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According to the Wikipedia article on the Tits alternative,

a group $G$ is said to satisfy the Tits alternative if for every subgroup $H$ of $G$ either $H$ is virtually solvable or $H$ contains a nonabelian free subgroup.

It is claimed in the article that the Grigorchuk group does not satisfy the Tits alternative. However, no reference is given; there's only this link to the the article on the group, which does not itself have any mention of the Tits alternative.

So . . .

I need a reference, please, for the Grigorchuk group not satisfying the Tits alternative.

To be honest, I have somewhat black-boxed this group in my research. I need the reference for an introduction to my PhD.

Please help :)

NB: I might ask another, very similar question soon about the Thompson's group $F$, so if you have a reference for why that doesn't satisfy the Tits alternative either, you could save us both the time of writing up an answer/question.

Shaun
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  • Can you verify that Grigorchuk groups contain no nonabelian free subgroups? 2. Can you verify that Grigorchuk groups are non-solvable? 3. If both are too hard, read about growth of groups.
    • – Moishe Kohan Oct 08 '19 at 15:21
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    There is a MathOverflow answer discussing why Thompson’s group $F$ doesn’t satisfy the Tits alternative. There are some introductory notes on the group you may be interested in. – Santana Afton Oct 08 '19 at 16:13
  • This question might help you. – J.-E. Pin Oct 08 '19 at 16:25
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    It is a torsion group, so contains no nonabelian free groups, which is mentioned, and is not solvable(actually torsion solvable groups are finite). –  Oct 08 '19 at 16:32
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    Your definition of the Tits alternative is unnecessarily complicated. There is no need to mention subgroups. It satisfies it if either $G$ is virtually solvable or if $G$ has a nonabelian free subgroup. – Derek Holt Oct 08 '19 at 16:42
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    @PaulPlummer Do you mean finitely generated torsion solvable groups? I've been wondering if I'm missing something in my understanding of solvability – Robbie Lyman Oct 09 '19 at 02:02
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    @RyleeLyman Yes—my bad! –  Oct 09 '19 at 02:07
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    @DerekHolt I think the OPs def is fine, and is actually the one I see more frequently. I guess the reason being there are groups which contains nonabelian free groups but also subgroups which don't satisfy the Tits alternative in the sense you define. In the linear case, proved by Tits, this distinction isn't necessary as subgroups of linear groups are linear. –  Oct 09 '19 at 17:02