Quotient Group of Integers by Multiples

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

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

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

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

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

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

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

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

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

This is merely congruence of integers as defined in Congruence Modulo $m$.

Thus the quotient set $$\mathbb{Z} / m \mathbb{Z}$$ is $$\mathbb{Z}_m$$.

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

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