Skip to main content
10 events
when toggle format what by license comment
May 2, 2020 at 17:26 answer added YCor timeline score: 4
May 2, 2020 at 16:50 history edited Isaac CC BY-SA 4.0
edited title
May 2, 2020 at 13:35 history edited Isaac CC BY-SA 4.0
added 3 characters in body
May 1, 2020 at 22:41 comment added YCor Yes it has no free subgroup (as the whole group of continuous piecewise homographies of $P^1$ fixing $\infty$) — sorry I thought I wrote it
May 1, 2020 at 22:37 comment added Isaac Let us continue this discussion in chat.
May 1, 2020 at 22:24 comment added Isaac And how do you know it doesn’t contain a free group?
May 1, 2020 at 22:22 comment added YCor Then the group of piecewise homographies constructed by Monod and Lodha-Moore should work. Indeed I think each such group $G$ (which is finitely presented and non-amenable) has all its proper quotients metabelian (and $G''$ is simple), so it's not residually amenable. It's locally indicable, has no infinite Property T subgroup.
May 1, 2020 at 22:03 comment added Isaac What I really want is a finitely presented group that is not residually amenable and contains no free subgroup. Of course if the group has property T and is not residually finite then it is also not residually amenable.
May 1, 2020 at 21:55 comment added Isaac I want it not to be residually finite.
May 1, 2020 at 21:07 history asked Isaac CC BY-SA 4.0