From: mareg@csv.warwick.ac.uk (Dr D F Holt) Newsgroups: sci.math Subject: Re: Group theory Date: 1 Mar 1996 18:00:38 -0000 In article , gunnar@nada.kth.se (Gunnar Andersson) writes: >Is there anybody who know if this proposition is true? > "If G is a finitely generated group in which every element > has finite order, then G is finite." > >/ Gunnar Andersson, PhD student in Theoretical Computer Science > The proposition is false. It was first asked by Burnside in 1902, and so is known as the Burnside Problem. It remained unsolved until 1964, when Golod gave examples of finitely generated infinite p-groups for all primes p. There are now counterexamples known that are relatively easy to construct - I seem to remember they are described as permutations of subsets of the real line (possibly the diadic rationals, but I don't quite remember). In any case, I can dig out details of that if you are interested (or hopefully somebody else will know). A much more difficult question is, given a fixed integer n, are there finitely generated infinite groups in which each element has order dividing n. The answer to this is known to be yes for n=2,3,4 and 6. It is probably no for all other n, and has been proved no for large enough n (I think n more than a few hundred.) It was first proved to be no for sufficiently large odd n by Novikov and Adian in 1968 in a horrendously lengthy and intricate proof. Most of the original literature on this topic was in Russian, but I believe there are more accessible accounts available in English now. Again, hopefully someone else will supply more precise details if you need them. Another interesting type of group, which seems intuitively improbable, but which have been shown to exist for large enough primes p are the Tarski Monsters. These are infinite groups in which all nontrivial elements have order p, and the only subgroups of the group are of order p. Derek Holt. ============================================================================== From: kubo@abel.harvard.edu (Tal Kubo) Newsgroups: sci.math Subject: Re: Group theory Date: 2 Mar 1996 07:46:52 GMT In article <4h7e06\$cps@crocus.csv.warwick.ac.uk>, Dr D F Holt wrote: > >[Burnside problem] > >Most of the original literature on this topic was in Russian, but >I believe there are more accessible accounts available in English now. Kostrikin's book has been translated into English: Kostrikin, A. I. (Aleksei Ivanovich). Around Burnside. Ergebnisse der Mathematik und ihrer Grenzgebiete; 3. Folge, Bd. 20. Berlin; New York: Springer-Verlag, c1990. LC# QA171.K67413 1990 also: Vaughan-Lee, Michael. The restricted Burnside problem. Oxford science publications. London Mathematical Society monographs; new ser., 8. 2nd ed. Oxford: Clarendon Press; New York: Oxford University Press, 1993. LC# QA177.V38 1993 (n.b only the second edition describes Zelmanov's solution of RBP). ============================================================================== From: mareg@crocus.csv.warwick.ac.uk (Dr D F Holt) Newsgroups: sci.math Subject: Re: Probably a simple group theory question Date: 4 Sep 1998 10:58:18 GMT In article <35eed051.178484500@news.kodak.com>, mwdaly@pobox.com writes: >After many many years, I'm going through my old abstract algebra >textbooks, and I came across a problem that is driving me up a tree. >Here it is: > >Show that a group in which every a satisfies a^2 = 1 is abelian. What >if a^3 = 1 for every a? > Analysing groups with a^3 = 1 for all a is more difficult than for a^2 = 1, but is still reasonably easy. It turns out that such groups are nilpotent of class at most 3 (so they are not necessarily abelian). The largest group B(r,3) with r generators has order 3^m(r), where m(r) = (r choose 3) + (r choose 2) + r so, for example, m(3)=7. One reference for the proof is the chapter on the Burnside Problem in M. Hall, 'The Theory of Groups'. This is of course a special case of the well-known Burnside Problem. Let B(r,n) be the 'largest' group with r generators such that a^n=1 for all elements a. (This is well-defined even when B(r,n) is infinite.) Then B(r,n) is known to be finite for n=2,3,4 and 6, and infinite for large enough n - about n>= 8000 is the best relaible result, although n > 117 has been claimed for odd n. So why don't you forget about n=3 and concentrate on the first unknown case which is B(2,5) - 2-generator groups with a^5=1 for all a. If you can decide whether these can be infinite or not, then you may not make the front page of the Press, but you will gain great acclaim in mathematical circles. It is known BTW that for all r and n there is a largest finite r-generator group of exponent n, but this does not imply that infinite groups do not exist. In the case of B(2,5) the largest finite group has order 5^34. Derek Holt,.