Proof that the number of Sylow subgroups divides the index

Proof

1. Let $N = N_G(P)$ be the normaliser 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 $g^{-1}Pg = P$.
2. For $x \in P$ we have $xPx^{-1} = P$, because $P$ is a subgroup. So $P \subseteq N \subseteq G$.
3. By the index formula for $P$ in $G$, for $N$ in $G$, and for $P$ in $N$, we get $|G| = [G : P]\,|P|$, $|G| = [G : N]\,|N|$ and $|N| = [N : P]\,|P|$.
4. Combining these, $[G : P]\,|P| = [G : N]\,[N : P]\,|P|$. Dividing by $|P| \geq 1$ gives $[G : P] = [G : N] \cdot [N : P]$.
5. By the formula for the number of Sylow subgroups, $n_p = [G : N]$.
6. So $[G : P] = n_p \cdot [N : P]$, where $[N : P]$ is a positive integer. Hence $n_p$ divides the index $[G : P]$.