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 nontrivial subgroup of $$\left({\Z, +}\right)$$ has the form $$n \Z$$.

Proof

 * 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 $$\N^*$$.

Thus $$H \cap \N^* \ne \varnothing$$.

From the Well-Ordering Principle, $$H \cap \N^*$$ 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 $$m \in H - n \Z$$.

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

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 $$0 < r < n$$.

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

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

Thus $$H - n \Z = \varnothing$$, thus from Subset Equivalences, $$H \subseteq n \Z$$.


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