Subgroup of Cyclic Group is Cyclic/Proof 1

Proof
Let $G$ be a cyclic group generated by $a$.

Let $H$ be a subgroup of $G$.

If $H = \left\{{e}\right\}$, then $H$ is a cyclic subgroup generated by $e$.

If $H \ne \left\{{e}\right\}$, then $a^n \in H$ for some $n \in \Z$ (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 $0 \le r < m$ by the Division Algorithm.

It follows that $a^n = a^{mq + r} = \left({a^m}\right)^qa^r$

and hence that $a^r = \left({a^m}\right)^{-q} a^n$.

Since $a^m \in H$ so is its inverse $\left({a^m}\right)^{-1}$, and all powers of its inverse by closure.

Now $a^n$ and $\left({a^m}\right)^{-q}$ 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 $0 \le r < m$, so $r = 0$.

Therefore $n = q m$ and $b = a^n = \left({a^m}\right)^q$.

We conclude that any arbitrary element $b = a^n$ of $H$ is generated by $a^m$.

So, by definition, $H = \left \langle {a^m}\right \rangle $ is cyclic.