Subgroups of Z/One generator/Fact/Proof

Proof

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

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

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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