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 Additive Group of Integers, $\left({m \Z, +}\right)$ is a subgroup of $\left({\Z, +}\right)$.

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

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