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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).

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up vote 17 down vote accepted

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.

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Mark, is the supposed proof contained in that Thompson F preprint, or is it something separate? –  Yemon Choi Feb 1 '12 at 2:35
Thanks Mark.-- Justin –  Justin Moore Feb 1 '12 at 2:43
@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. –  Mark Sapir Feb 1 '12 at 2:55
@Mark: thank you for the information. –  Yemon Choi Feb 1 '12 at 3:00

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