Cyclic Groups of Order pq
Then $G$ is cyclic.
We have that $1 + k p$ cannot divide $p$.
Then $1 + k p$ must divide $q$.
But as $q$ is prime, either:
- $1 + k p = 1$
- $1 + k p = q$
- $1 + k p = q \implies k p = q - 1 \implies p \divides q - 1$
Hence $1 + k p = 1$.
Thus there is only one Sylow $p$-subgroup of $G$.
Similarly, there is only one Sylow $q$-subgroup of $G$.
The intersection of these groups is $1_G$, so there are $q + p - 1$ elements in the union. Since $pq \geq 2q > q + p - 1$, there is a non-identity element in $G$ that is not in $H$ or $K$. Its order must be $pq$, so $G$ is cyclic.
- Group Direct Product of Cyclic Groups: a similar result which can often be confused with this one.