Subgroups of Additive Group of Integers

Theorem
Let $\left({\Z, +}\right)$ be the additive group of integers.

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

Every non-trivial subgroup of $\left({\Z, +}\right)$ has the form $n \Z$.

Proof
First we note that, from Additive Group of Integer Multiples, $\left({n \Z, +}\right)$ is an infinite abelian group.

Let $H$ be a non-trivial subgroup of $\left({\Z, +}\right)$.

Because $H$ is non-trivial:
 * $\exists m \in \Z: m \in H: m \ne 0$

Because $H$ is itself a group, $-m \in H$.

So either $m$ or $-m$ is positive and therefore in $\Z_{>0}$.

Thus $H \cap \Z_{>0} \ne \varnothing$.

From the Well-Ordering Principle, $H \cap \Z_{>0}$ has a least element, which we can call $n$.

It follows from Subgroup of Infinite Cyclic Group that $\forall a \in \Z: a n \in H$.

Thus, $n \Z \subseteq H$.

Now, suppose $m \in H \setminus n \Z$.

Then $m \ne 0$, and also $-m \in H \setminus n \Z$.

Assume $m > 0$, otherwise we consider $-m$.

By the Division Theorem, $m = q n + r$.

If $r = 0$, then $m = q n \in n \Z$, so $0 \le r < n$.

Now this means $r = m - q n \in H$ and $0 \le r < n$.

This would mean $n$ was not the smallest element of $H \cap \Z$.

Thus $H \setminus n \Z = \varnothing$, thus from Set Difference with Superset is Empty Set‎, $H \subseteq n \Z$.

Thus we have $n \Z \subseteq H$ and $H \subseteq n \Z$, thus $H = n \Z$.