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Let $G$ be a finitely generated profinite group, and $p$ a prime number. Suppose that there exists some open pro-$p$ subgroup $H \leq_o G$. Must $G$ have only finitely many maximal open subgroups?

Note that the case $H = G$ is trivial: A pro-$p$ group of rank $d$ has $\leq \frac{p^d-1}{p-1} < \infty$ maximal open subgroups simply because this is the number of subspaces of codimension $1$ in an $\mathbb{F}_p$-space of dimension $d$, and every maximal open subgroup of a pro-$p$ group is indeed a normal subgroup of index $p$.

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  • $\begingroup$ Yes, because the simple quotients of $G$ have bounded order. (Indeed, letting $K$ be a normal pro-$p$-subgroup of finite index, those finite quotients $S$ that are not of order $p$ have cardinal $\le [G,K]$ since the image of $K$ in $S$ has to be trivial.) $\endgroup$
    – YCor
    Commented May 31, 2015 at 17:49
  • $\begingroup$ @YCor I am asking about maximal subgroups which are not necessarily normal. $\endgroup$
    – Pablo
    Commented May 31, 2015 at 18:40
  • $\begingroup$ OK, my comment was for maximal normal subgroups. I answer below for maximal subgroups. $\endgroup$
    – YCor
    Commented May 31, 2015 at 19:46

2 Answers 2

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Yes, because $G$ is top. f.g. and its maximal subgroups have bounded index.

Let me prove the latter fact. Let $K$ be the core (=intersection of conjugates) of $H$, so $K$ is an open normal pro-$p$-subgroup. Let $M$ be a maximal open subgroup in $G$. Then either $MK=M$ or $MK=G$. In the first case, $K\subset M$, which bounds the index of $M$.

So now assume $MK=G$. Let $N$ be the core of $M$. Write $G'=G/N$ and also denote with primes the images of the various subgroups in $G'$; since $N\subset M$ we have $M'\neq G'$, so $M'$ is maximal in $G'$. Also $G'=M'K'$, so necessarily $K'$ is nontrivial. Let $A$ be the (nontrivial) center of the $p$-group $K'$; it is characteristic in the normal subgroup $K'$ and hence is normal in $G'$. Then $A\cap M'$ is normalized by both $M'$ and by $N'$, hence it is normal in $G'$. Since the core of $M'$ is trivial, we deduce that $A\cap M'=1$. By maximality of $M'$, we deduce that $G'=M'\ltimes A$. The centralizer of $A$ in $M'$ is normal in $G'$ and again since the core of $M'$ is trivial it has to be trivial. Thus $K'\cap M'=1$, and since $K'$ contains $A$ this forces $K'=A$. Thus $$|M'|=[G':K']\le [G:K] $$ Hence $M'$ has bounded cardinal. Since by maximality $M'$ acts irreducibly on $A$ (in the sense that $A\neq 0$ and $A$ has no nontrivial $M'$-invariant subgroup), $A$ is $p$-elementary, and has dimension $\le |M'|$, hence cardinal dividing $p^{[G:K]}$. Since the cardinal of $A$ equals the index of $M'$, we deduce that $[G:M]=[G':M']$ divides $p^{[G:K]}$, so is bounded.

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  • $\begingroup$ PS the proof can be extended to the case when $H$ has only finitely many possible isomorphism types of Jordan-Hölder simple factors. $\endgroup$
    – YCor
    Commented May 31, 2015 at 19:53
  • $\begingroup$ Your proof is very illuminating! In an answer below I give a different argument I have found. However, I do not see how you can handle the case of finitely many simple factors - the nontriviality of the center seems crucial in your argument. For example, how can you treat the case $|H|$ is divisible only by $2$ distinct primes? $\endgroup$
    – Pablo
    Commented Jun 8, 2015 at 8:55
  • $\begingroup$ If you include in the question the case of finitely many types of Jordan-Hölder simple factors I'll maybe try to add the generalized proof (indeed we don't have the centrality fact, but it's adaptable). $\endgroup$
    – YCor
    Commented Jun 8, 2015 at 9:46
  • $\begingroup$ I think that it is not adaptable: Let $\mathcal{C}$ be the variety of finite groups with order not divisible by primes greater than $3$. This variety is closed under every possible operation including extensions. By Burnside, $\mathcal{C}$ consists of solvable groups (with composition factors $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$). Take $G = H$ to be the free pro-$\mathcal{C}$ group on two generators. By Corollary 8.7.5 in Profinite Groups by Ribes and Zalesskii, the Frattini subgroup of $G$ is trivial, so $G$ has infinitely may maximal subgroups. $\endgroup$
    – Pablo
    Commented Jun 8, 2015 at 11:46
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    $\begingroup$ OK you're right. Explicitly: let $F(q)$ be a field of prime power order $q$ and $C(n)$ be a cyclic group of order $n$. Then for $n\ge 2$, $F(3^{2^n})$ has an element of multiplicative order $2^{n+1}$ acting $F(3)$-irreducibly on $F(3^{2^n})$. Then the semidirect product $C(2^{n+1})\ltimes F(3^{2^n})$ is generated by 2 elements, admits $C(2^{n+1})$ as a maximal subgroup. So lifting this to the free group on 2 generators in this variety, we get an explicit maximal open subgroup of index $2^{n+1}$ for any $n+1$. $\endgroup$
    – YCor
    Commented Jun 8, 2015 at 13:57
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Let me answer the question in the title, which is equivalent to the one appearing in the body. We may assume that $H \lhd_o G$ for otherwise we replace it with its core.The positive answer to the question follows from the following proposition:

Let $G$ be a profinite group, and let $H \lhd_c G$ be a closed normal subgroup. Then $\varPhi(H) \leq_c \varPhi(G)$.

Proof: Towards a contradiction, suppose that $M \leq_o G$ is a maximal subgroup not containing $\varPhi(H)$. Since $\varPhi(H)$ is a characteristic subgroup of $H$ and $H \lhd_c G$, we conclude that $\varPhi(H) \lhd_c G$, so $\varPhi(H)M = G$ as $M$ is maximal. Hence, for each $h \in H$ we have $h = am$ for some $a \in \varPhi(H) \leq H$ and $m \in M$. It follows that $m \in H$ as well, so $H = \varPhi(H)(H \cap M)$. A subgroup complements the Frattini subgroup only if it is the whole group, so $H \cap M = H$. Thus, $\varPhi(H) \leq_c H \leq_c M$ which is a contradiction. $\blacksquare$

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