Group of Order p q is Cyclic/Proof 2

Lemma
Let the Sylow $p$-subgroup of $G$ be denoted $P$.

Let the Sylow $q$-subgroup of $G$ be denoted $Q$.

We have that:
 * $P \cap Q = \set e$

where $e$ is the identity element of $G$.

Hence in $P \cup Q$ there are $q + p - 1$ elements.

As $p q \ge 2 q > q + p - 1$, there exists a non- identity element in $G$ that is not in $H$ or $K$.

Its order must be $p q$.

Hence, by definition, $G$ is cyclic.