Group of Order p q is Cyclic

Theorem
Let $G$ be a group of order $pq$, where $p, q$ are prime, $p < q$, and $p$ does not divide $q-1$.

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

So by Euclid's Lemma for Prime Divisors $1 + k p \in \left\{{1, p, q, p q}\right\}$.

That is, $k p \in \left\{{0, p - 1, q - 1, p q - 1}\right\}$.

As we have that $p \nmid q - 1$ we also have that $k p \ne q - 1$.

Thus it follows that $k = 0$ and thus $H$ is the only Sylow p-subgroup of $G$.

Similarly, there is only one Sylow q-subgroup of $G$.

Thus, by the a corollary to the Sylow theorems, $H$ and $K$ are normal subgroups of $G$.

Let $H = \left \langle x \right \rangle$ and $K = \left \langle y \right \rangle$.

To show $G$ is cyclic, it is sufficient to show that $x$ and $y$ commute, because then $\left|{x y}\right| = \left|{x}\right| \left|{y}\right| = p q$.

Since $H$ and $K$ are normal:


 * $x y x^{-1} y^{-1} = \left({x y x^{-1}}\right) y^{-1} \in K y^{-1} = K$

and


 * $x y x^{-1} y^{-1} = x \left({y x ^{-1} y^{-1}}\right) \in x H = H$

Thus $xyx^{-1}y^{-1} \in K \cap H = {1}$, and hence $xy = yx$.

Also see
Compare with the similar result Group Direct Product of Cyclic Groups, a similar result which can often be confused with this one.