Lemma 9.21.2. Let $E/F$ be a finite extension of fields. Then $E$ is Galois over $F$ if and only if $|\text{Aut}(E/F)| = [E : F]$.

**Proof.**
Assume $|\text{Aut}(E/F)| = [E : F]$. By Lemma 9.15.9 this implies that $E/F$ is separable and normal, hence Galois. Conversely, if $E/F$ is separable then $[E : F] = [E : F]_ s$ and if $E/F$ is in addition normal, then Lemma 9.15.9 implies that $|\text{Aut}(E/F)| = [E : F]$.
$\square$

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