Galois group of the compositum of two Galois extensions

Theorem 1.

Let $E$ and $F$ be Galois extensions of a field $K$. Then:

1. The intersection $E\cap F$ is Galois over $K$.
   

2. The compositum $EF$ is Galois over $K$. Moreover, the Galois group $\mathrm{Gal}(EF/K)$ is isomorphic to the subgroup $H$ of the direct product $G=\mathrm{Gal}(E/K)\times \mathrm{Gal}(F/K)$ given by:

   $$H=\{(\sigma ,\psi ):{\sigma |}_{E\cap F}={\psi |}_{E\cap F}\}$$

   i. e. $H$ consists of pairs of elements of $G$ whose restrictions to $E\cap F$ are equal.

Corollary 1.

Let $E$ and $F$ be Galois extensions of a field $K$ such that $E\mathrm{\cap }F\mathrm{=}K$. Then $E\mathit{}F$ is Galois over $K$ and the Galois group is isomorphic to the direct product:

$$\mathrm{Gal}(EF/K)\cong \mathrm{Gal}(E/K)\times \mathrm{Gal}(F/K).$$