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Apr 3, 2016 at 16:15 vote accept Stefan Kohl
Apr 3, 2016 at 16:15 answer added Stefan Kohl timeline score: 10
Feb 19, 2013 at 13:43 comment added YCor @Agol: evidence for what?
Feb 19, 2013 at 9:20 history edited Stefan Kohl
Added tag 'open-problem'.
Feb 19, 2013 at 2:36 comment added Ian Agol Some evidence for: arxiv.org/abs/1105.0719
Feb 18, 2013 at 21:05 comment added YCor Btw here's part of the argument for Erschler's result: take for granted that there exists a f.g. group $G$ with a central subgroup $Z$ which is an infinite-dimensional vector space over the field on $p$ elements, such that $G/Z$ has subexponential growth (actually in the example, $p=2$ and $G$ is the 1st Grigorchuk group). If $Z'$ is a hyperplane of $Z$ s.t. $G/Z'$ is residually finite, then there exists a normal subgroup of finite index $N$ such that $Z'=N\cap Z$. There are countably many such $N$ and hence such $Z'$. On the other hand, there are uncountably many $Z'$, so one $G/Z'$ isn't RF.
Feb 18, 2013 at 20:49 comment added Stefan Kohl @Ives: I have edited the question, providing the reference to Erschler's result.
Feb 18, 2013 at 20:47 history edited Stefan Kohl CC BY-SA 3.0
Added the reference pointed out by Mark Sapir on the negative answer regarding the stronger assertion.
Feb 18, 2013 at 20:08 comment added YCor @Stefan: you could maybe edit the question accordingly. On the other hand, it remains an open question whether there exists a nontrivial f.g. group of subexponential growth with no nontrivial finite quotient. (The derived subgroups of full topological groups of infinite minimal subshifts, which are infinite, f.g., amenable and simple, have free subsemigroups by Matui-2011 so have an exponential growth).
Feb 18, 2013 at 17:15 comment added user6976 Not every finitely generated group of subexponential growth isresidually finite (Erschler). The answer to the first question is not known.
Feb 18, 2013 at 17:03 history asked Stefan Kohl CC BY-SA 3.0