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

Let $n_0,\ldots,n_{r-1}$ be a finite pairwise-coprime list of positive integers and let $N:=\prod_{i<r}n_i$. The map $\Phi:\mathbb Z/N\longrightarrow\prod_{i<r}\mathbb Z/n_i,\qquad[x]_N\longmapsto([x]_{n_i})_{i<r},$ 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 $m,n\ge1$, then in the external direct product $\operatorname{ord}(g,h)=\operatorname{lcm}(m,n).$ (If $g$ and $h$ have finite orders $m$ and $n$, then $\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))$ in $G\times H$).