Proving a subgroup of a Galois group is normal

Question

"Let $N/K$ be a finite Galois extension with Galois group $G = Gal(N/K)$.

Let $M$ be an intermediate field of the extension, let $E = Gal(N/M)$, and let $F = \cap_{\sigma \in G} σEσ^{-1}$ .

Show that $F$ is a normal subgroup of $G$"

Where do I even begin here?

I know for all $\sigma \in G$, $F \subset \sigma E \sigma ^{-1}$, so $\sigma ^{-1} F \sigma \subset E$, so $F$ is a subgroup of $E$ under conjugation? Is this true?

Now how do I prove $F$ is normal?

Answer 1

This is just an exercise n group theory: take a group $\;G\;$ , a subgroup $\;H\le G\;$ and form what's called the core of $\;H\;$ , merely: the subgroup $\;\bigcap_{x\in G}xHx^{-1}\;$ . You can read about this in the web, but we can also do as follows:

Let $\;X:=H\backslash G\;$ be the set of left cosets of $\;H\;$ in $\;G\;$ , and define an action of $\;G\;$ on $\;X\;$ as follows: $\;g\cdot xH:=(gx)H\;$ . It's easy to check this is indeed an action. As usual, this action determines a homomorphism $\;\phi:G\to Sym_X\cong S_n\;$ , when $\;n=[G:H]\;$ , by the rule $\;\phi g(xH):=\;(gx)H$ (you only have to convince yourself that the above actually determines a permutation on $\;Sym_X\;$ for any $\;g\in G$ ).

Well, it is now a nice exercise to prove:

$$\begin{align*}(1)\;\;&\ker\phi=\bigcap_{x\in G}xHx^{-1}\\{}\\(2)\;\; &\ker\phi\;\;\text{is the maximal normal subgroup of $\;G\;$ contained in $H$}\end{align*}$$

Give a try to the above two claims