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



Is residual finiteness a quasi isometry invariant for f.g. groups?
No: let $Q$ be a non-abelian 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 (non-labeled) Cayley graph with $Q\wr\mathbf{Z}$.


Also, Burger-Mozes groups are QI to products of 2 free groups, but I guess this example was mentioned various times on this site.


Also, various finite-by-RF 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](https://www.sciencedirect.com/science/article/pii/S0021869303006410)) produced many examples too in the context of branched groups.


One more recent example: Adrien Le Boudec ([arXiv link](https://arxiv.org/abs/1802.04736)) proved that if $C$ is a nontrivial finite group and $F$ a finitely generated non-abelian free group, then $C\wr F$ (which is residually finite if $C$ is abelian) is quasi-isometric to some finitely generated simple group. The latter also shows that having finite amenable radical is not a QI-invariant.


