If $K,E$ are subfields of $\Omega/F$ then $KE/F$ is a finite Galois imply $K/K\cap E$ is Galois? Let $\Omega/F$ be a field extension and $K,E$ be two subfields of
$\Omega/F$. Assume that $KE/F$ is a finite Galois.
I have a theorem in my lecture notes that claim $\text{Gal}(KE/E)\cong \text{Gal}(K/K\cap E)$,
while it is easy to see that $KE/F$ is Galois $\implies KE/E$ is
Galois I can not understand why $K/K\cap E$ is Galois (it is separable
since $KE/F$ is, but I don't see why it's normal).
Why $K/K\cap E$ is Galois ?
 A: You need to assume that $K$ is Galois over $K\cap E$; if that is the case, then the isomorphism drops out of the Galois correspondence and the Isomorphism Theorems.
We can replace $F$ with $K\cap E$, so that we are in the following situation:


*

*$KE/F$ is finite Galois;

*$K\cap E = F$.


If $G=\mathrm{Gal}(KE/F)$, let $M$ be the subgroup corresponding to $K$ and $N$ the subgroup corresponding to $E$. Then $M\cap N=\{e\}$ (since $KE$ is the field "on top"), and $\langle M,N\rangle = G$.
Now, $K$ is Galois over $F$ if and only if $M$ is normal in $\langle M,N\rangle$. 
In particular, if $K$ is Galois over $F$, then $M\triangleleft \langle M,N\rangle$, so $\langle M,N\rangle = MN$. Thus, $N\cap M$ is normal in $N$, and  by the isomorphism theorems we have that
$$\frac{G}{M} =\frac{MN}{M} \cong \frac{N}{N\cap M} \cong N.$$
Now, $\frac{G}{M}\cong\mathrm{Gal}(K/F)$; and $N=\mathrm{Gal}(KE/E)$; so we get the isomorphism if $K$ is Galois.
For an example showing that the given conditions do not imply that $K$ is Galois over $K\cap E$, let $F=\mathbb{Q}$, $K=\mathbb{Q}[\sqrt[3]{2}]$, and $E=\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive cubic root of unity; then $K\cap E=\mathbb{Q}$. Then $KE$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, hence is Galois. Even though $KE$ is Galois over $E$, $K$ is not Galois over $\mathbb{Q}$, so you cannot have the claimed isomorphism.
