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 a 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.