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I have two questions about residually amenable groups:

  1. Is every finitely presented amenable group residually elementary amenable?
  2. Given $n$, is the free Burnside group of exponent $n$ on two generators residually amenable?

Regarding 1., I know that Grigorchuk constructed an example of a finitely presented amenable group that is not elementary amenable, but I was unsure if it could be residually elementary amenable. Regarding 2., I believe it is an open question whether or not the free Burnside group on two generators is sofic, so if the answer to 2. is known, I'm guessing that it would be in the negative.

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    $\begingroup$ (1) No, there's a finitely presented ascending HNN extension of Grigorchuk's group, and it's an isolated group (it has a unique minimal nontrivial normal subgroup— more precisely every proper quotient is metabelian [Sapir-Wise]), so it's not residually elementary amenable. See details and references §5.7 in my 2007 paper with Guyot and Pitsch on isolated groups (arXiv link). $\endgroup$
    – YCor
    Commented May 8, 2020 at 22:31
  • $\begingroup$ For (2) this group does not exist: it depends on an exponent and the answer depends on the exponent. However I think that the only cases for which the answer is known are those for which it's known to be finite (exponent $1,2,3,4,6$). That is, my guess is that even for large odd $n$ the failure of residual amenability is unknown, but this should be confirmed by specialists. $\endgroup$
    – YCor
    Commented May 8, 2020 at 22:34
  • $\begingroup$ Thanks for the answer to (1). For (2), yes, I really did mean this family of groups. $\endgroup$
    – Isaac
    Commented May 9, 2020 at 3:05
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    $\begingroup$ Benjamin Weiss asked which free Burnside groups are sofic. jstor.org/stable/25051326 $\endgroup$
    – Ian Agol
    Commented May 9, 2020 at 3:39
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    $\begingroup$ @HJRW ok, I've just done so. $\endgroup$
    – YCor
    Commented Oct 2, 2021 at 11:08

3 Answers 3

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(From my initial comments)

  1. No, there's a finitely presented ascending HNN extension of Grigorchuk's group, and it's an isolated group (it has a unique minimal nontrivial normal subgroup— more precisely every proper quotient is metabelian [Sapir-Wise]), so it's not residually elementary amenable. See details and references §5.7 in my 2007 J. Algebra paper with Guyot and Pitsch on isolated groups ArXiv link.

  2. As confirmed by Mark Sapir for each given exponent and number $\ge 2$ of generators the question of residual amenability of the given Burnside group is open, except in the few cases where it's known to be locally finite (exponent $\le 4$ and $6$).

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Regarding "Regarding 1". (The) Grigorchuk group is residually finite. A non-residually finite amenable but not elementary amenable group was constructed by Anna Erschler in Not residually finite groups of intermediate growth, commensurability and non-geometricity. I do not know if her group is residually elementary amenable. Regarding 2: There is a conjecture due to Y. Shalom (see The algebraization of Kazhdan's property (T). International Congress of Mathematicians. Vol. II, 1283–1310, Eur. Math. Soc., Zürich, 2006.) that free Burnside groups have Kazhdan Property (T). Since infinite finitely generated Burnside groups are not residually finite (Zelmanov), Shalom's conjecture would imply that infinite finitely generated Burnside groups are not residually amenable. The Shalom conjecture has been disproved in Osajda, Damian Group cubization. With an appendix by Mikaël Pichot. Duke Math. J. 167 (2018), no. 6, 1049–1055.

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    $\begingroup$ Actually the disproof of this conjecture takes 3 lines after the state of knowledge in 2006... but wasn't observed then (Pichot made the observation), possibly because of the focus on prime exponent. Namely if there is an infinite f.g. group $G$ of exponent dividing $n$ and $r$ generator, then the wreath product $C_k\wr G$ has exponent dividing $kn$ and is generated by $r+1$ elements. And doesn't have Property T (easy) for $k\ge 2$. So the free Burnside on $r+1$ generators and exponent $kn$ doesn't have T for any $k\ge 2$. $\endgroup$
    – YCor
    Commented Jun 8, 2020 at 22:11
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    $\begingroup$ @YCor: All complains about Shalom's conjecture should be addressed to Shalom. $\endgroup$
    – user158834
    Commented Jun 8, 2020 at 22:22
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    $\begingroup$ it's not a complaint, it's a complement of information, addressed to the reader. In any case as the conjecture seems highly false, it seems quite irrelevant to the question (a positive solution to the conjecture would have provided a negative answer to (2)). $\endgroup$
    – YCor
    Commented Jun 8, 2020 at 22:24
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    $\begingroup$ What is "highly false" as opposite to simply "false"? It was a non-disproved conjecture till 2018. Your 3-line proof repeats (word-by-word) the appendix of Pichot in the paper by Osajda, except that his appendix has references and in general is respectful to Osajda. $\endgroup$
    – user158834
    Commented Jun 8, 2020 at 22:39
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    $\begingroup$ "Highly false" is subjective, so I won't argue. Yes Osajda first did it, so what? my comment is not a publication. I was among the ones who tried to disprove this conjecture and didn't manage to do so, although I was aware of the failure of Property T for wreath products, so I felt stupid when I heard Pichot's argument. $\endgroup$
    – YCor
    Commented Jun 8, 2020 at 22:44
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About 2), It is unknown if every bounded torsion amenable f.g. group is finite. If that was true then for free Burnside groups residually amenable would be equivalent to residually finite. By Zelmanov's result, for large enough exponents, the free Burnside group with more than 1 generators is not residually finite.

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