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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.

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  • $\begingroup$ Correct me if I'm wrong, but did you not use the finiteness assumption to apply Galois Correspondence? $\endgroup$
    – Andrew
    Jul 19 '15 at 0:40
  • $\begingroup$ How do you an $\alpha$ with this property exists? $\endgroup$
    – Bernard
    Jul 19 '15 at 0:44
  • $\begingroup$ @ Bernard. It's clear it exists. If $F$ is any subgroup of $K$ with $E$ not contained in $F$ and $F$ not contained in $E$ then $E-F$ is nonempty. Also, there has to be an $\alpha$ in $E$ not contained in any proper subfield of $E$ since $E$ is strictly larger than any proper subfield. $\endgroup$
    – TuoTuo
    Jul 19 '15 at 0:47
  • $\begingroup$ @ Andrew If it is the case that the Fundamental Theorem of Galois Theory is only true for finite extensions, then yes, I suppose I did. $\endgroup$
    – TuoTuo
    Jul 19 '15 at 0:48
  • $\begingroup$ $H$, being a subgroup of $G$, represents an intermediate field $F'$ with $F\subseteq F' \subseteq K$. Since these are all Galois extensions, then $F'$ is generated by a single element over $F$ (by the primitive element theorem). This is your $\alpha$. $\endgroup$
    – Arthur
    Jul 19 '15 at 0:53
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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.

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$E=K^H$, 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: $$H=\{\sigma\in G\mid \sigma(\alpha)=\alpha\}.$$ 

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