The cyclic group of order six in elementary-divisor and invariant-factor forms

Given: The objects and hypotheses in the example.

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).

For every finite abelian group $G$ there is a unique list $1<n_1\mid \cdots \mid n_r$ such that $G\cong C_{n_1}\times \cdots \times C_{n_r}$. Moreover $\mathrm{\mid }G\mathrm{\mid }=n_1\cdots n_r$. The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

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).

For every $n\in \mathbb{N}$, view $n$ as its canonical nonnegative integer and put $n\mathbb{Z}:=\{nk:k\in \mathbb{Z}\}$. Then the left cosets of $n\mathbb{Z}$ in $(\mathbb{Z},+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}=(\mathbb{Z}\mathrm{/}n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in \mathbb{N}$, the congruence-class group $(\mathbb{Z}\mathrm{/}n,+)$ is the quotient group $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}$).