---
tag: 04RG02S6BA
kind: statement
role: proposition
name: Cyclic groups are abelian
msc: [20A05]
about: [0KVGH0J5B3, 082GHNPJ53]
uses: [0B85Q41RP6, 00DRK90Z6K, 071DDFVG4J]
formal:
  - system: lean4
    library: mathlib
    decl: IsCyclic.isMulCommutative
    match: variant
    adapter: Math2.Match.cyclic_comm
sources:
  - url: https://en.wikipedia.org/wiki/Cyclic_group
    locator: Additional properties
---

## Statement

Let $G$ be a [[cyclic group|0KVGH0J5B3]].
Then $G$ is [[abelian|082GHNPJ53]].

## Remarks

If $g$ generates $G$, then any two elements are $g^a$ and $g^b$ for integers $a, b$, and $g^a g^b = g^{a+b} = g^b g^a$ by [[the laws of exponents|00DRK90Z6K]].
The converse is false: the Klein four-group is abelian but not cyclic.
The group $(\mathbb{Q}, +)$ is abelian but not cyclic.

## Explanations

### L1 human

All elements of a cyclic group are powers of one element.
Two powers of the same element always commute, because the exponents just add.
Thus the order of the factors never matters in a cyclic group.

### L2 agent

A cyclic group is generated by one element $g$. Each element is a power $g^m$ for some integer $m$. The result says that such a group is abelian.

The reason is that powers of one element add their exponents:

1. Let $x = g^m$ and $y = g^n$.
2. By the laws of exponents, $xy = g^{m+n}$ and $yx = g^{n+m}$.
3. Addition of integers is commutative, so $xy = yx$.

A worked example in $\mathbb{Z}/6\mathbb{Z}$ under addition, with generator $1$:

1. Write $2 = 1 + 1$ and $5 = 1 + 1 + 1 + 1 + 1$.
2. Then $2 + 5$ and $5 + 2$ are both seven copies of $1$.
3. Seven copies of $1$ give $7 \bmod 6 = 1$. Thus $2 + 5 = 5 + 2 = 1$.

The converse is false. The Klein four-group $V = \{e, a, b, c\}$ is abelian, but each element other than $e$ has order $2$. Thus no element generates the four elements, and $V$ is not cyclic. The group $(\mathbb{Q}, +)$ is another abelian group that is not cyclic.

A useful consequence is a test for groups that are not cyclic. If a group is not abelian, then it is not cyclic. For example, $S_3$ is not abelian, so it is not cyclic.

Each subgroup of a cyclic group is also cyclic. This fact needs a different proof, by division with remainder.