---
tag: PZ9CPKHXVZ
kind: definition
name: One-relator group
msc: ["20F05", "20F06"]
defines: 68AFPHR4GN
uses: [G5S1M9, 0JGEQXQ4AC, 0KTW26C80J]
sources:
  - url: https://en.wikipedia.org/wiki/One-relator_group
---

## Definition

Let $G$ be a [[group|G5S1M9]].
$G$ is a *one-relator group* if it has a [[presentation|0JGEQXQ4AC]] $\langle S \mid r \rangle$ with a single relator $r$.
Here $r$ is an element of the [[free group|0KTW26C80J]] $F(S)$ on the set $S$.
That is, $G$ is isomorphic to the quotient of $F(S)$ by the normal subgroup generated by $r$.

## Remarks

The relator may be taken cyclically reduced, since a cyclic permutation of $r$ is a conjugate of $r$ and generates the same normal subgroup.
A 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 $n \ge 2$.
In that case, if $s$ is chosen not to be a proper power itself, then $s$ has order exactly $n$ in $G$.

Example: $\Z^2$ is a one-relator group.
The 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.

Non-example: $\Z/2 \times \Z/2$ is not a one-relator group.
If $\langle S \mid r \rangle$ 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 $d \ge 0$.
The group $\Z/2 \times \Z/2$ is abelian, so it equals its abelianization.
It is finite, so $k - 1 = 0$, and then it would be cyclic, which it is not.
A presentation with $S$ infinite gives an abelianization with infinitely many free factors $\Z$, so it fails as well.