Bijection from Divisors to Subgroups of Cyclic Group

Theorem

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

Let $S = \set {m \in \Z_{>0}: m \divides n}$ 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:

$\phi: m \to \gen {a^{n / m} }$

where $\gen {a^{n / m} }$ 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 $\gen {a^{n / m} }$ of $G$ with $a$ elements.

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

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.8$