Intermediate field, normal closure and Galois group Let $K/F$ be Galois with $G=Gal(K/F)$ and let $L$ be an intermediate field. Let $N\subseteq K$ be the normal closure of $L/F$. If $H=Gal(K/L)$ show that $Gal(K/N)=\cap_{\sigma\in G}\sigma H\sigma^{-1}$. (Exercise $8$, page $60$, Field and Galois Theory, Patrick Morandi.)
Help me a hint to prove it. Thanks a lot.
 A: Here is my proof.
For convenience, let $M$ denote the group $\cap_{\sigma\in G} \sigma H\sigma^{-1}$. The notation $M(a)=a$ means that $a$ is fixed under the action of each element in $M$.
First we prove that $Gal(K/ N) \supseteq M$.
It suffices to prove that $N$ is fixed by the group $M$.
For any $a\in N$, since $N$ is the normal closure of $L/F$,
there exists some $\sigma\in G$ such that $\sigma^{-1} (a) \in L$.
Hence $H\sigma^{-1}(a)=\sigma^{-1}(a)$ because $L$ is fixed by $H$.
Therefore, we see that $\sigma H\sigma^{-1}(a) = a,$
hence $M(a)=a$ because $M$ is contained in  $\sigma H\sigma^{-1}$ by definition.
Note that $a$ is arbitrary in $N$, hence $N$ is fixed by $M$.
Next we prove that $Gal(K/ N) \subseteq M$.
Suppose $\tau\in Gal(K/N)$, for any $\sigma\in G$,
$\tau\in\sigma H\sigma^{-1}\Leftrightarrow \sigma^{-1}\tau\sigma\in H\Leftrightarrow \sigma^{-1}\tau\sigma$ fixes $L$. 
So we only need to prove that 
$\sigma^{-1}\tau\sigma$ fixes $L$.
$\forall a\in L$,
we have $\sigma(a)\in N$ since $N$ is the normal closure of $L/F$.
Because $\tau$ fixes $N$,
we deduce that
$\sigma^{-1}\tau\sigma(a) = \sigma^{-1}\sigma(a)=a$,
which means that $\sigma^{-1}\tau\sigma$ fixes $L$, or $L$ is fixed by $\sigma^{-1}\tau\sigma$.
We are done.
