# Profinite completions

I call a profinite group $G$ Noetherian, if evrey ascending chain of closed subgroups is eventually stable. A standart argument shows that every closed subgroup of a Noetherian profinite group is finitely generated.

A profinite group $G$ is called just-infinite if every nontrivial $M \lhd_c G$ is open.

Let $K$ be a profinite Noetherian just-infinite group. Must $K$ be the profinite completion of some residually finite group $R$?

• Since there are responses below, this must be well known, but I don't know the notation: what does $\triangleleft_c$ mean? Characteristic subgroup? Compact subgroup? – LSpice Jul 7 '14 at 21:54
• closed normal subgroup. – Pablo Jul 8 '14 at 7:51

The answer is yes in general.

Since $K$ is finitely generated, by the Nikolov-Segal theorem it coincides with its own profinite completion. So you simply may take $R=K$.

Perhaps, you also want $R$ to be finitely generated. In this case the answer is no.

If $K$ is a non-soluble p-adic analytic pro-$p$ group, then it has polynomial subgroup growth. However, finitely generated groups of polynomial subgroup growth are virtually soluble (see the book "Subgroup Growth" of Lubotzky and Segal).

• I wanted $R$ to be f.g as well, as you have said. Why is the answer no in this case? I didn't understand... – Pablo Jul 7 '14 at 16:00
• @Pablo: The profinite completion of a virtually soluble group is also virtually soluble and so it is soluble if it is also pro-p. – Andrei Jaikin Jul 7 '14 at 16:29
• Ok, great! I've got it. Do you know whether a f.g noetherian just-infinite profinite group can satisfy schreier's index formula? – Pablo Jul 7 '14 at 20:28
• @Pablo: $\mathbb Z_p$ – Andrei Jaikin Jul 8 '14 at 7:32
• Lubotzky and v.d.Dries give an example of a profinite group satisfying the formula with every prime appearing with finite multiplicity in its order (hence, sylow subgroups are finite, and the cohomology is $\infty$). Another strategy would be to take a free (f.g) profinite group $F$ and pick an open normal subgroup $N$ of index 2, and then take $M$ to be the maximal pro-$3$ kernel of $N$ (intersection of all open normal subgroups of $N$ of index a power of 3). Then, $F/M$ satisfies the formula (since $N/M$ does) and is not projective (since it has a finite subgroup of order $2$). – Pablo Jul 8 '14 at 8:05

It is a completely open problem whether Noetherian pro-$p$ group has finite rank, i.e., there is a bound on the number of generators of closed subgroups, which is equivalent to being $p$-adic analytic. As one can classify all just-infinite $p$-adic analytic pro-$p$ group if indeed any Noetherian pro-$p$ group is $p$-adic analytic, then all one needs to do for pro-$p$ groups is to go over the list. My gut feeling would be that the answer for $p$-adic analytic pro-$p$ groups is yes, but I am not 100% sure.