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  "body": "\n## Counterexample\n\nThe converse of [[Lagrange's theorem|3C53TD]] would say this. If $G$ is a finite group and $d$ is a positive divisor of $|G|$, then $G$ has a subgroup of order $d$.\nThis is false.\nLet $A_4$ be the group of the even permutations of $\\{1, 2, 3, 4\\}$, under composition.\nIt has order $12$, and $6$ divides $12$, but $A_4$ has no subgroup of order $6$.\n\n## Remarks\n\nThe elements of $A_4$ are the identity, the three products of two disjoint transpositions, and the eight $3$-cycles.\nA subgroup $H$ of order $6$ would have index $2$, and thus it would contain the square of each element of $A_4$.\nThe square of each $3$-cycle is a $3$-cycle, and each $3$-cycle is such a square. So $H$ would contain all eight $3$-cycles, which is too many.\nThe group $A_4$ is the smallest group for which the converse fails.\n\n## Explanations\n\n### L1 human\n\nLagrange's theorem says which sizes a subgroup can have, but it does not promise that each allowed size occurs.\nThe group $A_4$ has $12$ elements, and $6$ is an allowed size, but no subgroup has $6$ elements.\nFor cyclic groups the converse is true, so a counterexample must be a group that is not cyclic.\n",
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    "name": "Converse of Lagrange's theorem fails for the alternating group A4",
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        "url": "https://en.wikipedia.org/wiki/Lagrange's_theorem_(group_theory)"
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  "markdown": "https://math2.tamnd.com/node/0PDBVPP233.md",
  "msc": [
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  "name": "Converse of Lagrange's theorem fails for the alternating group A4",
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  "url": "https://math2.tamnd.com/wiki/converse-of-lagranges-theorem-fails-for-the-alternating-group-a4/",
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