TITLE: Proving Cauchy's theorem (group theory)
QUESTION [5 upvotes]: Let $G$ be a finite group and $p$ a prime number. Prove that $G$
  contains an element of order $p$ if $p$ divides $|G|$. (Hint: reduce
  to the situation where $G$ is cyclic and of order $p^r$)

Suppose $|G|=p^r m$ where $p\nmid m$. By considering a Sylow $p$-subgroup, we have a subgroup of order $p^r$. But how can I reduce to where the group is cyclic? (I'm trying to follow the hint in the problem)
Thank you.

REPLY [3 votes]: Here is an easy proof of this due to McKay. Consider the set
$$ S = \{(a_1,\ldots,a_p) \in G^p : a_1\cdots a_p = e\}. $$
Also, define $A = \{x \in G : x^p = e\}$.
Clearly $|S| = |G|^{p-1}$. Every non-constant vector in $S$ has $p$ shifts (including itself) also in $S$, and so $p \mid |S| - |A|$. Since $p \mid |G|$, we conclude that $p \mid |A|$, and in particular $|A| \geq p$ (since $e \in A$). Apart from $e$, all other elements in $A$ have order $p$.