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Preliminary remark: I'm mainly interested in an answer (or link to ressources) in the specific context of the first Grigorchuk group, but I believe that it may be of some interest to state the question in the general setting.

Let $G$ be a finitely generated residually finite group endowed with the profinite topology and $\hat G$ it's profinite completion. When have to applications $$\Theta\colon Sub_{cl}(G)\to Sub_{cl}(\hat G), \quad H\mapsto \bar H$$ and $$\Psi\colon Sub_{cl}(\hat G)\to Sub_{cl}(G), \quad K\mapsto K\cap G$$ By Nikolov and Segal, when restricted to finite index subgroups, these maps are lattice isomorphisms which preserves the index, why for general subgroups we only have $$\Psi\cdot\Theta=Id\quad\text{and}\quad \Theta\cdot\Psi(K)\leq K$$

Question 1 Is it possible that $\Psi$ and $\Theta$ are lattice isomorphisms?

Remark Since $\hat G$ depends only on the lattice of finite index subgroups of $G$, this is equivalent to ask if it is possible to find $G$ and $H$ (both finitely generated residually finite) with isomorphic lattice of finite index subgroups but with non-isomorphic lattices of closed subgroups.

A positive answer to question $1$ may be too much to ask, so we will focus on a more restrictive situation. Suppose that $G$ is moreover a $p$-group, then $\hat G$ is a pro-$p$ group and every closed subgroup of infinite index in $G$ (respectively in $\hat G$) is contained in a maximal closed subgroup of infinite index of $G$ (respectively of $\hat G$) (nb. such a subgroup is maximal among closed subgroups of infinite index, not necessarily maximal among all proper closed subgroups). Called such maximal subgroups MCI.

Question 2 Let $M<G$ be a MCI. Is $\Theta(M)=\bar M$ a MCI of $\hat G$?

Question 3 Let $M<\hat G$ be a MCI. Is $\Psi(M)=M\cap G$ a MCI of $G$?

Final remark: as said before, I am particularly interested in answers to (or references concerning) the above questions in the particular case of the first Grigorchuk group. Some remarkable properties of this group that may be relevant for the discussion.

  1. It is a $2$-group
  2. It is a regular branch group
  3. It is just infinite and it's completion is strongly just infinite
  4. Every subgroup of $\hat G$ is either of finite index or of uncountable index
Improve this question
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  • $\begingroup$ Note: When $G$ is a $p$-group, its profinite completion is pro-$p$, and the theorem that finite index subgroups of f.g. pro-$p$ groups are open is an old theorem of Serre. $\endgroup$
    – YCor
    Commented Apr 10, 2018 at 12:03
  • $\begingroup$ (4) sounds surprising to me, what's a reference? $\endgroup$
    – YCor
    Commented Apr 10, 2018 at 12:09
  • $\begingroup$ Question 3 has a negative answer even for $\mathbf{Z}^2$: there are MCI subgroups in the profinite completion whose intersection with $\mathbf{Z}^2$ is reduced to $\{0\}$. $\endgroup$
    – YCor
    Commented Apr 10, 2018 at 12:12
  • $\begingroup$ @YCor The reference for (4) is "On strongly just infinite profinite branch groups" from Le Maître and Wesolek where they proved that for G a profinite branch groups the following are equivalent (among other properties) i) G is strongly just infinite ii) every countable index subgroup of G is open $\endgroup$
    – PHL
    Commented Apr 10, 2018 at 12:29
  • $\begingroup$ Also, thank you for the $\mathbf Z^2$ example $\endgroup$
    – PHL
    Commented Apr 10, 2018 at 12:32

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