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  "body": "\n## Statement\n\nLet $G$ be a finite [[group|G5S1M9]] and let $p$ be a prime number.\nThen the number $n_p$ of [[Sylow p-subgroups|0053CQE9Y2]] of $G$ satisfies $n_p \\equiv 1 \\pmod p$.\n\n## Remarks\n\nThe proof lets a Sylow $p$-subgroup $P$ act by conjugation on $\\operatorname{Syl}_p(G)$, and $P$ is the only fixed point.\nTogether with [[the divisibility condition|0M3PYJ3TSE]], this often forces $n_p = 1$.\nIn $S_4$, $n_2 = 3$ and $n_3 = 4$, and both numbers are $\\equiv 1$ modulo the prime.\n\n## Explanations\n\n### L1 human\n\nThe number of Sylow $p$-subgroups is one more than a multiple of $p$.\nThis simple fact, together with a divisibility condition, leaves only a few possible values.\nOften the only possible value is $1$, and then the Sylow subgroup is normal.\n",
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  "name": "Third Sylow theorem",
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