---
tag: XWEMJ7T2XV
kind: problem
name: Burnside problem
msc: ["20F50", "20F05"]
uses: [082DWWXMRJ, JZNQTTMG58, 0ZDW0H0VX4, 0KTW26C80J, 0WE2NHZJ7B]
about: [JZNQTTMG58, 082DWWXMRJ]
sources:
  - url: https://en.wikipedia.org/wiki/Burnside_problem
same_as:
  wikidata: Q852206
  wikipedia: Burnside problem
history:
  posed_by: "William Burnside"
  first_stated: "1902"
---

## Problem

1. General Burnside problem: let $G$ be a [[finitely generated group|082DWWXMRJ]] in which every element has finite [[order|0WE2NHZJ7B]]. Must $G$ be finite?
2. Bounded Burnside problem: let $m \ge 1$ and $n \ge 1$ be integers. Let $B(m, n)$ be the quotient of the [[free group|0KTW26C80J]] $F_m$ of rank $m$ by the normal subgroup generated by all $w^n$ with $w \in F_m$. For which $m$ and $n$ is $B(m, n)$ finite?

## Remarks

The group $B(m, n)$ is the free Burnside group: every group with $m$ generators and [[exponent|0ZDW0H0VX4]] dividing $n$ is a quotient of it.
So the bounded problem asks whether all groups with $m$ generators and exponent $n$ are finite of bounded order.
Burnside posed the problem in 1902.

The general problem has a negative answer.
Golod and Shafarevich built infinite finitely generated $p$-groups in 1964, and the [[Grigorchuk group|M0WAT1VQZH]] is a later explicit example.

For the bounded problem, $B(m, n)$ is finite for $n = 2, 3, 4, 6$ and all $m$.
For instance $B(m, 2) \cong (\Z/2)^m$, since $x^2 = 1$ for all $x$ forces the group to be abelian.
Novikov and Adian proved that $B(m, n)$ is infinite for $m \ge 2$ and all odd $n$ large enough, and later work extended this to large even $n$.
[[Tarski monster groups|TE19BGFFF1]] are infinite $2$-generated groups of large prime exponent.
Whether $B(2, 5)$ is finite is open.

## Explanations

### L1 human

In a finite group, every element returns to the identity after finitely many repetitions.
Burnside asked about the converse: if a group is generated by finitely many elements and every element returns to the identity, must the group be finite?
The answer is no in general, and even if all elements satisfy $x^n = 1$ for one fixed $n$, the answer is no for many large $n$.
The question for $2$ generators and $n = 5$ is still open.