(Thanks to Harry Richman for bringing this example up.)

The free group and the right-angled Coxeter group have the same Cayley graph—the infinite 4-valent tree—; hence they are quasi-isometric. We note here that quasi-isometries are *not* required to group homomorphisms, and indeed the one we have here—an “identity map” of sorts (in an admittedly very loose and imprecise sense) between Cayley graphs—is not.

By results of Stallings and Dunwoody involving ends of groups ad accessibility (although Chapter 20 of Drutu-Kapovich’s draft manuscript seems to be the only reference I can find for this online), free groups are quasi-isometrically rigid, and hence, since finite-index subgroups of free groups are free, the RACG is virtually free (!)

With a little more thought we can explicitly exhibit a finite-index free subgroup: the subgroup of the RACG generated by *ab*,* ac* and *ad* is a nonabelian free group on three generators; we can verify it has finite index via a covering space argument.

Now I wonder if there is a finite-index -subgroup of the RACG as well, so that the commensurability need not involve passing to finite-index subgroups on both sides. It seems like there shouldn’t be, although I can’t quite prove this yet.

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