Number of Sylow p-Subgroups is Index of Normalizer of Sylow p-Subgroup

Theorem

Let $P$ be a Sylow p-subgroup of a group $G$, $N_G (S)$ its normalizer, and $n_p$ be the number of Sylow p-subgroups in $G$. Then $[ G : N_G (S) ] = n_p$.

Proof

By Number of Distinct Conjugate Subsets is Index of Normalizer, $[ G : N_G (S) ] = n_p$.

$\blacksquare$