$K/F$ Galois Extension, $H \leq G$, where $G$ the Galois group, then there is $\alpha \in K$ s.t. $H=\{\sigma \in G : \sigma \alpha = \alpha\}$.

Question

Let $K/F$ be a finite Galois extension of fields with Galois group $G$. Let $H$ be a subgroup of $G$. Then there is $\alpha \in K$ such that $H=\{\sigma \in G : \sigma \alpha = \alpha\}$.

My proof is this:

Consider the fixed field $E$ of $H$. Let $\alpha \in E$ be such that if $\alpha \in E' \subseteq K$ then $E\subseteq E'$. Then $H = \{\sigma \in G : \sigma \alpha = \alpha \}$ since if $\sigma \in H$, then sigma fixes $\alpha$. Conversely, if $\sigma \in G$ fixes alpha, then by choice of $\alpha$, $\sigma$ is contained in some subgroup $H'$ with fixed field $E'$ containing $\alpha$ and $E' \supseteq E$. In particular, since $E \subseteq E' \iff H' \subseteq H$, we have $\sigma \in H$. The result follows.

Does this work? I feel like I might be missing something since I (apparently) did not use the finiteness assumption on the field extension.

People have posted (better) solutions but I'm still interested in knowing whether what I came up with is correct.

Answer 1

You used the assumption of finiteness of the extension when you assumed that exist $ \alpha $ as you choose. The existence of such $ \alpha $ is ensured by the Primitive element theorem and not necessary true for infinite extensions. From then on, your solution is absolutely correct.

Answer 2

$E=K^H$, being a subextension of the Galois extension $K/F$ is a finite separable extension of $K$, hence by the Primitive element theorem, there exists $\alpha\in E$ such that $E=K(\alpha)$. Now, it isresults from the Galois correspondence that: $$H=\{\sigma\in G\mid \sigma(\alpha)=\alpha\}.$$