0
$\begingroup$

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 $p$.

$\endgroup$
  • $\begingroup$ 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 May 31 '15 at 17:49
  • $\begingroup$ @YCor I am asking about maximal subgroups which are not necessarily normal. $\endgroup$ – Pablo May 31 '15 at 18:40
  • $\begingroup$ OK, my comment was for maximal normal subgroups. I answer below for maximal subgroups. $\endgroup$ – YCor May 31 '15 at 19:46
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 May 31 '15 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 Jun 8 '15 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 Jun 8 '15 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 Jun 8 '15 at 11:46
  • 1
    $\begingroup$ 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 Jun 8 '15 at 13:57
1
$\begingroup$

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. $\blacksquare$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.