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I recently asked this question about whether or not every subgroup of a profinite group is contained in a maximal subgroup. The answer is no, and a counterexample is given by $\mathbb{Z}<\mathbb{Z}_p$. I've noticed that $\mathbb{Z}$ is dense in $\mathbb{Z}_p$, but $\mathbb{Z}\neq\mathbb{Z}_p$. In particular, this means that $\mathbb{Z}$ is not closed in $\mathbb{Z}_p$. Is it true that every closed subgroup of a profinite group is contained in a maximal subgroup?

Note: A maximal subgroup must be either closed and nondense or open and dense: Indeed, if it is closed and dense, then it is the whole group, and if it is open and nondense, then its closure lies above it.

Alex Byard
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    You mean "closed and non-dense or non-closed and dense". There may be subgroups which are neither closed nor open! – Alex Kruckman Jul 14 '23 at 22:30
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    Please note that open subgroups are always closed, so open and dense implies being the whole group, which means it can't be maximal. I don't know what the general answer to your question is at the moment, but if I were to guess I would say it's no. However in the specific case of the p-adic integers atleast, if I'm not mistaken, all of its closed subgroups are of the form $p^n \mathbb{Z}_p$ for $n \ge 0$. So in that particular case, the answer is positive. – Pedro Lourenço Jul 15 '23 at 12:29

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Suppose $N$ is a proper closed subgroup of the profinite group $G$. Then $G/N$ is non-trivial and profinite (as I pointed out here), so it has a maximal subgroup $M$ of finite index (as Eric explained here). Letting $\pi\colon G\to G/N$ be the quotient map, $\pi^{-1}(M)$ is a proper subgroup of $G$ of finite index containing $N$. Let $M'$ be a subgroup of $G$ of minimal finite index among those proper subgroups containing $N$. Then $M'$ is a maximal subgroup of $G$.

Alex Kruckman
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