---
tag: 0BDJBH8B4Z
kind: statement
role: theorem
name: Order of the Galois group of a Galois extension equals its degree
msc: [12F10]
uses: [0DS2F7GK05, 0XVR72M51M, 0QM98DNWVW]
sources:
  - url: https://en.wikipedia.org/wiki/Galois_extension
    locator: Characterization of Galois extensions
  - url: https://en.wikipedia.org/wiki/Galois_group
    locator: Cardinality of the Galois group and the degree of the field extension
---

## Statement

Let $L/K$ be a finite [[Galois extension|0DS2F7GK05]].
Then
$$|\operatorname{Gal}(L/K)| = [L : K].$$

## Remarks

Conversely, a finite extension with $|\operatorname{Aut}(L/K)| = [L : K]$ is Galois. So the [[general bound|0QM98DNWVW]] is an equality exactly for Galois extensions.
For $\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}$, the Galois group has order $4$: an automorphism can change the signs of $\sqrt{2}$ and of $\sqrt{3}$ independently.
For $\mathbb{F}_{p^n}/\mathbb{F}_p$, the Galois group is cyclic of order $n$, generated by the [[Frobenius automorphism|0VP4TBVJ19]].

## Explanations

### L1 human

For a Galois extension, counting symmetries is the same as computing the degree.
This links the size of a group to the size of a field extension.