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  "body": "\n## Proof\n\nIn 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*:\n\n$$\\sum_{g \\in G}|X^g| = \\#\\{(g,x)\\in G\\times X \\mid g \\cdot x = x\\} = \\sum_{x \\in X} |G_x|.$$\n\nHere $X^g = \\{ x \\in X : g\\cdot x = x\\}$ 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 $x \\in X$.\n\nThe orbit-stabilizer theorem says that for each $x \\in X$ there is a natural [[bijection|0GG9TVW90M]] between the orbit $G \\cdot x = \\{ g\\cdot x : g \\in G\\}$ and the set of left cosets $G / G_x$. [[Lagrange's theorem|3C53TD]] implies that\n\n$$|G \\cdot x| = [G : G_x] = |G| / |G_x|.$$\n\nThe sum may therefore be rewritten as\n\n$$\\sum_{x \\in X} |G_x| = \\sum_{x \\in X} \\frac{|G|}{|G \\cdot x|} = |G| \\sum_{x \\in X}\\frac{1}{|G \\cdot x|}.$$\n\nWriting $X$ as the disjoint union of its orbits in $X/G$ gives\n\n$$|G| \\sum_{x \\in X}\\frac{1}{|G \\cdot x|} = |G| \\sum_{A\\in X/G} \\sum_{x\\in A} \\frac{1}{|A|} = |G| \\sum_{A\\in X/G} 1 = |G|\\cdot|X/G|.$$\n\nPutting everything together gives the desired result:\n\n$$\\sum_{g \\in G}|X^g| = |G| \\cdot |X/G|.$$\n\nThis 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 $G_x = Z_G(x)$.\n",
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