Number of $p$-local subgroups of a group

Question

A subgroup of a finite group is '$p$-local' if it is the normalizer of some Sylow $p$-subgroup. I want to prove that the number of $p$-local subgroups of a group is congruent to $1$ modulo $p$. I know that the index of a normalizer of a Sylow $p$-subgroup is congruent to $1$ modulo $p$ but how this is related with the statement? Any help would be great!

Answer 1

Use the fact that all Sylow $p$-subgroups are conjugate and the fact that $N_G(H^ {g})=N_G(H)^{g}$ for any subgroup $H$ of $G$ and $g \in G$. Let $P \in Syl_p(G)$. Then the number of $p$-local subgroups is the number of different conjugates of $N_G(P)$, which equals index$[G:N_G(N_G(P))]$. Now finally note (a well-known fact) that the $N_G(N_G(P))=N_G(P)$. Hence the number of local $p$-subgroups is index$]G:N_G(P)]$ and this is $\equiv 1$ mod $p$.

Answer 2

Let me give another answer that is maybe easier. Let $\mathscr{N}=\{N_G(P):P \in Syl_p(G)\}$. Define a map $\phi: Syl_p(G) \rightarrow \mathscr{N}$ by $\phi(P)=N_G(P)$. Then surely $\phi$ is surjective. But it is also injective: suppose $\phi(P)=\phi(Q)$, so $N_G(P)=N_G(Q)$. Hence $Q \subseteq N_G(P)$ and $Q$ is a Sylow $p$-subgroup of $G$, so of $N_G(P)$. But $P$ is the unique Sylow $p$-subgroup of $N_G(P)$ and it follows that $P=Q$. Hence $\phi$ is bijective and $\#\mathscr{N}=\# Syl_p(G)$.