Is the property of not containing the free group on two generators invariant under quasi-isometry? Amenability is, so if there is a counterexample it is also a solution to the von Neumann-Day problem (which of course already has a solution).


1 Answer 1


It is a famous open problem. Akhmedov in MR2424177 claimed he could prove that the answer is "no". No proof exists, so I guess he discovered a gap in his argument.

  • $\begingroup$ Mark, is the supposed proof contained in that Thompson F preprint, or is it something separate? $\endgroup$
    – Yemon Choi
    Feb 1, 2012 at 2:35
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    $\begingroup$ @Yemon: That is separate. The paper MR2424177 (see MathSci) actually contains the claim, but proves a much weaker (still nice, though!) result where "free subgroups" are replaced by "free subsemigroups" or "no non-trivial law". He says that the "big example" will be in the sequel of that paper but the sequel never happened. $\endgroup$
    – user6976
    Feb 1, 2012 at 2:55
  • $\begingroup$ @Mark: thank you for the information. $\endgroup$
    – Yemon Choi
    Feb 1, 2012 at 3:00
  • $\begingroup$ @YemonChoi do you know if that "or" can be taken for two separate statements, or for one; i.e. are "no free subsemigroups" QI-invariant and "no non-trivial law" QI-invariant, or is "no free subsemigroups and no non-trivial law" QI-invariant? (sorry for asking, I don't have access to the paper) $\endgroup$
    – ARG
    Apr 21, 2021 at 11:04
  • $\begingroup$ @ARG I'm afraid I never looked at the paper which Mark mentions, and I don't have immediate access to it although I can probably get hold of it through my university's VPN or similar if you need $\endgroup$
    – Yemon Choi
    Apr 21, 2021 at 13:00

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