Show that $L_N$ is the Galois closure of $K$

Question

Let $L$ be a Galois extension of $F$ and $K$ be an intermediate field of $L|F$. Let $N =\bigcap_{g∈Gal(L|F )} gGal(L|K)g^{−1}$. I wish to show that $L_N$ is the Galois closure of $K$. I know a Galois closure of $L/F$ is a field $L$ that is a Galois extension of $F$ and is minimal in that respect but I am not sure how to prove. Let $H$ be $Gal(L|K)$ Suppose $\tau\in H. $So $\sigma^{-1}\tau\sigma$ fixes $K$. For any $\sigma\in G$, $\tau\in\sigma H\sigma^{-1}\Leftrightarrow \sigma^{-1}\tau\sigma\in H\Leftrightarrow \sigma^{-1}\tau\sigma$ fixes $K$. So, $\forall a\in K$, we have $\sigma(a)\in N$? Not sure if my notation right even so feel free to use different notations~

Answer 1

First of all note that $N \lhd \text{Gal}(L/K)$, as $N$ is the normal core of the group $\text{Gal}(L/K)$. In fact this is the largest normal subgroup of $\text{Gal}(L/K)$, containing every other normal subgroup of it. It's not hard to prove this and I recommend you doing this if you haven't done it already.

Anyway back to the problem. We have that $K \subseteq L$ is a Galois extension. Also by the Fundamental Theorem of Galois Theory we have that $K \subseteq L_N \subseteq L$. Also as $N \lhd \text{Gal}(L/K)$ we have that the extension $K \subseteq L_N$ is a Galois extension.

Now it remains to show it's the minimal Galois extension. Assume that another field $M$ is the Galois closure of $K$. Then we have $\text{Gal}(M/K)$ is a normal subgroup of $\text{Gal}(L/K)$. However all such subgroups are contained in $N$ and thus we have that:

$$\text{Gal}(M/K) \subseteq N \implies L_N \subseteq M$$

However $M$ is Galois closure of $K$ and so we must have that $M \subseteq L_N$. Combining these two results we obtain what we wanted.