Integer Multiples under Addition form Infinite Cyclic Group

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

Then $$\left({n \mathbb{Z}, +}\right)$$ is an infinite abelian group.

Proof
Taking the group axioms in turn:

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

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

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

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

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

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

G3: Inverses
Let $$x \in n \mathbb {Z}$$.

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

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