# Is residual finiteness a quasi isometry invariant for f.g. groups?

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 $$G'$$?

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) 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 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.

• More generally if G and A are finite groups of the same cardinality with G non-abelian 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. Mar 2, 2019 at 14:19
• 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 torsion-free also fails to be QI-invariant, since $\mathbf{Z}\wr\mathbf{Z}$ is QI to $D_\infty\wr\mathbf{Z}$ and the latter is not virtually torsion-free, having infinite torsion subgroups.
– YCor
Mar 2, 2019 at 14:30