Re: [math-fun] Interesting unsolved problem in group theory

Huh? I'm confused. Wouldn't B(2,2) be the free group on {a,b} modulo {a^2,b^2}, so contain all square-free words? But that's clearly infinite, with an element ab of infinite order. --Michael Kleber kleber@brandeis.edu Dan Asimov said:

I learned from some recent postings on sci.math.research that if F(r) is a free group (on r >= 2 generators), and if G^n denotes the subgroup generated by all nth powers of elements of a group G, then it's unknown whether the quotient group,

                                B(r,n) = F(r) / F(r)^5 is finite or infinite.

It is known that B(r,n) is always finite for n = 2,3,4, and 6 (cf. Marshall Hall's book, "Group Theory"), as well as for sufficiently large n (apparently independent of r).

--Dan