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 an 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
Taking the group axioms in turn:

G0: Closure
Let $$x, y \in n \Z$$.

Then $$\exists p, q \in \Z: x = n p, y = n q$$.

So $$x + y = n p + n q = n \left({p + q}\right)$$ where $$p + q \in \Z$$.

Thus $$x + y \in n \Z$$ and so $$\left({n \Z, +}\right)$$ is closed.

G1: Associativity
$$+$$ is associative in $$\left({n \Z, +}\right)$$ by dint of associativity of integer addition.

G2: Identity
$$0 = n 0 \in n \Z$$, so $$\left({n \Z, +}\right)$$ has an identity.

G3: Inverses
Let $$x \in n \Z$$.

Then $$\exists p \in \Z: x = n p$$ and so $$-x = n \left({-p}\right) \in \left({n \Z, +}\right)$$.

Commutativity
$$+$$ is commutative in $$\left({n \Z, +}\right)$$ by dint of commutativity of integer addition.

Countably Infinite
From the description of the set of integer multiples:
 * $$n \Z = \left\{{\ldots, -3n, -2n, -n, 0, n, 2n, 3n, \ldots}\right\}$$

there is an obvious bijection to the set of integers $$\Z$$, which is itself a countably infinite set.

Cyclic
Its cyclic nature is shown in Subgroup of Infinite Cyclic Group.