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  "body": "\n## Proof\n\nWe let a Sylow subgroup act by conjugation on the set of all Sylow subgroups and count fixed points modulo $p$.\n\n1. Write $|G| = p^n m$ with $n \\geq 0$ and $p \\nmid m$. Let $\\Omega = \\operatorname{Syl}_p(G)$ be the set of [[Sylow $p$-subgroups|0JWZX6FKSH]], so $|\\Omega| = n_p$. By the [[first Sylow theorem|0V98WY94B8]], $\\Omega$ contains some $P$.\n2. For $g \\in G$ and $Q \\in \\Omega$, the [[conjugate subgroup|0D6WHTTH7T]] $gQg^{-1}$ is a subgroup of order $|Q| = p^n$, since $x \\mapsto gxg^{-1}$ is a bijection. So $gQg^{-1} \\in \\Omega$.\n3. So $G$ acts on $\\Omega$ by $g \\cdot Q = gQg^{-1}$, as in the [[action by conjugation|0P8CY9V437]]. Restrict this action to $P$.\n4. The [[order of each element|08V7JHS3QZ]] of $P$ divides $p^n$, so it is a power of $p$. So $P$ is a finite [[$p$-group|06DAH3VS6M]]. The [[fixed point lemma|098D2E7DEH]] gives $n_p = |\\Omega| \\equiv |\\Omega^P| \\pmod p$, where $\\Omega^P$ is the set of fixed points.\n5. The subgroup $P$ is a fixed point, because $xPx^{-1} = P$ for each $x \\in P$.\n6. Let $Q \\in \\Omega^P$. Then $xQx^{-1} = Q$ for each $x \\in P$. So $P \\subseteq N$, where $N = N_G(Q)$ is the [[normaliser|0AT1ZSNM25]] of $Q$ in $G$.\n7. The set $N$ is a subgroup of $G$. Indeed, if $gQg^{-1} = Q$ and $hQh^{-1} = Q$, then $(gh)Q(gh)^{-1} = Q$, and $g^{-1}Qg = Q$ follows from $Q = gQg^{-1}$.\n8. Also $Q \\subseteq N$, and $Q$ is normal in $N$ by the definition of $N$.\n9. By [[Lagrange's theorem|3C53TD]], $|N|$ divides $p^n m$, and $p^n = |Q|$ divides $|N|$. So $|N| = p^n m'$ with $m' \\mid m$, hence $p \\nmid m'$.\n10. So $P$ and $Q$ are both subgroups of $N$ of order $p^n$, that is, Sylow $p$-subgroups of $N$.\n11. By the [[second Sylow theorem|09GCQ5STVJ]] in the group $N$, there is $g \\in N$ with $P = gQg^{-1}$.\n12. Since $g \\in N = N_G(Q)$, we have $gQg^{-1} = Q$. So $P = Q$.\n13. By steps 5 and 12, $\\Omega^P = \\{P\\}$, so $|\\Omega^P| = 1$.\n14. By step 4, $n_p \\equiv 1 \\pmod p$.\n\n## Idea\n\nA Sylow subgroup $P$ acts on the set of Sylow subgroups by conjugation. A fixed point $Q$ is normalised by $P$, and inside the normaliser of $Q$ both are Sylow subgroups, so they are conjugate there, hence equal. With exactly one fixed point, the fixed point lemma gives $n_p \\equiv 1 \\pmod p$.\n",
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