The Fundamental Theorem of Galois Theory states that in a Galois extension $K/F$, there is a bijection of subfields $E$ of $K$ containing $F$ and the subgroupsn $H$ of $G$. For any subgroup $N \leq\operatorname{Aut}(K/F)$, let $F^N$ denote the field containing $F$ fixed by $N$. I want to show the following:
Prove that if $N, M \leq\operatorname{Aut}(K/F) $ in a galois extension $K/F $, the fixed field of $\langle N, M \rangle $ is $F^N \cap F^M $
I managed to show that $\langle N, M \rangle \subset\operatorname{Aut}(K/F^N \cap F^M) $. However, I am having difficulty showing the other side of the inclusion. My difficulty is that given $\sigma \in\operatorname{Aut}(K/F^N \cap F^M)$ we have to show that somehow it's a finite product of elements of $N$ and $M$, or equivalently that $\sigma $ is contained in any subgroup of $\operatorname{Aut}(K/F) $ containing both $N $ and $M $. And we cannot conclude hastily that $\sigma $ fixes either $F^N $ or $F^M $.
The Fundamental theorem states that the bijection is also inclusion-reversing (if $H \subset H' \subset \text{Aut}(K/F)$ then $F^H \supset F^{H'}$ and if $F \subset E_1 \subset E_2 \subset K$ then Aut$(K/E_2) \subset$ Aut$(K/E_1)$).
Therefore we can prove the claim in the following way: $\langle N,M \rangle$ fixes all elements in $F^N \cap F^M$ because each element of $N$ and each element of $M$ do so. Therefore $F^{\langle N,M \rangle} \supset F^N \cap F^M$. On the other hand the fixed field of $\langle N,M \rangle$ has to be a sub-field of $F^N$ and $F^M$ because $\langle N,M \rangle$ is a bigger subgroup of Aut$(K/F)$ than $N$ and $M$. The fixed field $F^{\langle N,M \rangle}$ must therefore also be a subset of $F^N \cap F^M$. This gives the desired equality.