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  "body": "\n## Statement\n\nLet $p$ be a prime number and let $G$ be a [[group|G5S1M9]] of order $p^2$.\nThen $G$ is [[abelian|082GHNPJ53]].\n\n## Remarks\n\nBy the [[nontrivial centre theorem|0RZCTT30H5]] and [[Lagrange's theorem|3C53TD]], $|Z(G)|$ is $p$ or $p^2$.\nIf $|Z(G)| = p$, then $G/Z(G)$ has order $p$, so it is cyclic, and [[this criterion|0AH9Q6YNMT]] gives a contradiction.\nThus $Z(G) = G$, and $G$ is isomorphic to $\\mathbb{Z}/p^2\\mathbb{Z}$ or to $\\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$.\nFor $p^3$ the result is false: the [[dihedral group|0JZ27DMJ5D]] $D_4$ and the [[quaternion group|0NRJGW864D]] have order $8$ and are not abelian.\n\n## Explanations\n\n### L1 human\n\nA group of order $p^2$ has a centre with more than one element.\nIf the centre were not the whole group, the quotient by the centre would be cyclic, and that is impossible.\nSo all elements commute.\n",
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