Group of Order p q is Cyclic

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

Thus, by Normal Sylow $p$-Subgroup is Unique, $H$ 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:
 * $\left\vert{x y}\right\vert = \left\vert{x}\right\vert \left\vert{y}\right\vert = p q$

where $\left\vert{x}\right\vert$ denotes the order of $x$ in $G$.

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


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

Now suppose $a \in H \cap K$.

Then:
 * $\left|{a}\right| \mathrel \backslash p \land \left|{a}\right| \mathrel \backslash q \implies \left|{a}\right| = 1 \implies a = 1$

Thus:
 * $x y x^{-1}y^{-1} \in K \cap H = 1$

Hence $x y = y x$ and the result follows.

Also see

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