Suppose we have an extension of fields $F ⊆ E ⊆ L ⊆ K$ and let $G = Aut(K/F)$ with subgroups $1 ≤ J ≤ H ≤ G$. We have the functions $G(L) = Aut(K/L)$ (subgroup generated by L) and $F(H) = \{α ∈ K : ϕ(α) = α$ for all $ϕ ∈ H\}$ (fixed field of H).
I want to prove the inclusions $G(L) ≤ G(E)$ and $F(H) ⊆ F(J)$.
Could I start with:
Let $\alpha \in G(L)$ and $\phi \in E$, then $\phi (\alpha)=\alpha$
So $\alpha \in G(E)$ .
But I feel like this doesn't make sense. Any help is greatly appreciated.
if any mapping fixes a set $E$ elementwise, then a fortiori it fixes any subset of $E$ elementwise.
this simple observation is the basis of both the inclusions you mention. so you mainly need to convince yourself (using the terms 'smaller' and 'larger' in the sense of inclusions) that a smaller field is fixed by a larger group, and that a smaller group fixes a larger field.
this reversal is sometimes described by saying that the Galois connection between groups and fields is contravariant