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  "body": "\n## Proof\n\nWe split $X$ into orbits and reduce their sizes modulo $p$. Let $e$ be the identity of $G$.\n\n1. First, $\\abs{G}$ is a power of $p$. Let $q$ be a prime that divides $\\abs{G}$. By [[Cauchy's theorem|0JM7PA288B]], some $g \\in G$ has order $q$.\n2. Since $G$ is a [[$p$-group|06DAH3VS6M]], the order of $g$ is $p^s$ for some $s \\geq 0$. So $q = p^s$, and as $q$ is prime, $q = p$.\n3. So $p$ is the only prime factor of $\\abs{G}$. By the [[fundamental theorem of arithmetic|084HH1NDCB]], $\\abs{G} = p^r$ for some $r \\geq 0$.\n4. Let $d$ be a positive divisor of $p^r$ and $q$ a prime dividing $d$. Then $q \\mid p^r$, so $q \\mid p$ by [[Euclid's lemma|051VYKA1EX]], and $q = p$. As in step 3, $d$ is a power of $p$.\n5. By the [[orbit size formula|0S19CWFZM2]], the size of each [[orbit|00F2JFA41J]] divides $\\abs{G} = p^r$. By step 4, each orbit has size $p^j$ for some $j \\geq 0$.\n6. An orbit $G \\cdot x$ has size $1$ if and only if $G \\cdot x = \\set{x}$, since $x = e \\cdot x$ lies in it. This means $g \\cdot x = x$ for all $g$, that is, $x$ is a [[fixed point|08JF1K4DW1]].\n7. So the orbits of size $1$ are the sets $\\set{x}$ with $x \\in X^G$. Every other orbit has size $p^j$ with $j \\geq 1$, which is divisible by $p$.\n8. The [[orbits partition|0ZC7FYE0NZ]] $X$, and $X$ is finite. So $\\abs{X}$ is the sum of the orbit sizes, which is $\\abs{X^G}$ plus a sum of multiples of $p$.\n9. Hence $\\abs{X} \\equiv \\abs{X^G} \\pmod p$.\n\n## Idea\n\nThe orbits cut $X$ into pieces whose sizes divide the order of the group, which is a power of $p$. Every piece of size more than one is then a multiple of $p$ and disappears modulo $p$. Only the one-point orbits, the fixed points, are left.\n",
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