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I happened upon the question if there are infinite groups which don't have any infinite abelian subgroups. I am pretty sure that this is true, but couldn't come up with any examples. So I am looking for an example.

My thoughts until now:

The group can only have elemets of finite index or else it has the integers as a subgroup (free group generated by element with infinite index, which is abelian).

It is probably not going to be an algebraic group by a theorem of Pillay on the existence of infinite abelian subgroups of o-minimal groups. So maybe one has to do something analytic, maybe a Liegroup?

Edit: I am interested in any subgroups, not necessairily proper subgroups, therefore this already excludes all infinite abelian groups as counterexamples.

linkja
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    One quick example is a Tarski monster, which has no infinite proper subgroups at all. – Steve D Jan 07 '25 at 10:25
  • Does $\langle a,b,\dots|a^2,b^2,\dots\rangle$ work? – Karl Jan 07 '25 at 10:32
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    @Karl but doesn't $ab$ have infinite index? – linkja Jan 07 '25 at 10:44
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    @SteveD why don't you post your comment as an answer? – ckefa Jan 07 '25 at 10:57
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    See https://en.wikipedia.org/wiki/Tarski_monster_group – Nicky Hekster Jan 07 '25 at 13:38
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    This question is similar to: How to prove ''any proper subgroup of $G={z \in \Bbb C~:~z^{2^n}=1,~n=0,1,2,3,...}$ is finite and cyclic''. More precisely: that post answers your question. – Anne Bauval Jan 07 '25 at 16:01
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    I must say I strongly disagree with this question being closed as a duplicate, and am thus voting to reopen. First, because it could only be a duplicate if we make the assumption that by "subgroup", the author really means "proper subgroup". Second, because this question is, IMHO, simply not a duplicate. This question is (much) broader than the question it supposedly duplicates, in particular, it can be answered with other examples that are completely unrelated to the other question. – sTertooy Jan 07 '25 at 19:28
  • Well, I don't agree. If X has an answer over at Y, we should close X as a duplicate of Y. This also has been recently confirmed by moderator Xander. But yeah not all mods follow this procedure, apparently, let alone the rest of the users. – Martin Brandenburg Jan 07 '25 at 22:14
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    Just to be clear: this question does not have an answer at the linked question, unless we make the (so far) baseless assumption the OP meant "proper subgroup". – Steve D Jan 07 '25 at 22:18
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    @MartinBrandenburg I'm interested in hearing the reasoning behind your (and Xander's) stance. To not clutter up the comments section of this question any further (and to perhaps gather more opinions and viewpoints), I've posted a question on Meta about this. – sTertooy Jan 07 '25 at 23:31
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    To return to the question, I don't know whether the Grigorchuk group has infinite abelian subgroups, but the centralizers of its elements have been investigated, so the answer is probably known. – Derek Holt Jan 08 '25 at 08:51

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