Suppose I have fields $F \leq L_1 < L_2 \leq E$, where $E/F$ is separable, $[E : F] = p^n$ for a prime number $p$, and $L_2 / L_1$ is Galois and of degree $p$. In addition, I have a subfield $F \leq K \leq E$. Is it true then that $L_2 \cap K / L_1 \cap K$ is Galois (and of degree $1$ or $p$)?
I have proven this in the analogous case for groups, where you replace Galois-ness with normality. However, that proof relies on the Second Isomorphism Theorem, and I cannot find a way of adapting it to the case of fields. As a partial result, $E/F$ is separable, so $L_2 \cap K / L_1 \cap K$ is separable as well, so we only need to show it is also normal to get Galois.
Edit: This is my attempt at solving a more general question I asked here, so it may turn out that this statement is false if this isn't the correct approach.
$L_2 = \mathbb{Q}(\sqrt[3]{2},\zeta_3),L_1 = \mathbb{Q}(\zeta_3)$, $K= \mathbb{Q}(\sqrt[3]{2})=L_2\cap K, L_1 \cap K = \mathbb{Q}$ isn't a counter-example because you asked $[L_2:K \cap L_1] = p^m$
Assume $L_2 \cap K/ L_1\cap K$ is of degree $p$ . With $L_3/K \cap L_1$ its Galois closure, then $Gal(L_3/K \cap L_1)$ embeds in $S_p$ (permutation group) thus $[L_3:K \cap L_1] \ | \ p!$. But since $[E:L_3 :L_1 \cap K: F]$ is a power of $p$, then so is $[L_3:K \cap L_1]$ and $L_3 = L_2 \cap K$.
Thus it leaves us with the problem of classifying Galois extensions $L_3/L_1 \cap K$ of degree $p^{3+m}$ with a non-Galois subextension $L_2 \cap K/L_1 \cap K$ of degree $p^{2+l}$