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Is there any elementary argument showing that there exist uncountably many distinct quasi-isometry classes of elementary amenable groups? How about solvable groups?

For amenable groups it follows from the result of Grigorchuk (proved in the 80's) stating that there are uncountably many groups of intermediate growth with pairwise incomparable growth functions.

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Doesn't it follows from our paper on lacunary hyperbolic groups? The elementary amenable lacunary hyperbolic groups corresponding to sufficiently different sequences of parameters will be not quasi-isometric because their asymptotic cones corresponding to certain sequences of parameters will not be bi-Lipschitz equivalent (one cone will be a tree while another one won't).

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    $\begingroup$ @Denis: You are too young to be forgetting your own theorems. :) $\endgroup$
    – Bill Johnson
    Commented Dec 25, 2011 at 20:41
  • $\begingroup$ @Mark: I think it does not follow directly, although one can probably prove it this way. One problem is that there exist 2 elementary amenable groups G, H as in our paper (corresponding to different parameters) such that for any scaling sequence and any ultrafilter, the corresponding cone of at least one of G, H is an R-tree. So any third group cannot be distinguished from both G and H by means of asymptotic cones. It is not obvious to me how to choose those "sufficiently different sequences of parameters" to avoid this problem. $\endgroup$
    – Denis Osin
    Commented Dec 25, 2011 at 22:10
  • $\begingroup$ @Denis: It does not seem to be difficult: each group should have a sequence of relations of certain fast growing lengths while in all other groups this sequence misses all relations. I think it is easy. A solvable example is more difficult. One would have to deal with Abels' group or Kharlampovich's group and its factor-groups over central subgroups. Perhaps again asymptotic cones can help. But it is not clear. $\endgroup$
    – user6976
    Commented Dec 25, 2011 at 22:25
  • $\begingroup$ @Mark: Of course, each our group has a sequence of relations of certain fast growing lengths. But we do not know anything about the geometric shape of these relations in the Cayley graph, so it is not clear how to show that the corresponding asymptotic cone is not an R-tree. If we could do this, we could probably also decide whether or not all cones of our groups have cut points (which is an open problem in our paper). $\endgroup$
    – Denis Osin
    Commented Dec 25, 2011 at 22:30
  • $\begingroup$ @Denis: to any f.g. group $G$, you can associate the set $N(G)$ of ultrafilters $\omega$ such that $Cone(G,\omega,(1/n))$ is a real tree. So $N(G)$ is a QI-invariant of $G$. I think the idea is to show that elementary amenable groups achieve continuum many sets $N(G)$. It's not obvious (and I haven't checked) but I guess you can do this. $\endgroup$
    – YCor
    Commented Dec 25, 2011 at 22:33
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The affirmative answer in given by Yves de Cornulier and Romain Tessera in arXiv:1203.4696.

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