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  "body": "\n## Statement\n\nLet $G$ be a [[group|G5S1M9]] with [[centre|0DZ93493VN]] $Z(G)$.\nIf the [[quotient group|0KYX2N69J0]] $G/Z(G)$ is [[cyclic|0KVGH0J5B3]], then $G$ is [[abelian|082GHNPJ53]].\n\n## Remarks\n\nLet $gZ(G)$ generate $G/Z(G)$; then each element of $G$ is $g^k z$ with $k \\in \\mathbb{Z}$ and $z \\in Z(G)$, and two such elements commute.\nIn that case $G = Z(G)$, so in fact $G/Z(G)$ is trivial.\nA consequence is that $G/Z(G)$ is never cyclic of order greater than $1$. For example, the centre of a group never has prime index.\n\n## Explanations\n\n### L1 human\n\nIf the elements of $G$ are powers of one element, up to central elements, then all elements commute.\nPowers of one element commute with each other, and central elements commute with everything.\nSo the quotient by the centre cannot be a nontrivial cyclic group.\n",
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