Subgroup of Cyclic Group is Cyclic

Theorem
A subgroup of a cyclic group is cyclic.

Proof 1
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 $0 \leq r < m$ 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 $0 \leq r < m$, so $r = 0$.

Therefore $n = q m$ 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 = \left \langle {a^m}\right \rangle $ is cyclic.

Proof 2
Let $G$ be a cyclic group generated by $a$.

Finite Group
Let $G$ be finite.

By Bijection from Divisors to Subgroups of Cyclic Group there are exactly as many subgroups of $G$ as divisors of the order of $G$.

As each one of these is cyclic by Subgroup of Cyclic Group whose Order Divisor‎, the result follows.

Infinite Group
By Morphism from Integers to Group, an infinite cyclic group is isomorphic to $\left({\Z, +}\right)$.

So all we need to do is show that any subgroup of $\left({\Z, +}\right)$ is cyclic.

Suppose $H$ is a subgroup of $\left({\Z, +}\right)$.

From Subgroup of Integers is Ideal and Ring of Integers is Principal Ideal Domain, we have that:
 * $\exists m \in \Z_+^*: H = \left({m}\right)$

where $\left({m}\right)$ is the principal ideal of $\left({\Z, +, \times}\right)$ generated by $m$.

But $m$ is also a generator the subgroup $\left({m}\right)$ of $\left({\Z, +}\right)$, as:
 * $n \in \Z: n \circ m = n \cdot m \in \left\langle{m}\right\rangle$

Hence the result.