Corollary 5

Let E be a subfield of N containing K. For E to have finite degree over K it is necessary and sufficient that g(E) should be open in $\Gamma$. When this is so, the index $(\Gamma : g(E))$ is finite and equal to [E : K].

For $g(E)$ to be open it is necessary and sufficient that there should exist a subextension F of N, of finite degree over K, such that in the notation of V, p. 60.

g(E) contains $U_F(\mathrm{Id}_N) = g(F)$. The relation $g(E) \supset g(F)$ is equivalent to $E \subset F$ by Cor. 1 (V, p. 68), whence the first assertion of Cor. 5.

Suppose that $[E : K]$ is finite. Let $\Omega$ be an algebraic closure of $K$ containing $N$ as subextension (V, p. 23, Th. 2) and let $\mathscr{H}$ be the set of $K$-homomorphisms of $E$ into $\Omega$. Every element of $\mathscr{H}$ is induced by a $K$-automorphism of $\Omega$ (V, p. 52, Prop. 1), and since $N$ is quasi-Galois over $K$, the mapping $\sigma \mapsto \sigma|_E$ of $\Gamma$ into $\mathscr{H}$ is surjective. For $\sigma$ and $\sigma'$ in $\Gamma$ to have the same restriction to $E$ it is necessary and sufficient that $\sigma^{-1}\sigma' \in g(E)$, whence Card $A? = (\Gamma : g(E))$. Finally since $E$ is an etale algebra over $K$, we have Card $\mathscr{H} = [E : K]$ (V, p. 32, Prop. 4), so in conclusion we have $(\Gamma : g(E)) = [E : K]$.