Burnside problem

Importance: Outstanding ✭✭✭✭
Author(s): Burnside, William
Subject: Group Theory
Keywords:
Recomm. for undergrads: no
Posted by: dlh12
on: April 11th, 2008
Conjecture   If a group has $ r $ generators and exponent $ n $, is it necessarily finite?

It is possible to define the $ <em>free Burnside group</em> $ $ B(r,n) $ to be the group generated by $ x_{1}, \ldots, x_{r} $ with relations $ w^{n}=1 $ where $ w $ ranges over every word in the generators. There is a universality property: Any homomorphism $  \phi : G \to H $ where $ H $ has r generators and exponent dividing $ n $ can be written as a composition of a homomorphism $  \psi : G \to B(r,n)  $ with a homomorphism $  \pi : B(r,n) \to H $. Some cases of this are known: $ B(1,n) $ is a cyclic group of order $ n $, for any positive integer $ n $. $ B(r,1) $ is trivial for any positive integer $ r $. $ B(r,2) $ is isomorphic to the Cartesian product of $ r $ cyclic groups of order $ 2 $, for any positive integer $ r $. This is because the relations make it easy to prove that the generators commute. $ B(r,3) $ is a finite group, and its order is \[    3^{r + \binom{r}{2} + \binom{r}{3}}. \] $ B(r,6) $ is a finite group, and its order is \[   2^{1 + (r-1)3^{r + \binom{r}{2} + \binom{r}{3}} }3^{1 + (r-1)2^{r + \binom{r}{2} + \binom{r}{3}} }. \] $ B(r,4) $ is a finite group for any positive integer $ r $. The order is known for $ r $ up to $ 5 $: \[ |B(1,4)| = 2^{2} \] \[ |B(2,4)| = 2^{12} \] \[ |B(3,4)| = 2^{69}  \] \[ |B(4,4)| = 2^{422} \] \[ |B(5,4)| = 2^{2728} \] $ B(r,n) $ is known to be infinite for sufficiently large $ r $ and odd $ n \geq 665 $, as well as $ r > 1 $ and $ n \geq 10^{48} $ divisible by $ 2^{9} $.

Burnside_problem \[</p>
<p>\] Burnside Problem -- from Wolfram MathWorld

Bibliography



* indicates original appearance(s) of problem.