Group of Order p q is Cyclic

Theorem
If $$G \ $$ is 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 $$1+kp \ $$ and divides $$pq \ $$, so $$1+kp = 1, p, q, \ $$ or $$pq \ $$.

By this result, and the fact that $$p \ $$ does not divide $$q-1 \ $$, 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 $$|xy|=|x| |y| = pq \ $$.

Since $$H \ $$ and $$K \ $$ are normal,

$$xyx^{-1}y^{-1} = (xyx^{-1})y^{-1} \in Ky^{-1} = K$$

and

$$xyx^{-1}y^{-1} = x(yx^{-1}y^{-1}) \in xH = H$$

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