If $G$ is a group, $P$ is a 5-sylow group of $G$, $N_{G}(P)$ is the normalizer of $P$ in $G$ then: $[G : N_{G}(P)] \Bigm| [G : P]$
Why is this true?
If $G$ is a group, $P$ is a 5-sylow group of $G$, $N_{G}(P)$ is the normalizer of $P$ in $G$ then: $[G : N_{G}(P)] \Bigm| [G : P]$
Why is this true?
Note that $P \le N_G(P)$, so the order of $P$ divides the order of $N_G(P)$ by Lagrange's theorem. If $|N_G(P)| = m|P|$, then $m[G:N_G(P)] = m\frac{|G|}{|N_G(P)|} = m\frac{|G|}{m|P|} = \frac{|G|}{|P|} = [G:P]$. Thus, your statement is in fact general for any subgroup of $G$ and its respective normalizer.