Let $E/K$ be a Galois extension and let $F$ be an intermediate field such that $K\subseteq F\subseteq E$. Then $E/F$ is a Galois extension too and $H=\mbox{Gal}(E/F)$ is a closed subgroup of $G=\mbox{Gal}(E/K)$. Then we have two topologies for $H$: The subspace topology by considering $H$ as a subspace of $G$ (which I call the subgroup topology) and the Krull topology by considering $H$ as the Galois group of the extension $E/F$. In both topologies, $H$ is clearly a profinite group.
I believe that both topologies on $H$ have to be the same. Let $$ \mathscr{V} = \{L : L/F \text{ is a finite Galois extension and } F\subseteq L\subseteq E\}. $$ Then $\mathscr{V}$ is an open neighborhood basis at $1$ of $H$. In order to prove that the subgroup topology on $H$ coincides with the Krull topology on $H$, it suffices to prove that $\mathscr{V}$ is an open neighborhood basis at $1$ in the subgroup topology.
It is easy to prove that given $U$, an open neighborhood of $1$ in the subspace topology of $H$, there exists a finite Galois extension $L/F$ contained in $E$ such that $\mbox{Gal}(E/L)\subseteq U$, so in order to conclude we only need to prove that the sets $\mbox{Gal}(E/L)$, with $L/F$ a finite Galois extension, are open in the subgroup topology.
I wanted to prove that we can write $$ \mbox{Gal}(E/L) = \mbox{Gal}(E/F)\cap \mbox{Gal}(E/M) = \mbox{Gal}(E/FM) $$ where $M/K$ is a finite Galois extension contained in $E$. But in this case this implies (by Galois correspondence) that $L=FM$, and I don't believe that this always works (or maybe it does but I don't know how to prove it).
So if you can give me a hint telling me how I can prove that both topologies are the same (or a counterexample if it is not true), that would be awesome.
Maybe a proof using the charaterization of Galois group as a projective limit would be interseting too (and I would love to know such a proof!).