Subgroup of Cyclic Group is Cyclic

Theorem: A subgroup of a cyclic group is cyclic.

Proof:

Let $$G$$ be a cyclic group generated by $$a$$ and let $$H$$ be a subgroup of $$G$$. If $$H = \{e\}$$ then $$H$$ is a cyclic subgroup generated by $$e$$. If $$H \neq \{e\}$$, then $$a^n \in H$$ for some positive integer n (since every element in $$G$$ has the form $$a^n$$ and $$H$$ is a subgroup of $$G$$). Let $$m$$ be the smallest positive integer such that $$a^m \in H$$.

Consider an arbitrary element $$b$$ of $$H$$. Since $$H$$ is a subgroup of $$G$$, $$b = a^n$$ for some $$n$$. Find integers $$q$$ and $$r$$ such that $$n = mq + r$$ with $$0 \leq r < m$$ by the Division Algorithm. It follows that $$a^n = a^{mq + r} = (a^m)^qa^r$$ and hence that $$a^r = (a^m)^{-q}a^n.$$ Since $$a^m \in H$$ so is its inverse $$(a^m)^{-1}$$, and all powers of its inverse by closure. Now $$a^n$$ and $$(a^m)^{-q}$$ are both in $$H$$, thus so is their product $$a^r$$ by closure. However, $$m$$ was the smallest positive integer such that $$a^m \in H$$ and $$0 \leq r < m$$, so $$r = 0$$. Therefore $$n = qm$$ and $$b = a^n = (a^m)^q.$$

We conclude that any arbitrary element $$b = a^n$$ of $$H$$ is generated by $$a^m$$ so $$H = \langle a^m \rangle $$ is cyclic.

Reference: John B. Fraleigh, A first course in abstract algebra, seventh edition, Addison Wesley (2003).