Let $K/F$ be a Galois extension. Also, let $E_1$ and $E_2$ be two Galois extensions of $F$ in $K$ with $E_1E_2 = K$ and $E_1 \cap E_2 = F$. Prove that $G = \text{Gal}(K/F) \cong \text{Gal}(K/E_1) \times \text{Gal}(K/E_2)$.
I think this is just an application of results of the Fundamental theorem of Galois theory (and maybe the Chinese Remainder Theorem?), but I'm uncertain on how to prove it.
Since we know $E_1/F$ and $E_2/F$ are Galois extensions, we have $\text{Gal}(K/E_1)$ and $\text{Gal}(K/E_1)$ are normal subgroups of $G$. We get $\text{Gal}(E_1/F) \cong G/\text{Gal}(K/E_1)$ and similarly $\text{Gal}(E_2/F) \cong G/\text{Gal}(K/E_2)$. I don't know what to do from here, or if these statements if even help. Thoughts?