MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.