Let $P \in Syl_{p}(A_{p+1})$, where $A_{p+1}$ is the alternating group on $p+1$ elements. Show that $N_{A_{p+1}}(P)$ has order $p(p-1)/2$.
Attempt: I'm supposed to consider all elements of order $p$. But, I'm not sure where to go with this. A bird's eye view of the proof would be really helpful.
On
Hint: A $p$-Sylow $P$ subgroup of $A_{p+1}$ is cyclic of order $p$ (for $p>2$). We can assume that $P=\langle \sigma \rangle$, where $\sigma=(1\,2\,\dots\,p)$. Then a permutation $\tau$ is in the normalizer of $P$ if and only if $\tau\sigma\tau^{-1}\in P$, which means that $\tau$ only permutes elements $1,2,\dots p$.
The degree of $N_{A_{p+1}}(P)$ in $A_{p+1}$ is the number of Sylow $p$-subgroups of $A_{p+1}$. Every Sylow $p$-subgroup must contain only elements of order $p^k$ for some $k \in \mathbb N$, but the order of any element must divide the order of the group, and the only $p$th power that divides $\frac{(p+1)!}2$ is $p$ itself. Therefore, each Sylow $p$-subgroup must contain only elements of order $p$. Distinct subgroups of order $p$ may only intersect at the identity, so if $M$ is the number of elements of order $p$, $\frac{M}{p-1}$ is the number of Sylow $p$-subgroups. The order of $N_{A_{p+1}}(P)$ is just the order of $A_{p+1}$ divided by the degree of $N_{A_{p+1}}(P)$, so if you can find how many elements of order $p$ there are you should have your answer.