[math-fun] Interesting unsolved problem in group theory
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
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
Sorry, forget that, clearly I'm an idiot... On Thursday, January 30, 2003, at 09:29 AM, Michael Kleber wrote:
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
At 09:29 AM 1/30/03 -0500, Michael Kleber wrote:
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.
I think Michael is confused because he has misread Dan A's definition of G^2. If G is <a,b>, G^2 is not <a^2,b^2>. For example, <a,b>^2 contains abab, while <a^2,b^2> doesn't. If I'm doing this right, <a,b>/<a,b>^2 has four elements and is isomorphic to Z[2]^2 = V. This problem feels to me like a grown-up version of the problem of stutter-free strings that we beat to death about a decade ago on this list.
--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
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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).
1992 S I Ivanov proved that B(r, n) is infinite for r > 2 and n >= 13. See A History of the Burnside Problem at http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Burnside_problem.html --Edwin
participants (4)
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Allan C. Wechsler -
asimovd＠aol.com -
Edwin Clark -
Michael Kleber