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  "body": "\n## Description\n\nLet $p$ be a prime.\nA *Tarski monster group* for $p$ is an infinite [[group|G5S1M9]] $G$ in which every [[subgroup|0X1JXT47GA]] other than $\\set{1}$ and $G$ is cyclic of order $p$.\nOl'shanskii proved, with [[small cancellation|3A1D37TVF6]] methods, that a Tarski monster group exists for every prime $p \u003e 10^{75}$.\n\n## Remarks\n\nLet $g \\ne 1$ in $G$.\nIf $\\langle g \\rangle = G$, then $G \\cong \\Z$, which has the subgroup $2\\Z$ of infinite order; this is impossible.\nSo $\\langle g \\rangle$ is a proper subgroup other than $\\set{1}$, and $g$ has order $p$.\nTwo elements $x, y$ that do not lie in one subgroup of order $p$ generate $G$, since $\\langle x, y \\rangle$ is neither trivial nor cyclic of order $p$.\nSo $G$ is generated by $2$ elements and has [[exponent|0ZDW0H0VX4]] $p$, and it answers the bounded [[Burnside problem|XWEMJ7T2XV]] in the negative for this $p$.\n\n$G$ is [[simple|0YMYMNQ0QZ]].\nLet $N \\ne \\set{1}$ be a [[normal subgroup|0CB0DX5BVV]] with $N \\ne G$; then $N$ has order $p$.\nTake $x \\notin N$; then $\\langle x \\rangle N$ is a subgroup of order at most $p^2$, so it is proper since $G$ is infinite.\nIt contains $N$ and $x$, so it is not cyclic of order $p$, a contradiction.\n\n## Explanations\n\n### L1 human\n\nIn a Tarski monster, every subgroup except the trivial one and the whole group has exactly $p$ elements.\nSo there are no medium-sized subgroups at all: any two elements that are not in the same small subgroup generate the whole infinite group.\nGroups of this kind were long expected not to exist, which is why they are called monsters.\n",
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