---
tag: KHTCR3VFST
kind: definition
name: Finitely presented group
msc: [20F05]
defines: TJKANTKASZ
uses: [G5S1M9, 0JGEQXQ4AC, 0KTW26C80J, 6XM4CCVTKE, 08MG7H8JW5, 082DWWXMRJ]
sources:
  - url: https://en.wikipedia.org/wiki/Presentation_of_a_group
---

## Definition

Let $G$ be a [[group|G5S1M9]].
For a set $S$ and a subset $R$ of the [[free group|0KTW26C80J]] $F(S)$, let $N(R)$ be the [[normal closure|6XM4CCVTKE]] of $R$ in $F(S)$.
The group $G$ is *finitely presented* if it is [[isomorphic|08MG7H8JW5]] to $\langle S \mid R \rangle = F(S) / N(R)$ for some finite $S$ and finite $R$.

## Remarks

The pair $(S, R)$ is a finite [[presentation|0JGEQXQ4AC]] of $G$.
Example: $\Z^2 \cong \langle a, b \mid a b a^{-1} b^{-1} \rangle$; the map $a \mapsto (1, 0)$, $b \mapsto (0, 1)$ induces the isomorphism, since the quotient is abelian and generated by two elements, and $\Z^2$ is the free abelian group on them.
Every finite group $G$ is finitely presented, with $S = G$ and $R = \set{ g h k^{-1} : g, h, k \in G, \ gh = k \text{ in } G}$.
Non-example: the additive group $\Q$ is not finitely presented; it fails already because it is not [[finitely generated|082DWWXMRJ]], since a subgroup generated by finitely many fractions has bounded denominators.
A finitely generated group need not be finitely presented; the [[lamplighter group|4GNPFMT7DZ]] $\Z/2 \wr \Z$ is such an example.
If $G$ is finitely presented, then for every finite generating set $S$ of $G$ there is a finite set of relations $R$ with $G \cong \langle S \mid R \rangle$.