Given a finite extension $E/k$ of a field $k$, how do I prove the following? $|\operatorname{Gal}(E/k)| \text{ divides } [E:k].$ Thanks.
Galois group of finite extensions
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$\begingroup$
abstract-algebra
galois-theory
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0@ÁlvaroLozano-Robledo, I am sorry. I was mistaken. Gal just means Aut. – 2012-03-19
1 Answers
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The first thing that comes to mind is the following. It's a theorem of Artin that if $G$ is a finite group of automorphisms of $E$ then $E$ is Galois over the fixed field $E^G$ with Galois group $G$. See Corollary 3.5 and the surrounding paragraphs in Milne's notes. Since \[ [E : k] = [E : E^G][E^G : k] = |G|[E^G : k] \] this gives the result.
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0@ Dylan, that was nice. Thank you. – 2012-03-19