Every abelian group of order n is cyclic if and only if n is squarefree

Given: The objects and hypotheses in the statement.

A positive integer $n$ is squarefree if no square of a prime divides $n$. Equivalently, every exponent in its canonical prime factorisation is $0$ or $1$. The integer $1$ is squarefree by the empty factorisation. (Squarefree positive integers).

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

If $G$ is finite abelian and $\mathrm{\mid }G\mathrm{\mid }={\prod }_{i<r}{p}_i^{a_i}$ is its prime factorisation, then the subgroups $G(p_i)$ form an internal direct product of $G$. Thus $G\cong {\prod }_{i<r}G(p_i)\mathrm{.}$ For the trivial group, this is the empty product. (A finite abelian group is the internal direct product of its primary components).

Let $n_0,\dots ,n_{r-1}$ be a finite pairwise-coprime list of positive integers and let $N:={\prod }_{i<r}{n}_i$. The map $\mathrm{\Phi }:\mathbb{Z}\mathrm{/}N⟶{\prod }_{i<r}\mathbb{Z}\mathrm{/}{n}_i,[x{]}_N⟼([x{]}_{n_i}{)}_{i<r},$ is a bijection. It preserves addition, multiplication, $[0]$, and $[1]$ componentwise. For the empty list, $N=1$ and both sides have one element. (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

Let $\iota :\mathbb{N}\to \mathbb{Z}$ be the canonical embedding. If $g\in G$ and $h\in H$ have finite orders $m,n\ge 1$, then in the external direct product $\mathrm{ord}(g,h)=\mathrm{lcm}(m,n)\mathrm{.}$ (If $g$ and $h$ have finite orders $m$ and $n$, then $\iota (\mathrm{ord}(g,h))=\mathrm{lcm}(\iota (m),\iota (n))$ in $G\times H$).