Number of Elements of Order p in Group of Order pq is Multiple of q

Theorem
Let $p$ and $q$ be distinct prime numbers.

Let $G$ be a non-abelian group of order $p q$.

Then the number of elements of $G$ of order $p$ is a multiple of $q$.

Proof
Let $x$ be an element of $G$ of order $p$.

From Center of Non-Abelian Group of Order pq is Trivial:
 * $p \notin \map Z G$

where $\map Z G$ denotes the center of $G$.

As $x \notin \map Z G$:
 * $\map C x \subsetneq G$

where $\map C x$ is the centralizer of $x$.

From Order of Element divides Order of Centralizer:
 * $p \divides \order {\map C x}$

Then from Lagrange's Theorem, it follows directly that:
 * $\order {\map C x} = p$

By Number of Conjugates is Number of Cosets of Centralizer:
 * $\card {\conjclass x} = \index G {\map {C_G} x}$

where $\conjclass x$ denotes the conjugacy class of $x$.

It follows that:
 * $\card {\conjclass x} = q$

and so the element of $G$ of order $p$ come in sets whose cardinality is a divisor of $q$.