Integer Multiples under Addition form Subgroup of Integers

Theorem
Let $\struct {\Z, +}$ denote the additive group of integers.

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

Then $\struct {n \Z, +}$ is a subgroup of $\struct {\Z, +}$.

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 $\struct {\Z, +}$.

Also see

 * Definition:Additive Group of Integer Multiples