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.
$\blacksquare$