Bijection from Divisors to Subgroups of Cyclic Group

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

Let $S = \left\{{m \in \Z_{>0}: m \backslash n}\right\}$ 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 Finite Cyclic Group is Determined by Order, 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.