The following is proposition 7.10 in Milne's Fields and Galois Theory:

Let $G$ be a group of automorphisms of a field $E$, and let $F=E^G$ (ie. $F$ is fixed field of $G$). If $G$ is compact and the stabilizer of each element of $E$ is open in $G$ (wrt Krull Topology), then $E$ is a Galois extension of $F$ with Galois group $G$.

The author starts by considering a finite subset of $E$ say $\{x_1,x_2, \cdots x_n\}$. Then using compactness, he argues that orbit of each $x_i$ is finite. Let $N$ be the intersection of all the conjugates of $H_i$'s where $H_i$ is stabilizer of $x_i$. Then by previous observations, its straightforward to see that $N$ is a normal open subgroup of $G$. Then the following claim is made:

The fixed field of $N$ is the subfield of $E$ that is generated over $F$ by the elements of orbit of $x_i$'s.

I am able to see that field generated over $F$ by the elements of orbit of $x_i$'s is a subset of the fixed field of $M$ but I can't show the converse. I tried many times. One approach which seemed promising was for show $E$ was algebraic extension of $F$, then take an arbitrary element $\alpha \in M$ ($M$ is fixed field of $N$) and consider its minimal polynomial $m_{\alpha}$ over $F$ and using similar arguments from finite Galois Theory, I showed $m_{\alpha}$ splits in $M$ into distinct linear factors. Thus $M$ is Galois over $F$. However this goes nowhere (and moreover I found a flaw in my reasoning where I showed $E$ is algebraic over $F$).

I would really appreciate a hint.

PS: I'm referring to version 5.10 of JS Milne's Fields and Galois Theory. The proposition is in pg. 97.



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