Subgroups of Additive Group of Integers

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

Let $n \Z$ be the additive group of integer multiples of $n$.

Every non-trivial subgroup of $\struct {\Z, +}$ has the form $n \Z$.

Proof
First we note that, from Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is an infinite cyclic group.

From Cyclic Group is Abelian, it follows that $\struct {n \Z, +}$ is an infinite abelian group.

Let $H$ be a non-trivial subgroup of $\struct {\Z, +}$.

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:
 * $H \cap \Z_{>0} \ne \O$

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

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

Thus:
 * $n \Z \subseteq H$


 * $\exists m \in \Z: m \in H \setminus n \Z$
 * $\exists m \in \Z: m \in H \setminus n \Z$

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 $H \cap \Z$.

Hence, by Proof by Contradiction, there can be no such $m \in H \setminus n \Z$.

Thus:
 * $H \setminus n \Z = \O$

Thus from Set Difference with Superset is Empty Set‎:
 * $H \subseteq n \Z$

Thus we have $n \Z \subseteq H$ and $H \subseteq n \Z$.

Hence:
 * $H = n \Z$