# Cyclic Groups of Order p q

## Contents

## 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 1

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:

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

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

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

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:

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

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

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

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

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.

$\blacksquare$

## Proof 2

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:

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

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

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

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 $G$.

Let the Sylow $p$-subgroup of $G$ be denoted $P$.

Let the Sylow $q$-subgroup of $G$ be denoted $Q$.

We have that:

- $P \cap Q = \set e$

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

Hence in $P \cup Q$ there are $q + p - 1$ elements.

As $p q \ge 2 q > q + p - 1$, there exists a non- identity element in $G$ that is not in $H$ or $K$.

Its order must be $p q$.

Hence, by definition, $G$ is cyclic.

$\blacksquare$

## Also see

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

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \alpha$