Why is $Gal(\Omega/k)$ a topological group under the Krull topology?

Question

For an infinite Galois extension $\Omega/k$ the Krull topology on $G:=Gal(\Omega/k)$ is defined by taking as a basis for the neighbourhood of an element $\sigma \in G$ all cosets of the form $\sigma Gal(\Omega/K)$ as $K$ runs through all finite Galois extensions of $k$ which are contained in $\Omega$. In showing that the group operations of $G$ turn it into a topological group one has to show that the multiplication is continuous. This amounts to proving that for any two elements $\sigma,\tau \in G$ and a finite extension $K$ as above one has $$ \sigma Gal(\Omega/K)\tau Gal(\Omega/K) \subseteq \sigma \tau Gal(\Omega/K).$$ This is then equivalent to showing that $\tau^{-1}Gal(\Omega/K)\tau \subseteq Gal(\Omega/K)$ but I cannot think of a proof of this. It would hold if we knew that the subgroup $Gal(\Omega/K)\subset G$ is normal but I also do not see why this is obvious.

Answer 1

This follows from the fact that $K/k$ is Galois. Let $K$ be the splitting field of some polynomial $f$. Then $\tau \in G$ must permute the roots of $f$, showing that $\tau(K) = K$. Thus, for an arbitrary $y\in K$ and $g\in \operatorname{Gal}(\Omega/K)$ , we have $\tau(y) \in K$ so that $\tau^{-1}(g\tau(y)) = \tau^{-1}(\tau(y)) = y$. This shows that $\operatorname{Gal}(\Omega/K)$ is normal in $G$, as desired.