---
tag: 0423TXR9AN
kind: statement
role: theorem
name: Subgroups of a finite cyclic group correspond to divisors of its order
aliases: [fundamental-theorem-of-cyclic-groups]
msc: [20K01]
uses: [0B85Q41RP6, 0KB1ND7XDS, 0VVPQNVM8S]
sources:
  - url: https://en.wikipedia.org/wiki/Cyclic_group
    locator: Subgroups
---

## Statement

Let $G$ be a [[cyclic group|0KVGH0J5B3]] of finite order $n$, generated by $g$.
For each positive divisor $d$ of $n$, $G$ has exactly one [[subgroup|0X1JXT47GA]] 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|3C53TD]] says, and here every divisor occurs.