For the set $X=\{g\in G:g^p=1\}$, show that $p$ divides $|X|$.

Question

Let $G$ be a finite group and $p$ be a prime divisor of $|G|$. Consider the set $X=\{g\in G:g^p=1\}$. Show that $p$ divides $|X|$.

My attempt: Consider the action of $G$ on $X$ by conjugation. Then ${\rm Stab}_g=C_G(g)$ for all $g\in X$. I'm stuck here.

Answer 1

$X$ comprises all and only the elements of $G$ of order $p$, plus the identity. Such elements are grouped in $m$ trivially intersecting subgroups each of order $p$. Therefore: \begin{alignat}{1} |X| &= m(p-1)+1 \\ &=mp-(m-1) \\ \tag1 \end{alignat} But $m\equiv 1\pmod p$, because the number $n_k$ of subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$ for every $k=0,1,\dots,k_{\text{max}}$ (see e.g. here), and in particular for $k=1$. Therefore, $m=n_1\equiv 1\pmod p$, and hence from $(1)$ $p\mid |X|$.