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 $\struct {\Z, +}$.

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

Suppose $H$ is a subgroup of $\struct {\Z, +}$.

From Subgroup of Integers is Ideal and Ring of Integers is Principal Ideal Domain, we have that:
 * $\exists m \in \Z_{> 0}: H = \ideal m$

where $\ideal m$ is the principal ideal of $\struct {\Z, +, \times}$ generated by $m$.

But $m$ is also a generator of the subgroup $\ideal m$ of $\struct {\Z, +}$, as:
 * $n \in \Z: n \circ m = n \cdot m \in \gen m$

Hence the result.