# 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 $$H = \left\{ {{a^m}} \right\}$$.

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__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 $$G$$.