# $P\in Syl_p(G)\Rightarrow|G:N_G(P)|=|Syl_p(G)|$

Let $G$ be a finite group s.t. $|G|=p^rm$, where $(p,m)=1$.

Let then $P$ be a $p$-Sylow subgroup of $G$, i.e. $P\le G$ with $|P|=p^r$.

We want to show that $|G:N_G(P)|$ is the number of $p$-Sylow subgroups of $G$.

I made several attempts but nothing good came out.

Any help/answer/hint will be appreciated so much; thank you all.

• Have you done Sylow's theorem, which may be read as "$G$ transitively permutes its Sylow $p$-subgroups by conjugation"? What is the stabilizer of one of these, $P$ say? – Geoff Robinson May 30 '14 at 15:30
• Actually, it is true for all subgroups i.e $|G:N_G(H)|$ is equal to number of the conjugeta of $H$ in $G$ and since all sylow$p$ subgroups are conjugate the result follows. – mesel May 30 '14 at 15:58

The sylow subgroups are all coniugates, i.e. they form an unique orbit of the action $$g \cdot P \mapsto gPg^{-1} \ \ g \in G$$
The second Sylow theorem says that all Sylow subgroups are conjugate. So you are looking for the number of distinct $g^{-1}Pg$. And $$g^{-1}Pg=h^{-1}Ph$$ if and only if $$(gh^{-1})^{-1}Pgh^{-1}=P$$ or $$gh^{-1} \in N(P)$$
Let $P$ be a $p$-Sylow subgroup of $G$, then $gPg^{-1}$ is also a $p$-Sylow subgroup. Furthermore, as all $p$-Sylow subgroups are conjugate, all the $p$-Sylow subgroups are generated in this way. Now note that $gPg^{-1} = hPh^{-1}$ if and only if $h^{-1}g \in N_G(P)$. Therefore $$\operatorname{Syl}_p(G) = \{gPg^{-1} \mid g \in G/N_G(P)\},$$ so $$\mid\operatorname{Syl}_p(G)\mid = |G/N_G(P)| = [G : N_G(P)].$$