Absolute Galois group $\text{Gal}(\overline{K}/K)$ of any number field $K$ has a non-open subgroup of any prime index $p$?

Question

Let $K$ be a number field, and let $p$ be a prime number. Does $G = \text{Gal}(\overline{K}/K)$ necessarily have a subgroup of index $p$ that is not open?

Answer 1

Let $K$ be a number field and $G = \text{Gal}(\overline{K}/K)$ its absolute Galois group. The open subgroups of the profinite group $G$ are precisely the "lifts" of the identity $1\in\text{Gal}(L/K)$ (more precisely, kernels of projection maps $G\to \text{Gal}(L/K)$) for finite Galois extensions $L$, and there are countably many $K$-field extensions of any given degree, since $\mathbb{Q}$, hence also $K$, is countable. Thus, $G$ has countably many open subgroups.

Fix a prime $p$. We claim that $G$ has uncountably many subgroups of index $p$, so that at least one — in fact, uncountably many must be non-open, as desired.

To prove this, let $\omega\in \overline{K}$ be a primitive $p$th root of unity. We inductively construct a sequence$$q_1,\,q_2, \,\ldots\in\mathbb{Z}$$of prime integers such that for every $i \ge 1$, the prime $q_i$ is not a $p$th power in $$K\left(\omega,q_1^{1/p},\ldots,q_{i-1}^{1/p}\right),$$where for $i=1$, we just mean $K(\omega)$; for instance, we can just take any $q_i$ not dividing the discriminant of$$K\left(\omega,q_1^{1/p},\dots,q_{i-1}^{1/p}\right),$$so that any $\mathfrak{q} \mid q_i$ is unramified, and in particular $q_i$ is not a $p$th power. Then$$\left[K\left(\omega,q_1^{1/p},\dots,q_i^{1/p}\right):K\left(\omega,q_1^{1/p},\dots,q_{i-1}^{1/p}\right)\right] = p$$ for $i\ge1$.

Set$$K' = K\left(\omega,q_1^{1/p},q_2^{1/p},\dots\right) \subset \overline{K}.$$Since $\omega\in K'$, we have $K'/K$ Galois, with Galois group$$G' := \text{Gal}(K'/K) \simeq C_{p-1}\times \prod_{i\ge1} C_p \text{ (isomorphism of abelian groups);}$$any $K$-automorphism of $K'$ is uniquely determined by an "independent" choice of values at $\omega$, which has $p-1$ conjugates, and the $q_i^{1/p}$, which each have $p$ conjugates. The infinite product$$\prod_{i\ge1} C_p$$is an infinite-dimensional $\mathbb{F}_p$-vector space, so $G'$ has uncountably many subgroups of index $p$, one for each of uncountably (by a standard diagonalization argument) many $1$-dimensional subspaces of the vector space.

Finally, if$$\pi: G = \text{Gal}(\overline{K}/K)\to \text{Gal}(K'/K) = G'$$denotes the projection induced by the inverse limit — which in the case of Galois groups is given by restriction of $K$-automorphisms via $\iota_{K'\subset\overline{K}}$, since $K$-automorphisms of $\overline{K}$ are still $K$-automorphisms of $K'$ — then$$\left[G:\pi^{-1}(H')\right] = \left[G':H'\right]$$for any finite-index subgroup $H'$ of $G'$ — if$$\bigsqcup x'_j H' = G'$$and $x_j\in G$ is any "$\pi$-lift" of $x_j'$, then$$\bigsqcup x_j \pi^{-1}(H') = \bigsqcup \pi^{-1}(x'_j H') = \pi^{-1}(G') = G$$— so we are done.