This is a really classical result, but this proof is just so magical that I have to include it.

**Lemma.** *Let be a finite group, and let be a prime dividing . Then has an element of order .*

*Proof.* Consider the action of on

given by

Then , since the orbits of size all have size (and the fixed points are exactly the union of the orbits of size 1). The right hand side is as . Thus,

But there is a bijection

The former trivially contains the element 1, hence (since its size is divisible by ) it has to contain some other element . This element will then have (exact) order .

**Remark.** Of course, the result also follows from the much stronger Sylow theorems (of which we only need the existence statement).