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  "body": "\n## Proof\n\nWe count the set $F = \\set{(g, x) \\in G \\times X : g \\cdot x = x}$ in two ways. Both $G$ and $X$ are finite, and $X/G$ denotes the set of orbits.\n\n1. For fixed $g$, the pairs $(g, x)$ in $F$ correspond to the points $x \\in X^g$, the [[points fixed by $g$|08JF1K4DW1]]. Summing over $g$ gives $\\abs{F} = \\sum_{g \\in G} \\abs{X^g}$.\n2. For fixed $x$, the pairs $(g, x)$ in $F$ correspond to the elements $g$ of the [[stabiliser|0RGGF8ZS6D]] $G_x$. Summing over $x$ gives $\\abs{F} = \\sum_{x \\in X} \\abs{G_x}$.\n3. By the [[orbit size formula|0S19CWFZM2]], $\\abs{G_x} = \\abs{G} / \\abs{G \\cdot x}$ for each $x$. So $\\abs{F} = \\abs{G} \\sum_{x \\in X} \\frac{1}{\\abs{G \\cdot x}}$.\n4. The [[orbits partition|0ZC7FYE0NZ]] $X$. So we can split the sum in step 3 into one sum for each [[orbit|00F2JFA41J]] $O \\in X/G$.\n5. If $x \\in O$, then $G \\cdot x = O$, since each point lies in exactly one orbit and $x \\in G \\cdot x$. So the points of $O$ contribute $\\sum_{x \\in O} \\frac{1}{\\abs{O}} = 1$.\n6. Hence $\\sum_{x \\in X} \\frac{1}{\\abs{G \\cdot x}} = \\abs{X/G}$, and $\\abs{F} = \\abs{G} \\cdot \\abs{X/G}$.\n7. Comparing with step 1, $\\sum_{g \\in G} \\abs{X^g} = \\abs{G} \\cdot \\abs{X/G}$. Dividing by $\\abs{G} \\geq 1$ gives $\\abs{X/G} = \\frac{1}{\\abs{G}} \\sum_{g \\in G} \\abs{X^g}$.\n\n## Idea\n\nCount the pairs \"group element, point it fixes\" once by group elements and once by points. Each orbit of size $k$ has stabilisers of size $\\abs{G}/k$ at its $k$ points, so it contributes exactly $\\abs{G}$ to the count. Dividing by $\\abs{G}$ counts the orbits.\n",
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