Order of Galois Group Equals Degree of Extension

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

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

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

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

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

Then:
 * $\left[{L : K}\right] = \deg \left({m_\alpha}\right)$

Suppose $\sigma \in \operatorname{Gal} \left({L / K}\right)$.

Then:


 * $m_\alpha \left({\sigma \left({\alpha}\right)}\right) = \sigma \left({m_\alpha \left({\alpha}\right)}\right) = \sigma \left({0}\right) = 0$

since $m_\alpha$ has coefficients in $K$.

Therefore $\sigma \left({\alpha}\right)$ must be a root of $m_\alpha$.

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

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

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:


 * $K \left({\alpha}\right) \cong K \left[{x}\right] / \langle m_\alpha \rangle \cong K \left({\beta}\right)$

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

Thus:
 * $\left|{\operatorname{Gal} \left({L / K}\right)}\right| \ge \deg \left({m_\alpha}\right) = \left[{L : K}\right]$

from which our result follows.