If $H$ is a subgroup of a finite group of automorphisms of a field $K$, there is $a\in K$ such that $H=G_a$.

Question

My problem is stated below.

Let $K$ be a field and let $G\leq Aut(K)$ be a finite group. Prove that for any $H\leq G$, there exists $a\in K$ such that $$H=G_a=\{\varphi\in G :\varphi(a)=a\}.$$

I have been thinking of using Galois theory because this problem was an exam problem of the corresponding chapter. However, since we don't know anything about $K$, I don't know how to start. I will thank for any help!

Answer 1

The extensions $K/K^G$ (the subfield fixed by $G$) and $K/K^H$ are Galois thus separable, so $K^H/K^G$ is separable too, thus $K^H = K^G(a)$ and $$H = Gal(K/K^H) = Gal(K/K^G(a))=G_a$$