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I recently asked this question about whether or not profinite groups admit maximal subgroups: And indeed, profinite groups admit subgroups of finite index, so taking any minimum index subgroup containing the finite index subgroup gives a maximal subgroup of the profinite group.

Now I wonder if it's true that all subgroups of a profinite group are contained in a maximal subgroup, i.e., can you have infinite chains of infinite index subgroups which do not split off to maximal subgroups periodically as you go up the chain?

Alex Byard
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1 Answers1

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Consider the integers as an additive subgroup of the $p$-adic integers: $\mathbb{Z}\subset\mathbb{Z}_p$ is not contained in its maximal subgroup $M=p\mathbb{Z}_p$. Let's show this maximal subgroup is unique:

Suppose $M<\mathbb{Z}_p$ is maximal, so $\mathbb{Z}_p/M=\langle x+M\rangle\cong\mathbb{Z}/q\mathbb{Z}$ for some prime $q$. If $q\ne p$ then $\frac{1}{q}x+M$ would have order $q^2$ in the quotient, a contradiction, so $q=p$. This implies $p\mathbb{Z}_p\subseteq M$, which forces $M=p\mathbb{Z}_p$ by transitivity of index.

coiso
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