TITLE: Let $E$ be finite dim ext of a field $F$and let $G$ be a gp of $F$-automorphisms of $E$ s.t $[E:F]=|G|$ Show that $F$ is the fixed field of $G$.
QUESTION [2 upvotes]: Question: Let $E$ be finite dimensional extension of a field $F$and let $G$ be a group of $F$-automorphisms of $E$ such that $[E:F]=|G|$.  Show that $F$ is the fixed field of $G$.
$\DeclareMathOperator{\Aut}{Aut}$

My Thoughts: Let $K=\mathcal F(\Aut(E/F))$, thus we have $\Aut(E/K)\subseteq\Aut(E/F)$, but since $F\subseteq K$, we have $\Aut(E/F)\subseteq\Aut(E/K)$, so we get equality. So,
$$
|\Aut(E/F)|=[E:F]\geq[E:K]=|\Aut(E/K)|=|\Aut(E/F)|.
$$
Thus $[E:F]=[E:K]$ so $K=F$ and so $F=\mathcal F(\Aut(E/F))$, and since $E$ is finite dimensional and $[E:F]=|G|$, we have $F=\mathcal F(G)$.
But I am questioning my first sentence and my last sentence.  I am trying to do this without resorting to too much Galois Theory, but any help is greatly appreciated!  Thank you.

REPLY [0 votes]: We can use the standard result that if $G$ is a finite subgroup of group of automorphisms of any field $E$ then the the fixed field $E_G$ of $G$ has the property that $[E:E_G] =|G|$. The proof of this result involves showing that $[E:E_G]$ is finite and does not exceed $|G|$. Once we know that the extension $E/E_G$ is finite we can apply a little bit of Galois theory to get $[E:E_G] =|G|$.
We are given that $F\subseteq E_G$ and thus we get $F\subseteq E_G\subseteq E$. Now you can use the tower law to get $$[E_G:F] =\frac{[E:F]} {[E:E_G]} =\frac{|G|} {|G|} =1$$ and thus $E_G=F$.
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