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 $G'$?


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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.

  • $\begingroup$ 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. $\endgroup$ Mar 2, 2019 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 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. $\endgroup$
    – YCor
    Mar 2, 2019 at 14:30

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