Exercise 7.3.11

Proof. As $N=N_G(H)$ is the largest subgroup of $G=\mathrm{Gal}(L∕F)$ such that $H=\mathrm{Gal}(L∕K)$ is normal in $N$, since the Galois correspondance reverse inclusions, $L_N$ is the smallest subfield of $K=L_H$ such that the extension $L_N\subset K$ is normal. We give the details.

$\text{∙}$ Write $H=\mathrm{Gal}(L∕K)$. Then $L_H=K$. Since $H\subset N$, then $L_H\supset L_N$, so $L_N$ is a subfield of $K$.

$\text{∙}$ $H$ is a normal subgroup of $N=N_G(H)$. Therefore the extension $L_N\subset L_H=K$ is normal (Theorem 7.3.2).

$\text{∙}$ Let $F\subset M\subset K$ be an intermediate field, such that $M\subset K$ is a Galois extension.

Let $S=\mathrm{Gal}(L∕M)$. $S$ is a subgroup of $G=\mathrm{Gal}(L∕F)$ since $F\subset M\subset K$.

The extension $M\subset K$ is normal. Therefore the subgroup $H=\mathrm{Gal}(L∕K)$ is normal in $S=\mathrm{Gal}(L∕M)$ (Theorem 7.3.2). Since the normalizer $N=N_G(H)$ is the largest subgroup of $G$ with this property, we conclude $S=\mathrm{Gal}(L∕M)\subset N$, therefore $M=L_S\supset L_N$.

Conclusion: $L_N$ is the smallest subfield of $K$ such that $K$ is Galois over the subfield. □