Subgroups of a finite cyclic group correspond to divisors of its order

Statement

Let $G$ be a cyclic group of finite order $n$, generated by $g$. For each positive divisor $d$ of $n$, $G$ has exactly one subgroup of order $d$, namely $\langle g^{n/d} \rangle$. Each subgroup of $G$ has this form for some positive divisor $d$ of $n$.

Remarks

Thus the number of subgroups of $G$ is the number of positive divisors of $n$. For $d, d'$ dividing $n$, the subgroup of order $d$ is contained in the subgroup of order $d'$ exactly when $d$ divides $d'$. For example, $\mathbb{Z}/12\mathbb{Z}$ has six subgroups, of the orders $1, 2, 3, 4, 6, 12$.

Explanations

L1 human

In a cyclic group, the subgroups are as simple as possible: there is one for each divisor of the order. In the clock group $\mathbb{Z}/12\mathbb{Z}$, the subgroup of order $4$ is $\{0, 3, 6, 9\}$, generated by $12/4 = 3$. The order of each subgroup divides $12$, as Lagrange's theorem says, and here every divisor occurs.