Degree Equals Galois Group Order

Let $L/K$ be a finite field extension . Its Galois group is

i.e. the group of $K$-automorphisms of $L$.

Theorem (degree = group order for finite Galois extensions).
If $L/K$ is a finite Galois extension , then

where $[L:K]$ is the degree of the extension .

A useful companion fact is the inequality valid for any finite extension:

with equality if and only if $L/K$ is Galois (equivalently: finite, separable , and normal ; see separable + normal = Galois ).

Examples

1. Quadratic extensions.
   For $L=\mathbb{Q}(\sqrt2)$ over $K=\mathbb{Q}$, we have $[L:K]=2$. The two $K$-automorphisms are

   $$\sqrt2\mapsto \sqrt2 \quad\text{and}\quad \sqrt2\mapsto -\sqrt2,$$

   so $|\mathrm{Gal}(L/K)|=2=[L:K]$.

2. A Galois splitting field of degree 6.
   Let $L$ be the splitting field over $\mathbb{Q}$ of $x^3-2$. Then $L=\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ is finite Galois over $\mathbb{Q}$, and one finds

   $$[L:\mathbb{Q}]=6 \quad\text{and}\quad \mathrm{Gal}(L/\mathbb{Q})\cong S_3,$$

   so $|\mathrm{Gal}(L/\mathbb{Q})|=6=[L:\mathbb{Q}]$.

3. Contrast: a non-Galois finite extension.
   The simple extension $L=\mathbb{Q}(\sqrt[3]{2})$ has $[L:\mathbb{Q}]=3$, but it is not normal (it does not contain the other complex cube roots), so it is not Galois. In fact, any $\mathbb{Q}$-automorphism must send $\sqrt[3]{2}$ to another root of its minimal polynomial; only the real root lies in $L$, so $\mathrm{Aut}_{\mathbb{Q}}(L)$ is trivial and $|\mathrm{Aut}_{\mathbb{Q}}(L)|=1<3$.