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  "body": "\n## Definition\n\nLet $G$ be a [[group|G5S1M9]].\n$G$ is a *one-relator group* if it has a [[presentation|0JGEQXQ4AC]] $\\langle S \\mid r \\rangle$ with a single relator $r$.\nHere $r$ is an element of the [[free group|0KTW26C80J]] $F(S)$ on the set $S$.\nThat is, $G$ is isomorphic to the quotient of $F(S)$ by the normal subgroup generated by $r$.\n\n## Remarks\n\nThe relator may be taken cyclically reduced, since a cyclic permutation of $r$ is a conjugate of $r$ and generates the same normal subgroup.\nA one-relator group has an element of finite order other than $1$ if and only if $r$ is a proper power $s^n$ in $F(S)$ with $n \\ge 2$.\nIn that case, if $s$ is chosen not to be a proper power itself, then $s$ has order exactly $n$ in $G$.\n\nExample: $\\Z^2$ is a one-relator group.\nThe map $\\langle a, b \\mid a b a^{-1} b^{-1} \\rangle \\to \\Z^2$ with $a \\mapsto (1, 0)$ and $b \\mapsto (0, 1)$ is an isomorphism, since the relator says exactly that $a$ and $b$ commute.\n\nNon-example: $\\Z/2 \\times \\Z/2$ is not a one-relator group.\nIf $\\langle S \\mid r \\rangle$ with $S$ finite of size $k$ defined it, its abelianization would be $\\Z^k$ modulo the subgroup generated by one vector, which is $\\Z^{k-1} \\oplus \\Z/d$ for some integer $d \\ge 0$.\nThe group $\\Z/2 \\times \\Z/2$ is abelian, so it equals its abelianization.\nIt is finite, so $k - 1 = 0$, and then it would be cyclic, which it is not.\nA presentation with $S$ infinite gives an abelianization with infinitely many free factors $\\Z$, so it fails as well.\n",
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