Given that $f(x)=g(x)h(x)\in F[x]$ such that $E/F$ is a splitting field of $f(x)$ and $B/F$ and $C/F$ are splitting fields of $g(x)$ and $h(x)$ respectively and both contained in $E$ . $B\cap C=F$.
Then $$Gal(E/F) \cong Gal(B/F)\times Gal(C/F)$$ Now to arrive at this result I have to prove two equalities and ,combined, they will give the desired result .
They are
$$1)\ \ Gal(E/B)\cap Gal(E/C)=Gal(E/{B\lor C})$$ $$2)\ \ Gal(E/B)\lor Gal(E/C)=Gal(E/B\cap C)$$
For the first one if $$\sigma \in Gal(E/{B\lor C})$$ then $\sigma $ is an automorphism that acts as identity on $B\lor C$ which is the compositum of $B$ and $C$ and contains both of them so must act on $B$ and $C$ as identity . So $$\sigma \in Gal(E/B)\cap Gal(E/C)$$ and $$Gal(E/{B\lor C})\subseteq Gal(E/B)\cap Gal(E/C)$$
Conversely let $$\tau \in Gal(E/B)\cap Gal(E/C)$$ Then $\tau $ acts as identity on both $B$ and $C$ and hence on their compositum as compositum is generated by the union $B\cup C$ so $$\tau \in Gal(E/{B\lor C})$$ Thus $$ Gal(E/B)\cap Gal(E/C)\subseteq Gal(E/{B\lor C})$$
or, $$ Gal(E/B)\cap Gal(E/C)= Gal(E/{B\lor C})$$
Is there any flaw in my logic $?$
Then for the second one I am stuck . If $$\sigma \in Gal(E/B)\lor Gal(E/C)$$ then $$\sigma=\sum_{i=1}^{k}\gamma_{i} +\sum_{i=1}^{m}\tau_{i}\ \ where\ \ \gamma_{i}\in Gal(E/B)\ \ and\ \ \tau_{i}\in Gal(E/C)$$ Then clearly $\sigma$ fixes the elements that are in $B\cap C$ . So I have $$ Gal(E/B)\lor Gal(E/C)\subseteq Gal(E/B\cap C)$$
Here I have almost reached the end but I cannot prove the converse of this inequality . I need help with that .
Thanks.