Corollary 6

— For $i = 1, 2$ let $E_i$ be a subextension of $N$ and $\Gamma_i$ the Galois group of $N$ over $E_i$. The following conditions are equivalent: a) The group $\Gamma$ is the direct product of the subgroups $\Gamma_1$ and $\Gamma_2$. b) The extensions $E_1$ and $E_2$ are Galois over $K$, we have $E_1 \cap E_2 = K$ and

For $\Gamma$ to be the direct product of the subgroups $\Gamma_1$ and $\Gamma_2$ it is necessary and sufficient that the following conditions hold (I, p. 48, Prop. 15): (i) the subgroups $\Gamma_1$ and $\Gamma_2$ are normal in $\Gamma$; (ii) $\Gamma_1 \cap \Gamma_2 = \{ \varepsilon \}$, where $\varepsilon$ is the neutral element of $\Gamma$; (iii) $\Gamma = \Gamma_1 . \Gamma_2$.

Now (i) means that $E_1$ and $E_2$ are Galois over $K$ (Cor. 4). By Cor. 2, condition (ii) is equivalent to $N = K(E_1 \cup E_2)$, and finally if (i) and (ii) hold, $\Gamma_1 \Gamma_2$ is the least subgroup of $\Gamma$ containing $\Gamma_1 \cup \Gamma_2$; it is closed because $\Gamma_1$ and $\Gamma_2$ are compact and the mapping $(a, \tau) \mapsto \sigma \tau$ of $\Gamma_1 \times \Gamma_2$ into $\Gamma$ is continuous (Gen. Top., I, p. 63, Cor. 1). Cor. 2 now shows (iii) to be equivalent to $E_1 \cap E_2 = K$, and this proves the equivalence of a) and b).