 2004, Sciencedirect link) produced many examples too in the context of branched groups.
One more recent example: Adrien Le Boudec (arXiv link) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated nonabelian free group, then $C\wr F$ (which is residually finite if $C$ is abelian) is quasiisometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QIinvariant.

$\begingroup$ More generally if G and A are finite groups of the same cardinality with G nonabelian and A abelian and you take the Cayley graphs of $G\wr \mathbb Z$ and $A\wr \mathbb Z$ using the generating sets $ht$ where $h$ runs over the finite group and $t$ generates the infinite cyclic group then the Cayley graphs are isomorphic but one group is residually finite and the other is not. $\endgroup$ – Benjamin Steinberg Mar 2 at 14:19

$\begingroup$ Yep. One further generalization (due to Erschler too, IMRN 2000) is that if $H,L$ are bilipschitz f.g. groups and $C$ is another one, then $H\wr C$ and $L\wr C$ are QI (and actually bilipschitz). While, in this setting, and assuming $C$ infinite residually finite, $H\wr C$ is RF iff $H$ is abelian. (Gruenberg 1957). As a consequence, virtually torsionfree also fails to be QIinvariant, since $\mathbf{Z}\wr\mathbf{Z}$ is QI to $D_\infty\wr\mathbf{Z}$ and the latter is not virtually torsionfree, having infinite torsion subgroups. $\endgroup$ – YCor Mar 2 at 14:30