Group of Order p q is Cyclic

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
Let $H$ be a Sylow $p$-subgroup of $G$ and let $K$ be a Sylow $q$-subgroup of $G$.

By the Third Sylow Theorem, the number of Sylow $p$-subgroups of $G$ is of the form $1 + k p$ and divides $p q$.

We have that $1 + k p$ cannot divide $p$.

Then $1 + k p$ must divide $q$.

But as $q$ is prime, either:
 * $1 + k p = 1$

or:
 * $1 + k p = q$

But:
 * $1 + k p = q \implies k p = q - 1 \implies p \mathrel \backslash q - 1$

which contradicts our condition that $p$ does not divide $q - 1$.

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.

Also see

 * Group Direct Product of Cyclic Groups: a similar result which can often be confused with this one.