Does every countable, infinite, amenable, periodic group $G$ contain an infinite locally finite subgroup?


  1. I would be happy with an infinitely generated counterexample as long as it is countable.

  2. A counterexample cannot be elementary amenable, as elementary amenable periodic groups are locally finite.

  3. First Grigorchuk's group is not a counterexample: it is a 2-group, and hence it contains an infinite abelian group (by a result of Held, "On abelian subgroups of infinite 2-groups").

  4. Amenability of $G$ is essential, due to existence of Tarski monsters whose subgroups are finite cyclic (constructed by Olshanskii).

  5. Any infinite locally finite group contains an infinite abelian subgroup (see a paper of Hall-Kulatilaka here, so it is equivalent to ask whether $G$ contains an infinite abelian subgroup.

  6. This question was asked here in 2008, so perhaps it is an open problem.

  • 3
    $\begingroup$ certainly (1) does not help: indeed if your group is not locally finite, it contains an infinite f.g. subgroup and hence this subgroup also satisfies your requirements. $\endgroup$
    – YCor
    Jun 1, 2012 at 17:14
  • $\begingroup$ As far as I remember, Grigorchuk also constructs f.g. $p$-groups for $p \neq 2$ of intermediate growth. Could these serve as counterexamples? $\endgroup$ Jun 1, 2012 at 18:42
  • $\begingroup$ @Yves, thank you. @Ashot, I do not know the literature, which is why I ask here; there are of course other periodic amenable groups. $\endgroup$ Jun 1, 2012 at 19:10

1 Answer 1


@Ashot, Grigorchuk $p$-groups are branch, so one can easily construct infinite abelian subgroups as well. As generators one can just take elements from rigid stabilizers of different vertices, such that none of these vertices is a prefix of the other. So these groups don't provide counterexamples.

  • $\begingroup$ I meant adding this as a comment, but couldn't find the link to do it. $\endgroup$ Jun 2, 2012 at 12:05
  • $\begingroup$ @Dima: I think, you need reputation $\ge 50$ to comment. $\endgroup$
    – Misha
    Jun 2, 2012 at 13:08
  • $\begingroup$ @Dima: thank you, and welcome to mathoverflow. I think you can leave comments only after you earn enough reputation points; not sure how many is needed. $\endgroup$ Jun 2, 2012 at 13:08
  • $\begingroup$ @Dima: thanks for your comment. This gives an argument that does not involve Held's theorem. $\endgroup$ Jun 4, 2012 at 13:52

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