# Cyclic Groups of Order p q/Proof 1

## Theorem

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

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

Then $G$ is cyclic.

## Proof

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:

\(\displaystyle 1 + k p\) | \(=\) | \(\displaystyle q\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle k p\) | \(=\) | \(\displaystyle q - 1\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle p\) | \(\divides\) | \(\displaystyle q - 1\) |

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 Sylow $p$-Subgroup is Unique iff Normal, $H$ and $K$ are normal subgroups of $G$.

Let $H = \gen x$ and $K = \gen y$.

To show $G$ is cyclic, it is sufficient to show that $x$ and $y$ commute, because then:

- $\order {x y} = \order x \order y = p q$

where $\order x$ denotes the order of $x$ in $G$.

Since $H$ and $K$ are normal:

- $x y x^{-1} y^{-1} = \paren {x y x^{-1} } y^{-1} \in K y^{-1} = K$

and

- $x y x^{-1} y^{-1} = x \paren {y x ^{-1} y^{-1} } \in x H = H$

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

Then:

\(\displaystyle \order a\) | \(\divides\) | \(\displaystyle p\) | |||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle \order a\) | \(\divides\) | \(\displaystyle q\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \order a\) | \(=\) | \(\displaystyle 1\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a\) | \(=\) | \(\displaystyle e\) |

where $e$ is the identity of $G$.

Thus:

- $x y x^{-1}y^{-1} \in K \cap H = e$

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

$\blacksquare$