I am trying to prove the following statement: Let $L/K$ be a Galois extension with $Gal(L/K)\cong G$. Consider the subgroups $H,N \subset G$. Then $$L_{H\cap N}=L_H\cup L_N$$ where $L_H$ denotes the fixed field.
Here's what I have so far: the Fundamental Theorem of Galois Theory says that since $H,N \supset N\cap H$, $L_{N\cap H}\supset L_N,L_H\Rightarrow L_{N\cap H}\supset L_N\cup L_H$.
What I'm struggling to show is that $L_{N\cap H}\subset L_N\cup L_H$. If an element of $N\cap H$ fixes some element of $L$, surely there's no guarantee this would also be fixed by either all of $N$, or all of $H$?