# order of automorphism group and index of finite field extension

EDIT-1: Since I come up with a proof, I change this previously unanswered question into a proof-verification question.

Let $E / F$ be finite field extension. Define $G(E / F) = \{\phi: E \xrightarrow{\cong} E \mid \phi$ is $F$-automorphism$\}$. Define $\{E : F\}$ to be the number of $F$-embedding $\sigma: E \hookrightarrow \overline{F}$. We claim that $|G(E / F)|$ divides $\{E : F\}$.

Proof

Define $G = G(E/F)$ and $X = \{\sigma: E \hookrightarrow \overline{F} \mid \sigma$ is $F$-embedding$\}$. Observe that if $|G| = 1$, then there is nothing to proof. So WLOG we may assume $|G| > 1$, i.e. there exists a nontrivial $F$-automorphism.

Let $G \curvearrowright X$ be group action defined by $\phi \cdot \sigma = \sigma \phi^{-1}$ for all $\phi \in G, \sigma \in X$.

It is a well-defined function since pre-composing a $F$-embedding by a $F$-automorphism gives you another $F$-embedding. Next, we check it satisfies the axioms of group action.

Identity law: $$\operatorname{id}_E \cdot \sigma = \sigma \operatorname{id}_E^{-1} = \sigma$$ for all $\sigma \in X$

Associativity law:

$$(\phi'\phi) \cdot \sigma = \sigma (\phi'\phi)^{-1} = \sigma \phi^{-1}\phi'^{-1} = \phi' \cdot \sigma \phi^{-1} = \phi' \cdot (\phi \cdot \sigma)$$ for all $\phi, \phi' \in G$ and $\sigma \in X$

So it is indeed a well-defined group action.

Define $X_G = \{ \sigma \in X \mid \phi \cdot \sigma = \sigma$ for any $\phi \in G\}$ to be the fixed point set of the group action. We want to show $X_G = \emptyset$. Suppose not, say there is some $\sigma \in X_G$. Let $\phi$ be a nontrivial $F$-automorphism.

Then $$\phi \cdot \sigma = \sigma$$ By definition, $$\sigma \phi^{-1} = \sigma$$ Since $\sigma$ is injective, it is left cancellable, $$\phi^{-1} = \operatorname{id}_E$$ So $$\phi = \operatorname{id}_E$$

Contradict $\phi$ being nontrivial, so $X_G = \emptyset$.

Next, for each $\sigma \in X$, we define $G_\sigma = \{\phi \in G \mid \phi \cdot \sigma = \sigma\}$ to be the stabilizer of $\sigma$. We want to show all stabilizers are trivial.

Fix $\sigma \in X$. Suppose $\phi \in G_\sigma$.

Then $$\phi \cdot \sigma = \sigma$$ By definition, $$\sigma \phi^{-1} = \sigma$$ Since $\sigma$ is injective, it is left cancellable, $$\phi^{-1} = \operatorname{id}_E$$ So $$\phi = \operatorname{id}_E$$

Hence all stabilizers are trivial.

Finally, using the class equation, $$|X| = |X_G| + \sum_{G \cdot \sigma} \frac{|G|}{|G_\sigma|}$$ where $G \cdot \sigma$ is the orbit of $\sigma$

Using $X_G = \emptyset$ and $G_\sigma$ being trivial, $$|X| = \sum_{G \cdot \sigma} |G|$$

So $|G|$ divides $|X| = \{E : F\}$ as desired.

Is this proof correct?