Galois group of finite extensions

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.

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You mean, given a finite Galois extension ... – Álvaro Lozano-Robledo Mar 19 '12 at 13:33
@ÁlvaroLozano-Robledo Hm, have I blundered? If $E/k$ is Galois then it seems like (by definition, for some) we should have $|\operatorname{Gal}(E/k)| = [E : k]$. (Personally I would not write $\operatorname{Gal}$ for something that wasn't Galois, but it's not so hard to tell what is meant here.) – Dylan Moreland Mar 19 '12 at 14:48
If $E/k$ is not Galois, then how can you talk about $\operatorname{Gal}(E/k)$? – Álvaro Lozano-Robledo Mar 19 '12 at 15:00
I assumed that it was what I would write as $\operatorname{Aut}(E/k)$. – Dylan Moreland Mar 19 '12 at 15:10
@Álvaro: Many authors use $\mathrm{Gal}(E/k)$ to denote the group of automorphisms, whether or not the extension is assumed to be Galois, and define Galois extension as one in which the fixed field of $\mathrm{Gal}(E/k)$ is $k$ (or in the finite case, where $|\mathrm{Gal}(E/k)| = [E:k]$). – Arturo Magidin Mar 19 '12 at 15:46

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.