Proof of Burnside's lemma

Proof

In the proof of Burnside's lemma, the first step is to re-express the sum over the group elements g ∈ G as an equivalent sum over the set of elements x ∈ X:

$$\sum_{g \in G}|X^g| = \#\{(g,x)\in G\times X \mid g \cdot x = x\} = \sum_{x \in X} |G_x|.$$

Here $X^g = \{ x \in X : g\cdot x = x\}$ is the set of points of $X$ fixed by the element $g$ of $G$, whereas $G_x = \{ g \in G : g\cdot x = x\}$ is the stabilizer subgroup of $G$, consisting of those symmetries that fix the point $x \in X$.

The orbit-stabilizer theorem says that for each $x \in X$ there is a natural bijection between the orbit $G \cdot x = \{ g\cdot x : g \in G\}$ and the set of left cosets $G / G_x$. Lagrange's theorem implies that

$$|G \cdot x| = [G : G_x] = |G| / |G_x|.$$

The sum may therefore be rewritten as

$$\sum_{x \in X} |G_x| = \sum_{x \in X} \frac{|G|}{|G \cdot x|} = |G| \sum_{x \in X}\frac{1}{|G \cdot x|}.$$

Writing $X$ as the disjoint union of its orbits in $X/G$ gives

$$|G| \sum_{x \in X}\frac{1}{|G \cdot x|} = |G| \sum_{A\in X/G} \sum_{x\in A} \frac{1}{|A|} = |G| \sum_{A\in X/G} 1 = |G|\cdot|X/G|.$$

Putting everything together gives the desired result:

$$\sum_{g \in G}|X^g| = |G| \cdot |X/G|.$$

This is similar to the proof of the conjugacy class equation, which considers the conjugation action of $G$ on itself, that is, it is the case $X = G$ and $g \cdot x = gxg^{-1}$, so that the stabilizer of $x$ is the centralizer $G_x = Z_G(x)$.