Let $F/\mathbb{Q}$ be a finite galois extension, and $Aut(F/\mathbb{Q})$ be the corresponding galois group. What I want to show is that for all intermediate fields between $F$ and $\mathbb{Q}$, where $\mathbb{Q} \subset F_{1} \subset F_{2} \cdots \subset F_{m} = F$, if the field extensions are normal, then the corresponding subgroups of $Aut(F/\mathbb{Q})$ are also normal.
What I thought about was, because we can represent $F$ as a vector space over $\mathbb{Q}$, we can map every element in $Aut(F/\mathbb{Q})$ to matrices of $|F|$ dimension.
My question is, if all the field extensions are normal, can we describe elements of $Aut(F/\mathbb{Q})$ as diagonal matrices?
Because since all diagonal matrices are commutable, for any intermediate field, we would have $g\cdot Aut(F/F_{k}) = Aut(F/F_{k}) \cdot g$ for $g \in Aut(F/F_{k})$, hence $Aut(F/F_{k})$ is a normal subgroup of $Aut(F/F_{k})$.
What I'm currently not sure about is whether or not diagonal matrices can represent all possible automorphisms of $F$ that fixes $\mathbb{Q}$.
If this is a valid argument, is it because the normalcy of the field extensions implies that all automorphisms on the field can be written as a diagonal matrix?
If not, is it possible to prove the normalcy of subgroups using this line of thought?
No, in general the automorphisms of $F$ are not diagonalizable. In fact, most of them are not: If $n=[F:\mathbb{Q}]$ and $\sigma\in{\rm Gal}(F/\mathbb{Q})$ has eigenvalue $\lambda\neq 1$ with eigenvector $x$, then $x=\sigma^n(x)=\lambda^nx$, so $\lambda^n=1$, but the only roots of unity in $\mathbb{Q}$ are $\pm1$. That means that $x=\sqrt{a}$ for some nonsquare $a\in\mathbb{Q}$, so for example as soon as $[F:\mathbb{Q}]$ is odd, no nontrivial automorphism of $F$ is diagonalizable.
Also, note that ${\rm Gal}(F_{k+1}/F_k)$ is not at all a subgroup of ${\rm Gal}(F/\mathbb{Q})$.