Let \(E/F\) be a Galois field extension, and let \(K/F\) be an intermediate field of \(E/F\). Show that \(K\) is normal over \(F\) iff \(\mathrm{Gal}(E/K) \trianglelefteq \mathrm{Gal}(E/F)\).
Solution
Let \[ G=\operatorname{Gal}(E/F), \qquad H=\operatorname{Gal}(E/K). \]
1. For every \(g\in G\),\[ gHg^{-1}=\operatorname{Gal}(E/gK). \]
An automorphism \(\sigma\in G\) lies in \(gHg^{-1}\) iff \[ g^{-1}\sigma g\in H, \] which means that \(g^{-1}\sigma g\) fixes every element of \(K\). Equivalently, \(\sigma\) fixes every element of \(gK\). Thus the two subgroups are equal.
2. If \(K/F\) is normal, then \(H\trianglelefteq G\).
Because \(E/F\) is Galois, every \(g\in G\) is an \(F\)-automorphism of \(E\). Normality of \(K/F\) implies \[ gK=K. \] Therefore 1 gives \[ gHg^{-1}=\operatorname{Gal}(E/gK)=\operatorname{Gal}(E/K)=H \] for every \(g\in G\).
3. If \(H\trianglelefteq G\), then \(K/F\) is normal.
By the fundamental theorem of Galois theory, \[ K=E^H. \] For \(g\in G\), 1 and normality of \(H\) give \[ \operatorname{Gal}(E/gK)=gHg^{-1}=H=\operatorname{Gal}(E/K). \] The Galois correspondence is injective, so \[ gK=K \] for every \(g\in G\).
Since \(E/F\) is normal, every \(F\)-embedding of \(K\) into an algebraic closure extends to an \(F\)-automorphism of \(E\). The preceding equality shows that every such embedding sends \(K\) onto itself. Hence \(K/F\) is normal.
Thus \[ K/F\text{ is normal} \iff \operatorname{Gal}(E/K)\trianglelefteq\operatorname{Gal}(E/F). \]