Quotient Group of Integers by Multiples

Theorem
Let $$\left({\Z, +}\right)$$ be the Additive Group of Integers.

Let $$\left({m \Z, +}\right)$$ be the Additive Group of Integer Multiples of $$m$$.

Let $$\left({\Z_m, +_m}\right)$$ be the Additive Group of Integers Modulo $m$.

Then the Quotient Group of $$\left({\Z, +}\right)$$ by $$\left({m \Z, +}\right)$$ is $$\left({\Z_m, +_m}\right)$$.

Thus $$\left[{\Z : m \Z}\right] = m$$.

Proof
From Subgroups of the Integers, $$\left({m \Z, +}\right)$$ is a subgroup of $$\left({\Z, +}\right)$$.

From All Subgroups of Abelian Group are Normal, $$\left({m \Z, +}\right)$$ is normal in $$\left({\Z, +}\right)$$.

Therefore the quotient group $$\frac {\left({\Z, +}\right)} {\left({m \Z, +}\right)}$$ is defined.

Now $$\Z$$ modulo $$m \Z$$ is Congruence Modulo a Subgroup.

This is merely congruence of integers as defined in Congruence (Number Theory).

Thus the quotient set $$\Z / m \Z$$ is $$\Z_m$$.

The left coset of $$k \in \Z$$ is denoted $$k + m \Z$$, which is the same thing as $$\left[\!\left[{k}\right]\!\right]_m$$ from the definition of residue class.

So $$\left[{\Z : m \Z}\right] = m$$ follows from the definition of Subgroup Index.