Subgroups of Z/One generator/Fact/Proof/Exercise/Solution

A subset of the form $\mathbb{Z}d$ is a subgroup due to the distributive law. Let now $H\subseteq \mathbb{Z}$ be a subgroup. In case $H=0$, we can take $d=0$. Hence, we may assume that $H$ contains beside $0$ another element $x$. If $x$ id negative, then subgroup $H$ must contain its negative $-x$, , which is positive. This means that $H$ contains also a positive number. Let $d$ denote the smallest positive number in $H$. We claim $H=\mathbb{Z}d$. Here, the inclusion $\mathbb{Z}d\subseteq H$ is clear, as all (positive and negative) multiples of $d$ must belong to the subgroup. To show the inverse inclusion, let $h\in H$ be arbitrary. Due to the division with remainder, we have

$h=qd+r\text{ with }0\le r<d.$

Because of $h\in H$ and $qd\in H$, also $r=h-qd\in H$ holds. Because of the choice of $d$ and $r<d$, we must have $r=0$. This means $h=qd$. Therefore $h\in \mathbb{Z}d$, thus $H\subseteq \mathbb{Z}d$.

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