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  "body": "\n## Statement\n\nLet $G$ be a [[cyclic group|0KVGH0J5B3]].\nThen $G$ is [[abelian|082GHNPJ53]].\n\n## Remarks\n\nIf $g$ 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]].\nThe converse is false: the Klein four-group is abelian but not cyclic.\nThe group $(\\mathbb{Q}, +)$ is abelian but not cyclic.\n\n## Explanations\n\n### L1 human\n\nAll elements of a cyclic group are powers of one element.\nTwo powers of the same element always commute, because the exponents just add.\nThus the order of the factors never matters in a cyclic group.\n\n### L2 agent\n\nA 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.\n\nThe reason is that powers of one element add their exponents:\n\n1. Let $x = g^m$ and $y = g^n$.\n2. By the laws of exponents, $xy = g^{m+n}$ and $yx = g^{n+m}$.\n3. Addition of integers is commutative, so $xy = yx$.\n\nA worked example in $\\mathbb{Z}/6\\mathbb{Z}$ under addition, with generator $1$:\n\n1. Write $2 = 1 + 1$ and $5 = 1 + 1 + 1 + 1 + 1$.\n2. Then $2 + 5$ and $5 + 2$ are both seven copies of $1$.\n3. Seven copies of $1$ give $7 \\bmod 6 = 1$. Thus $2 + 5 = 5 + 2 = 1$.\n\nThe 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.\n\nA 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.\n\nEach subgroup of a cyclic group is also cyclic. This fact needs a different proof, by division with remainder.\n",
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  "name": "Cyclic groups are abelian",
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