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Importance: Outstanding Author(s): Burnside, WilliamSubject: Group
TheoryKeywords:Recomm. for undergrads: noPosted by: dlh12on: April
11th, 2008Conjecture If a group has $ r $ generators and exponent $
n $, is it necessarily finite?It is possible to define the $
\emph{free Burnside group} $ $ 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\[
\] Burnside Problem -- from Wolfram MathWorldBibliography*
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