Additive Group of Integer Multiples
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 Cyclic Group is Cyclic, $\left({n \Z, +}\right)$ is a cyclic group.
The final assertions follow from Subgroup of Infinite Cyclic Group.
$\blacksquare$
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.2$: Example $92$
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.4$: Example $99$
- John F. Humphreys: A Course in Group Theory (1996): $\S 5$: Example $5.7$