Galois group of $Gal(L/E_1E_2)$

Question

Let $L/K$ be a Galois extension and $E_1/K$ and $E_2/K$ two subextensions. Let $G_i = Gal(L/E_i)$ for $i=1,2$. Is it true that $G_1 \cap G_2 = Gal(L/E_1E_2)$? Surely $E_1E_2$ is fixed by $G_1 \cap G_2$ but I'm not sure whether it is exactly the fixed field of $G_1 \cap G_2$.

Answer 1

Suppose $\sigma$ fixes the composite $E_1E_2$. Then $\sigma$ fixes $E_1$, so $\sigma\in G_1$. Similarly, $\sigma$ fixes $E_2$, so $\sigma\in G_2$. Thus $\sigma\in G_1\cap G_2$. You already understand that $G_1\cap G_2$ fixes $E_1E2$, so the claim has been proven.