Center of Non-Abelian Group of Order pq is Trivial

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

Let $G$ be a non-abelian group of order $p q$ whose identity is $e$.

Then the center of $G$ is trivial:


 * $\map Z G = \set e$

Proof
From Center of Group is Normal Subgroup, $\map Z G$ is a normal subgroup of $G$.

By Lagrange's Theorem, the order of $\map Z G$ is either $1$, $p$, $q$ or $p q$.

Because $G$ is not abelian, $G \ne \map Z G$.

Hence $\order {\map Z G} \ne p q$.

From Quotient of Group by Center Cyclic implies Abelian:
 * $G / \map Z G$ cannot be a cyclic group which is non-trivial.

Then we have:
 * $\map C x \subset G$

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

Hence by Prime Group is Cyclic, $G / \map Z G$ cannot be cyclic of order $p$ or $q$.