Lemma 2

Let A be a subgroup of $\Gamma$. Let E be the field of invariants of A; then the Galois group of N over E is the closure of A in $\Gamma$.

The Galois group of N over E is closed in $\Gamma$ (Lemma 1) and contains A, hence it contains the closure $\bar{\Delta}$ of A. Let $\sigma$ be an E-automorphism of N and let $x_1, \ldots, x_n$ be in N. Since N is Galois over E (Lemma 1) there exist (V, p. 57, Prop. 2) a subextension $N_0$ of N, Galois of finite degree over E and containing $x_1, \ldots, x_n$. Let $A_0$ be the image of the subgroup A of $\mathrm{Gal}(N/E)$ under the restriction homomorphism of $\mathrm{Gal}(N/E)$ into $\mathrm{Gal}(N_0/E)$. Since $[N_0 : E]$ is finite, Dedekind's theorem (V, p. 27, Cor. 2) shows that $\mathrm{Gal}(N_0/E)$ is finite. Hence $A_0$ is finite, and since E is the field of invariants of $A_0$, we have $\Delta_0 =$

Gal(N_0/E) (V, p. 66, Th. 3). In particular, $\Delta_0$ contains the restriction of $\sigma$ to $N_0$. Therefore there exists $\tau \in A$ such that $\sigma$ and $\tau$ have the same restriction to $N_0$, whence $\sigma(x_1) = \tau(x_1), \ldots, \sigma(x_n) = \tau(x_n)$. It follows that $\sigma$ is a limit point of $A$ in $\Gamma$, and hence $\mathrm{Gal}(N/E) \subset \overline{\Delta}$.