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.