Bijection from Divisors to Subgroups of Cyclic Group

Theorem
Let $$G$$ be a cyclic group of order $$n$$ generated by $a$.

Let $$S = \left\{{m \in \Z^*_+: m \backslash n}\right\}$$ be the set of all divisors of $$n$$.

Let $$T$$ be the set of all subgroups of $$G$$

Let $$\phi: S \to T$$ be the mapping defined as:
 * $$\phi: m \to \left\langle{a^{n/m}}\right\rangle$$

where $$\left\langle{a^{n/m}}\right\rangle$$ is the subgroup generated by $a^{n/m}$.

Then $$\phi$$ is a bijection.

Proof
From Subgroup of Cyclic Group whose Order Divisor, there exists exactly one subgroup $$\left \langle {a^{n/m}} \right \rangle$$ of $$G$$ with $$a$$ elements.

So the mapping as defined is indeed a bijection by definition.