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

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~

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.