(Edit: The wikipedia article is correct. I messed up the notions normalizer and normal closure.)
Please take a look at the wikipedia article on Sylow's theorems here, more precisely, at the last bullet of Theorem 3.
It says that $n_p=\lvert G:N_G(P)\rvert$, where $P$ is a $p$-Sylow group in $G$, $N_G$ denotes the normalizer and $n_p$ is the number of $p$-Sylow groups in $G$.
Isn't this false? If $P$ is already normal, then we know that $n_p=1$ and $N_G(P)=P$. But $G:P$ is $m$ and not $1$ (where $\lvert G\rvert=p^n\cdot m$).
I would guess that $\frac{m}{n_p}=\lvert G:N_G(P)\rvert$ might be correct. What do you say?