Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence

For $x\in K$, the element $x$ is fixed by $\sigma H{\sigma }^{-1}$ exactly when ${\sigma }^{-1}(x)$ is fixed by $H$, exactly when ${\sigma }^{-1}(x)\in E$, and exactly when $x\in \sigma (E)$. Thus $K^{\sigma H{\sigma }^{-1}}=\sigma (E)$, and [L1] gives $\mathrm{Gal}(K\mathrm{/}\sigma (E))=\sigma H{\sigma }^{-1}$.

For the forward direction, if $H⊴G$, then step 1.1 gives $\sigma (E)=E$ for every $\sigma \in G$; every $F$-conjugate of an element of $E$ is obtained by extending its embedding to $K$ and hence lies in $E$, so $E\mathrm{/}F$ is normal, and it is separable as a subextension of the separable extension $K\mathrm{/}F$, hence Galois. For the reverse direction, if $E\mathrm{/}F$ is Galois, normality gives $\sigma (E)=E$ for every $\sigma \in G$, so step 1.1 and [L1] give $\sigma H{\sigma }^{-1}=H$ and [F1] gives $H⊴G$.

In the normal case, restriction $\rho :G\to \mathrm{Gal}(E\mathrm{/}F)$ is defined by step 2.1. Every $F$-automorphism of $E$ extends to an embedding of $K$ in an algebraic closure; normality of $K\mathrm{/}F$ makes the extension an element of $G$, so $\rho$ is surjective. Its kernel consists exactly of the automorphisms fixing $E$, namely $H$ by [L1]. The first isomorphism theorem therefore gives $G\mathrm{/}H\cong \mathrm{Gal}(E\mathrm{/}F)$; for $H=\{1\}$ and $H=G$ this yields the two endpoint quotients.