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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?

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If $H\leq K\leq G$, then $[G:H] = [G:K][K:H]$ in the sense of cardinality; in particular, if at least two of the terms are finite, then so is the third; here you have $H=P$ and $K=N_G(P)$. – Arturo Magidin Mar 20 '12 at 16:43
up vote 4 down vote accepted

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.

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Thank you very much! – bemyguest Mar 20 '12 at 9:46

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