From: boas@math.tamu.edu (Harold P. Boas)
Subject: Re: Burnside Problem
Date: 25 Feb 1999 12:44:37 -0600
Newsgroups: sci.math,sci.math.research
Keywords: Newer examples than Golod's to General Burnside Problem
"Markus Reitenbach" writes:
> The General Burnside Problem asks whether a finitely generated group in
> which every element has finite order is necessarily finite. The answer is
> no, and the first counterexample was constructed by Golod in 1964.
>
> Are there any newer counterexamples?
My colleague Jon McCammond (who has a preprint on "Burnside
Groups and Small Cancellation Theory" available on the web at
http://www.math.tamu.edu/~jon.mccammond/publications/ )
offers the following comment:
There are a number of more recent
(and elegant) examples which involve actions on rooted
uniformly branching trees (== wreath products). The main
class of examples is due to Grigorchuk (1980) and Gupta &
Sidki (1983). A nice write up is given in Gilbert
Baumslag's book "Topics in Combinatorial Group Theory"
(Birkhauser, 1993).
--
Harold P. Boas mailto:boas@math.tamu.edu
World-Wide Web URL: http://www.math.tamu.edu/~Harold.Boas/
--
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From: magidin@hilbert.berkeley.edu (Arturo Magidin)
Subject: Re: Burnside Problem
Date: 24 Feb 1999 18:54:54 GMT
Newsgroups: sci.math,sci.math.research
In article <01be5ffb$8560dba0$83693c86@Sproll913.wohnheim.uni-ulm.de>,
Markus Reitenbach wrote:
>The General Burnside Problem asks whether a finitely generated group in
>which every element has finite order is necessarily finite. The answer is
>no, and the first counterexample was constructed by Golod in 1964.
>
>Are there any newer counterexamples?
I suggest you take a look at:
N. Gupta, "On groups in which every element has finite order"
Amer. Math. Monthly 96 (1989) no. 4, pp. 297-308.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
==============================================================================
From: "A. Caranti"
Subject: Re: Help on fintely-generated groups with elements of finite order.
Date: Fri, 28 May 1999 17:40:27 +0200
Newsgroups: sci.math.research
J. Mayer wrote:
> Given a finitely generated group G such that the order of each element
> is numbers, I need to know, when such a group is itself finite, perhaps
> under additional conditions. The easiest case is of course for N=2,
[...]
> For N>2 this seems to be a bit harder and I am hoping that
> in this case as well the group is finite, but I have been unable to find
> a proof or to prove it myself (although I have ideas as to how one could
> generalize the proof for N=2) and I would be grateful for any help.
It is indeed harder. You have stated the so-called Burnside problem: is
the group B(r, n) finite? Here B(r, n) is the freest group on r
generators of exponent n. Deep results of Novikov, Adyan and Olshanskii
now tell you that B(r, n) is infinite if r > 1 and n > 665 is odd.
On the positive side, there is a biggest finite group R(r, n) in r
generators of exponent n. Zelmanov got the Fields medal for this result,
which gives a positive solution to the so-called restricted Burnside
problem.
Many authors have given important contributions to these problems.
Search MathRev for the long list of contributions, or ask me and I'll
send you a list of references.
Andreas
==============================================================================
From: mareg@lily.csv.warwick.ac.uk (Dr D F Holt)
Subject: Re: Help on fintely-generated groups with elements of finite order.
Date: 28 May 1999 16:27:21 GMT
Newsgroups: sci.math.research
In article <374E6EE0.566A7928@hadiko.de>,
"J. Mayer" writes:
>Given a finitely generated group G such that the order of each element
>is numbers, I need to know, when such a group is itself finite, perhaps
>under additional conditions. The easiest case is of course for N=2,
>because then we have x^2=1 for all x in G , which implies that the group
>is abelian and it is easy to "list" all elements: finite products of the
>generators, each generator occuring maximally once, and if it occurs
>then only to the first power- hence #G<=2^g, g being the number of
>generators. For N>2 this seems to be a bit harder and I am hoping that
>in this case as well the group is finite, but I have been unable to find
>a proof or to prove it myself (although I have ideas as to how one could
>generalize the proof for N=2) and I would be grateful for any help.
>
>Thanks in advance
>
>Jan Mayer
This is the well-known Burnside Problem. Let B(r,M) be the 'largest'
r-generator group in which x^M=1 for all x in the group.
Then, for r>1, B(r,M) is known to be finite only when M=2,3,4 or 6.
It is known to be infinite for large enough M - I think M larger than
a few hundred, but no doubt somebody else can supply latest results
and references.
B(2,5) is the smallest unknown case, and is one of the most challenging
open problems in group theory at present.
As for orders,
|B(r,2)| = 2^r,
|B(r,3)| = 3^m(r) with m(r) = (r choose 3) + (r choose 2) + r
|B(r,6)| = 2^a 3^(b + (b choose 2) + (b choose 3))
where a = 1 + (r-1)3^(r + (r choose 2) + (r choose 3))
and b = 1 + (r-1)2^r.
I don't think |B(r,4)| is known exactly, but for r=2,3,4 it
is 2^12, 2^69, 2^422, respectively.
Derek Holt.
==============================================================================
From: mareg@mimosa.csv.warwick.ac.uk ()
Subject: Re: Group Theory Question
Date: 29 Dec 1999 17:34:32 GMT
Newsgroups: sci.math.research
Keywords: which free Burnside groups are finite?
In article <842cga$55$1@nntp1.atl.mindspring.net>,
"Daniel Giaimo" writes:
> Let B(n,e) be the group with presentation
>
> I believe this is called the Free Burnside group on n generators and
>exponent e after his famous (false) conjecture that all such groups are
>finite? Anyway, my question is whether it is known for what values of n and
>e this group is finitely presentable. Also, is it known for what values
>this group is finite?
>
>--Daniel Giaimo
Let's assume that n and e are at least 2.
It is known that B(n,e) is infinite for all sufficiently large e -
certainly e > 8000 will do. It has also been shown more recently that,
for e > 8000, a group defined by any finite presentation in which all relators
are e-th powers contains elements of infinite order.
This shows that these B(n,e) cannot be defined by presentations using only
e-th powers, but I do not know whether they have been shown to be not
finitely presentable.
It is known that B(n,e) has a largest finite quotient for all n and e.
Apart from that, B(n,e) is known to be finite only when e=2,3,4 or 6.
In these cases it is of course finitely presentable, although if you only
allow e-th powers as relators, then in the case of e=6, the proven upper
bound on the number of 6th powers needed is something enormous - I
seem to remember it is more than 2^100, for B(2,6).
There is a huge amount of work going on on the first unknown case, B(2,5),
but nobody seems prepared to take a bet on whether it will turn out to
be finite or infinite.`
Derek Holt.