# Groups $G$ with a subgroup $H$ of finite index isomorphic to $G$

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?

• 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? – Benjamin Steinberg Feb 5 '15 at 2:37
• 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. – Ievgen Bondarenko Feb 5 '15 at 6:18
• 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. – YCor Feb 5 '15 at 9:45

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.