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 $|G|=n_1\cdots n_r$. The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

Let $n_0,\ldots,n_{r-1}$ 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).

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 $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ 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 $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).