Problem: Let $E$ be a finite, separable normal extension of $F$. Suppose that $F \subset L \subset E$ s.t. $L$ is also normal over $F$. Prove that the elements of $G(E/F)$ map $L$ onto itself.
Definition of Normal I'm Using: We have that $E$ is normal over $F$ iff (i) $E$ is finite dimensional over $F$ and (ii) $G(E/F)$ has $F$ as its fixed field.
Attempt:
Let $(E/F) = n$ and $(E/L) = m$. We have that $m \le n$.
Let $\sigma \in G(E/F)$.
If I could show that $\sigma$, when restricted to $L$, is a member of $G(L/F)$ then I would be done.
I'm not sure how here to proceed.