Theorem 11.17

Statement:

If $m$ is a square free integer,that is, $m$ is not divisible by the square of any prime, then every abelian group of order $m$ is cyclic.

Proof:

Let $G$ be an abelian group of square free order $m$. Then by Theorem 11.12, $G$ is isomorphic to

(1)
\begin{align} \mathbb{Z}_{(p_1)^{r_1}}\times \mathbb{Z}_{(p_2)^{r_2}}\times \dots \times \mathbb{Z}_{(p_n)^{r_n}} \end{align}

where $m=(p_1)^{r_1}(p_2)^{r_2}\dots (p_n)^{r_n}$. Since $m$ is square free, we must have all $r_i=1$ and all $p_i$ distinct primes. Corollary 11.6 then shows that $G$ is isomorphic to $\mathbb{Z}_{p_1p_2\dots p_n}$, so $G$ is cyclic.