Burnside problem

Importance: Outstanding ✭✭✭✭

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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 $2^{9}$ .