Group of Order p q is Cyclic/Lemma

Lemma for Group of Order $p q$ is Cyclic

Let $p, q$ be primes such that $p < q$ and $p$ does not divide $q - 1$.

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

There is:

exactly one Sylow $p$-subgroup of $G$.

exactly one Sylow $q$-subgroup of $G$.

Let $H$ be a Sylow $p$-subgroup of $G$.

Let $K$ be a Sylow $q$-subgroup of $G$.

By the Fourth 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:

$\ds 1 + k p$

$=$

$\ds q$

$\ds \leadsto \ \ $

$\ds k p$

$=$

$\ds q - 1$

$\ds \leadsto \ \ $

$\ds p$

$\divides$

$\ds q - 1$

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 $G$.

$\blacksquare$