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I am looking for simple examples of finitely generated residually finite group $G$ with a subgroup of finite index $H<G$ isomorphic to $G$. There are virtually nilpotent groups with this property, lamplighter groups $\mathbb{Z}_p\wr\mathbb{Z}$, Baumslag-Solitar groups. I am interested in "essentially different" examples.

Is there a just-infinite group $G$ with a subgroup of finite index isomorphic to $G$ and which is not virtually nilpotent?

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  • $\begingroup$ The main way I know of getting such things are self-replicating self-similar groups acting essentially freely on the boundary. Can such guys be just infinite? $\endgroup$
    – Benjamin Steinberg
    Feb 5, 2015 at 2:37
  • $\begingroup$ There are only a few known constructions of self-replicating self-similar groups acting essentially free. These are basically self-similar actions of lamplighter groups or Baumslag-Solitar groups and some generalizations. I suspect that the only hereditary just-infinite groups with isomorphic subgroup of finite index are $\mathbb{Z}$ and $D_{\infty}$. Also the Grigorchuk group does not contain a subgroup of finite index isomorphic to the whole group. $\endgroup$
    – Ievgen Bondarenko
    Feb 5, 2015 at 6:18
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    $\begingroup$ A closely related question, known to be open, is whether there exists a f.g. group $G$ with an endomorphism $f$ with image of finite index, with $\bigcap f^n(G)=1$, and $G$ not virtually nilpotent. $\endgroup$
    – YCor
    Feb 5, 2015 at 9:45

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I wanted to post this as a comment, but for some reason I coudn't. If your group is in addition profinite and not virtually-abelian the asnwer is no by Theorem E of [1]. The theorem says that a non-(virtually abelian) just infinite profinite group has no proper open subgroups isomorphic to the whole group.

Moreover I think that your group cannot be non-(virtually abelian) branch. In fact, the profinite completion of a just infinite branch group is just infinite and you should be able to apply Theorem E above.

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