Integer Multiples under Addition form Infinite Cyclic Group

Theorem
Let $n \Z$ be the set of integer multiples of $n$.

Then $\left({n \Z, +}\right)$ is a countably infinite cyclic group.

It is generated by $n$ and $-n$:
 * $n \Z = \left \langle n \right \rangle$
 * $n \Z = \left \langle {-n} \right \rangle$

Proof
Clearly $0 \in n \Z$ so $n \Z \ne \varnothing$.

Now suppose $x, y \in n \Z$.

Then $\exists r, s \in \Z: x = n r, y = n s$.

Also, $-y = - n s$.

So $x - y = n \left({r - s}\right)$.

As $r - s \in \Z$ it follows that $x - y \in n \Z$.

So by the One-Step Subgroup Test it follows that $\left({n \Z, +}\right)$ is a subgroup of the Additive Group of Integers $\left({\Z, +}\right)$.

From Integers Infinite Cyclic Group, $\left({\Z, +}\right)$ is a cyclic group.

So by Subgroup of a Cyclic Group is Cyclic, $\left({n \Z, +}\right)$ is a cyclic group.

The final assertions follow from Subgroup of Infinite Cyclic Group.