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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