Corollary 4

Let E be a subfield of N containing K and let $A = g(E)$. For E to be Galois over K it is necessary and sufficient that A should be a normal subgroup of $\Gamma$. When this is so, the restriction homomorphism of $\Gamma$ into $\mathrm{Gal}(E/K)$ defines by passage to quotients a topological group isomorphism of $\Gamma / \Delta$ onto $\mathrm{Gal}(E/K)$.

Since N is separable over K, the same is true of E (V, p. 36, Prop. 1). Therefore E is Galois over K if and only if it is quasi-Galois over K; this also means that $\sigma(E) = E$ for every K-automorphism $\sigma$ of N (V, p. 52, Prop. 1 and p. 54, Prop. 3). By Cor. 3 this is equivalent to $\sigma \Delta \sigma^{-1} = A$ for all $\sigma \in \Gamma$.

The restriction homomorphism $\varphi : \mathrm{Gal}(N/K) \to \mathrm{Gal}(E/K)$ is continuous and surjective (V, p. 60, Prop. 3) and its kernel is clearly equal to $A = \mathrm{Gal}(N/E)$. Since $\Gamma$ is compact, the homomorphism of $\Gamma / \Delta$ onto $\mathrm{Gal}(E/K)$ derived from $\varphi$ by passage to quotients is an isomorphism of topological groups (Gen. Top., I, p. 87, Cor. 2).