If $p\mid|G|$ then how many elements of order $p$ are there in $G$?

Question

Let $G$ be a finite group and $p$ be a prime such that $p\mid|G|$ , then obviously $G$ has an element of order $p$ (by Cauchy's theorem) ; I would like to know exactly how many elements of order $p$ are there in the group ? Please help

Answer 1

If $N_p$ is the number of subgroups of order $p$ in $G$ since any two of them intersect trivially, the total number $M(p)$ of elements of order $p$ is $$ M(p)=(p-1)N_p. $$ When $p$ divides exactly the order of $G$, i. e. $|G|=pt$ with $(p,t)=1$, the number $N_p$ can be obtained applying Sylow's theorems.

If a higher power of $p$ divides $|G|$ the situation is more complex, as one needs to determine how the $p$-Sylows intersect each other.

Answer 2

Something more can be said. Let's stick to the notation of Andrea Mori - let $M(p)$ denote the number of elements of order $p$ of $G$, then $$M(p) \equiv -1 \text { mod } p.$$ This follows easily from Sylow Theory or from the proof of James McKay of Cauchy's Theorem on the existence of elements of order $p$, if $p$ divides $|G|$.