Integer Multiples under Addition form Infinite Cyclic Group

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

Then $\struct {n \Z, +}$ is a countably infinite cyclic group.

It is generated by $n$ and $-n$:
 * $n \Z = \gen n$
 * $n \Z = \gen {-n}$

Hence $\struct {n \Z, +}$ can be justifiably referred to as the additive group of integer multiples.

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

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

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

Also, $-y = - n s$.

So $x - y = n \paren{r - s}$.

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

So by the One-Step Subgroup Test it follows that $\struct {n \Z, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$.

From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is a cyclic group.

So by Subgroup of Cyclic Group is Cyclic, $\struct {n \Z, +}$ is a cyclic group.

The final assertions follow from Subgroup of Infinite Cyclic Group is Infinite Cyclic Group.

Also see

 * Definition:Additive Group of Integer Multiples