Group of Order p q is Cyclic/Proof 2

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Theorem

Let $p, q$ be primes such that $p < q$ and $p$ does not divide $q - 1$.

Let $G$ be a group of order $p q$.


Then $G$ is cyclic.


Proof

Lemma

There is:

exactly one Sylow $p$-subgroup of $G$.
exactly one Sylow $q$-subgroup of $G$.

$\Box$


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.

$\blacksquare$