Proof that the number of Sylow subgroups is the index of the normaliser

Proof

We apply the orbit-stabiliser theorem to the action of $G$ by conjugation on its Sylow subgroups.

1. Let $\Omega = \operatorname{Syl}_p(G)$ be the set of Sylow $p$-subgroups of $G$, so $|\Omega| = n_p$.
2. For $g \in G$ and $Q \in \Omega$, the conjugate subgroup $gQg^{-1}$ has the same order as $Q$, because $x \mapsto gxg^{-1}$ is a bijection. So $gQg^{-1} \in \Omega$.
3. Thus $G$ acts on $\Omega$ by $g \cdot Q = gQg^{-1}$. This is the action by conjugation restricted to $\Omega$.
4. By the second Sylow theorem, each $Q \in \Omega$ equals $gPg^{-1}$ for some $g \in G$. So the orbit $G \cdot P$ is all of $\Omega$.
5. The stabiliser of $P$ is $\{g \in G : gPg^{-1} = P\}$, which is the normaliser $N_G(P)$.
6. By the orbit-stabiliser theorem, $|G \cdot P| = [G : N_G(P)]$.
7. By steps 4 and 6, $n_p = |\Omega| = [G : N_G(P)]$.