# Abelian Group of Semiprime Order is Cyclic

## Theorem

Let $p$ and $q$ be distinct prime numbers.

Let $G$ be an abelian group such that:

- $\order G = p q$

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

Then $G$ is cyclic.

## Proof

By Order of Element Divides Order of Finite Group, the order of elements of $G$ are all in $\set {1, p, q, p q}$.

We have that Identity is Only Group Element of Order 1.

Suppose $G$ were to contain more than $p - 1$ elements of order $p$.

Then by Order of Finite Abelian Group with $p+$ Order $p$ Elements is Divisible by $p^2$:

- $p^2 \divides \order G$

where $\divides$ denotes divisibility.

As $p^2 \nmid p q$ it follows that $G$ contains no more than $p - 1$ elements of order $p$.

Similarly it follows that $G$ contains no more than $q - 1$ elements of order $q$.

These, along with $e$, account for $\paren {p - 1} + \paren {q - 1} + 1 = p + q - 1$ elements.

Thus the number of elements of order $p q$ is:

- $p q - \paren {p + q - 1} = \paren {p - 1} \paren {q - 1}$

which is strictly positive.

The result follows from Group whose Order equals Order of Element is Cyclic.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $20$