# Subgroups of Cyclic Groups

Theorem 1: Every subgroup of a cyclic group is cyclic.

Proof: Let $G = \left\{ a \right\}$ be a cyclic group generated by $a$. Let $H$ be a subgroup of $G$. Now every element of $G$, hence also of $H$, has the form ${a^s}$, with $s$ being an integer. Let $m$ be the smallest possible integer such that ${a^m} \in H$. We claim that $H = \left\{ {{a^m}} \right\}$. For this it is sufficient to show that ${a^s} \in H$, then $s = mh$ for then ${a^s} = {\left( {{a^m}} \right)^h}$. Now, if $m$ does not divide $s$, then there exist integers $q$ and $r$ such that
$s = mq + r,\,\,\,0 \leqslant r < m$

Then ${a^s} = {a^{mq + r}} = {a^{mq}} \cdot {a^r}$ or ${a^r} = {a^s} \cdot {\left( {{a^{mq}}} \right)^{ – 1}}$

Since ${a^m} \in H$, it follows that ${a^{mq}} \in H$ and hence its inverse ${\left( {{a^{mq}}} \right)^{ – 1}} \in H$.

But ${a^s} \in H$ by supposition. Then from the above result it follows that ${a^r} \in H$, contrary to the choice of $m$ since $m$ was assumed to be the least positive integer such that ${a^m} \in H$. Therefore $r = 0$ and so $s = mq$. But then ${a^s} = {\left( {{a^m}} \right)^q}$

Thus every element ${a^s}$ of $H$ is of the form ${\left( {{a^m}} \right)^q}$. Hence $H = \left\{ {{a^m}} \right\}$.

Theorem 2: Every subgroup of an infinite cyclic group is infinite.

Proof: Let $G = \left\{ a \right\}$ be an infinite cyclic group. Let $H$ be a subgroup of $G$. Then by the preceding theorem, $H = \left\{ {{a^m}} \right\}$ where $m$ is the least positive integer such that ${a^m} \in H$. Now suppose, if possible, that $H$ is finite.

This implies that ${\left( {{a^m}} \right)^s} = e$ for some $s > 0$.

It follows that $a$ is of finite order and this in turn implies that $G$ is finite, contrary to the hypothesis. Hence $H$ must be an infinite cyclic subgroup of $G$.