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  "body": "\n## Proof\n\n1. Let $N = N_G(P)$ be the [[normaliser|0AT1ZSNM25]] of $P$ in $G$. It is a subgroup of $G$: if $gPg^{-1} = P$ and $hPh^{-1} = P$, then $(gh)P(gh)^{-1} = P$ and $g^{-1}Pg = P$.\n2. For $x \\in P$ we have $xPx^{-1} = P$, because $P$ is a subgroup. So $P \\subseteq N \\subseteq G$.\n3. By the [[index formula|04DC0SRN2K]] for $P$ in $G$, for $N$ in $G$, and for $P$ in $N$, we get $|G| = [G : P]\\,|P|$, $|G| = [G : N]\\,|N|$ and $|N| = [N : P]\\,|P|$.\n4. Combining these, $[G : P]\\,|P| = [G : N]\\,[N : P]\\,|P|$. Dividing by $|P| \\geq 1$ gives $[G : P] = [G : N] \\cdot [N : P]$.\n5. By the [[formula for the number of Sylow subgroups|09ZJ6CAK7S]], $n_p = [G : N]$.\n6. So $[G : P] = n_p \\cdot [N : P]$, where $[N : P]$ is a positive integer. Hence $n_p$ divides the [[index|04QEBF8S49]] $[G : P]$.\n\n## Idea\n\nThe number $n_p$ is the index of the normaliser of $P$. The normaliser sits between $P$ and $G$, and indices multiply along a chain of subgroups. So $n_p$ is a factor of $[G : P]$.\n",
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