Subgroup of Cyclic Group is Cyclic/Proof 2

Proof
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 Finite Cyclic Group is Determined by Order‎, 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:
 * $\exists m \in \Z_{> 0}: H = \left({m}\right)$

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.