Order of Galois Group Equals Degree of Extension

Theorem
Let $L/K$ be a Galois extension.

Then $|\operatorname{Gal}(L/K)| = [L : K]$.

Proof
Since $L/K$ is Galois, it is separable.

Thus, by the Primitive Element Theorem, there exists an $\alpha\in L$ such that $L = K(\alpha)$.

Let $m_\alpha\in K[x]$ be the minimal polynomial of $\alpha$ over $K$.

Then $[L : K] = \operatorname{deg}(m_\alpha)$.

Suppose $\sigma \in \operatorname{Gal}(L/K)$, then


 * $m_\alpha(\sigma(\alpha))= \sigma(m_\alpha(\alpha)) = \sigma(0) = 0$,

since $m_\alpha$ has coefficients in $K$. Therefore, $\sigma(\alpha)$ must be a root of $m_\alpha$.

Every element of $\operatorname{Gal}(L/K)$ is determined by its value at $\alpha$ by our assumption that $L=K(\alpha)$.

Therefore, $|\operatorname{Gal}(L/K)| \le \operatorname{deg}(m_\alpha) = [L : K]$.

Next, suppose $\beta$ is a root of $m_\alpha$. By the normality of $L/K$, we must have $\beta\in L$.

Then, by Abstract Model of Algebraic Extensions,


 * $K(\alpha) \cong K[x]/\langle m_\alpha \rangle \cong K(\beta)$.

Composing isomorphisms we have an automorphism of $L$ for each $\beta$.

Thus, $|\operatorname{Gal}(L/K)| \ge \operatorname{deg}(m_\alpha) = [L : K]$, from which our result follows.