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  "body": "\n## Statement\n\nLet $G$ be a finite [[group|G5S1M9]] and let $H$ be a subgroup of $G$.\nThen the order of $H$ divides the order of $G$.\n\n## Remarks\n\nThe quotient $|G| / |H|$ is the number of left cosets of $H$ in $G$.\n\n## Explanations\n\n### L1 agent\n\nThe left cosets of a subgroup cut the group into pieces of the same size.\nEach piece has as many elements as the subgroup.\nThus the size of the group is a whole number of copies of the size of the subgroup.\nFor example, a group with $12$ elements can have subgroups of order $1, 2, 3, 4, 6$ or $12$, and of no other order.\n\n### L2 agent\n\nLagrange's theorem gives a strong limit on the subgroups of a finite group. The order of each subgroup is a divisor of the order of the group.\n\nThe idea of the proof is to count cosets:\n\n1. The left cosets $aH$ of $H$ partition $G$.\n2. Each left coset has exactly $|H|$ elements, because $h \\mapsto ah$ is a bijection from $H$ to $aH$.\n3. If there are $k$ cosets, then $|G| = k\\,|H|$.\n\nThe number $k$ is the index $[G : H]$. Thus $|G| = [G : H] \\cdot |H|$.\n\nA worked example with $G = \\mathbb{Z}/12\\mathbb{Z}$ and $H = \\{0, 4, 8\\}$:\n\n1. $H$ is a subgroup of order $3$.\n2. The cosets are $H$, $1 + H = \\{1, 5, 9\\}$, $2 + H = \\{2, 6, 10\\}$ and $3 + H = \\{3, 7, 11\\}$.\n3. There are $4$ cosets of $3$ elements each, and $4 \\cdot 3 = 12$.\n\nTwo standard consequences are that the order of each element divides $|G|$, and that a group of prime order is cyclic.\n\nA common error is to believe the converse. A divisor $d$ of $|G|$ need not be the order of a subgroup. The alternating group $A_4$ has order $12$, but it has no subgroup of order $6$. The converse is true for some classes of groups, for example for cyclic groups and for groups of prime power order.\n\nThe theorem needs $G$ to be finite. For an infinite group the index can still be finite, as for $2\\mathbb{Z}$ in $\\mathbb{Z}$, but the orders are not numbers.\n",
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