Let $G$ be a finite group. $\forall n \in \mathbb{N}:\ f_n(G) := |\{x \in G | x^n = 1)$. Prove that if p is prime and $ p \bigg| |G|$ then $p \bigg| f_p(G)$.
Hint: Let $S \in \operatorname{Syl}_p(G) and \Omega = \{x \in G |x^p = 1\}$. Let $S$ act on $\Omega$ by conjugation and note that $\Omega \cap C_G(S)$ is a subgroup of S.
Well, usually I have something smart to say about how I approached the problem. But this time I'm stunned... I can say that there are at least $p-1$ elements in $S$, that for every $x\in\Omega$ they act upon the result is in $\Omega$. That because there's a p-order element in S, which creates a cyclic group with p-1 generators. But I don't think it's much of a use here.