Show that $|G(E/F)|$ divides $\{E:F\}$ 
Let $E/F$ be a finite field extension. Show that $|G(E/F)|$ divides
  $\{E:F\}$

Where $G(E/F)$ is the group of $F$-automorphism of $E$, $\{E:F\}$ is the index of $E/F$, i.e. number of $F$-embedding from $E$ to $\bar F$.
My idea is, let $E=F(\alpha_1,...,\alpha_n)$ then $\{E:F\}=$ (distinct roots of $\text{irr}(\alpha_1,F)$) ...$\left(\right.$distinct roots of $\text{irr}(\alpha_n,F)$ $\left. \right)$ and $|G(E/F)|=$ (number of conjugates of $\alpha_1$ in $E$) ... (number of conjugates of $\alpha_n$ in $E$) , so what I need to show is $\forall i$ number of conjugates of $\alpha_i$ in $E$ divides distinct roots of $\text{irr}(\alpha_i,F)$ , but I have no idea how to prove it.
Could you please give me some hints? Thank you.
 A: First, assume that $ E/F $ is separable, and let $ N/F $ be a Galois closure. If $ H $ is the subgroup of $ G(N/F) $ mapping $ E $ to itself, then there is a surjective group homomorphism $ H \to G(E/F) $ given by restricting to $ E $, with kernel $ G(N/E) $. It follows that
$$ |G(E/F)| = \frac{|H|}{[N : E]} $$
which divides
$$ \frac{|G(N/F)|}{[N : E]} = [E : F] $$
Now, let $ E/F $ be an arbitrary finite extension. Then, there is an isomorphism
$$ G(E/F) \to G(E^s/F) $$ 
where $ E^s $ is the maximal separable subextension of $ E/F $. By the separable case, we know that the order of this divides $ [E^s : F] $, which divides $ [E : F] $ by the tower law.
A: Well, there’s the very first theorem of Galois Theory, which says that if $\Gamma$ is a finite group of automorphisms of a field $K$, then the set of elements of $K$ fixed under all elements of $\Gamma$, sometimes denoted $K^\Gamma$, is a field, and $[K:K^\Gamma]=|\Gamma|$. Notice that there is no hypothesis of separability, but the upshot is that $K$ is separable over $K^\Gamma$.
This theorem applies. Just take $\Gamma$ to be your group of $F$-automorphisms of $E$. Then you have $F\subset E^\Gamma\subset E$, three fields, and so multiplicativity of the field degree shows that $|\Gamma|$ divides $E:F]$.
