Any subgroup $H$ (or ideal) in $\mathbb{Z}$ is of the form $(m) = m\mathbb{Z}$ for some $m\in\mathbb{Z}$

Question

Any subgroup $H$ (or ideal) in $\mathbb{Z}$ is of the form $(m) = m\mathbb{Z}$ for some $m\in\mathbb{Z}$.

Proof:

Suppose $H=\{0\}$, then $H=(0)$.

Now suppose $H\not=\{0\}$. Then there are positive integers in $H$ (take $a\in H$, if $a<0$ then take $-a\in H$ since H is a subgroup). Let $m$ be the smallest positive number in $H$ and let $n\in H$ be arbitrary, then we may write $$n=qm+r,\qquad \text{where } q,r\in\mathbb{Z},~ 0\le r<m.$$ Hence $r=n-qm\in H$ since $H$ is ideal. Thus $(*)r=0 \implies H\subset m\mathbb{Z}$. Note $am\in H$ for any $a\in\mathbb{Z}$ thus $m\mathbb{Z}\subset H$ and hence $H=m\mathbb{Z}$.

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$(*)$: How is this implication deduced?

Answer 1

If $r>0$, then $m$ would not be the smallest positive member of $H$.