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).