$K/F$ is an algebraic Galois extension. If the fix field of a subgroup $G$ of $Gal(K/F)$ is equal to $F$, is it true that $G=Gal(K/F)$?

Question

By a famous theorem in field theorey (Dedekind's Lemma), when $K/F$ is a finite field extension such that $K/F$ is Galois and $G$ is a subgroup of $Gal(K/F)$ such that the fix field of $G$ is equal to $F$, then $G=Gal(K/F)$. Now, Assume that $K/F$ is an algebraic field extension such that $K/F$ is Galois. Also, consider that $G$ is a subgroup of $Gal(K/F)$. If the fix field of $G$ is equal to $F$, is it true that $G=Gal(K/F)$?

Answer 1

I believe that this is false. Let $\Gamma = \textrm{Gal}(K/F)$. Then $\Gamma$ has a natural topology for which it is a profinite group. There is an order reversing bijection between intermediate fields of $K/F$ and closed subgroups of $\Gamma$, given by $E' \mapsto \textrm{Gal}(K/E')$. The following lemma follows directly from this fact, without having to know anything about how the topology is defined.

Lemma: Let $H$ be a subgroup of $\Gamma = \textrm{Gal}(K/F)$, and let $E$ be an intermediate field of $K/F$. Then $\textrm{Gal}(K/E)$ is the closure of $H$ in $\Gamma$ if and only if $E$ is the fixed field of $H$.

Now assume that $G$ is a subgroup of $\textrm{Gal}(K/F)$ whose fixed field is $F$. The lemma implies that the closure of $G$ is equal to $\textrm{Gal}(K/F)$. So to find a counterexample, it suffices to find a proper dense subgroup of $\textrm{Gal}(K/F)$.

I don't know to produce a proper dense subgroup $\textrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, but they almost certainly exist. For a more hands on example, take the Galois group of the algebraic closure of any finite field. The Galois group is isomorphic to

$$\prod\limits_p \mathbb{Z}_p$$

where $p$ runs through the prime numbers and $\mathbb{Z}_p$ is the completion of $\mathbb{Z}$ with respect to the $p$-adic topology. The diagonal embedding of $\mathbb{Z}$ into this group is dense.