Let $B(m,n)$ be the Burnside group on $m$ generators of exponent $n$. Suppose the class number - the number of conjugacy classes - of $B(m,n)$ is finite. Does it imply that $B(m,n)$ is finite?
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$\begingroup$ Since there are misunderstandings, the question is: "it it true that $\forall m,n$ conj(B($m,n$)) finite implies B($m,n$) finite". So a negative answer means "$\exists m,n$ conj(B($m,n$)) finite and B($m,n$) infinite". This is definitely unknown (very probably for large exponent, it is known that the number of conjugacy classes is infinite, but for intermediate exponents e.g. where finiteness of B($m,n$) is unknown, it's just unknown. $\endgroup$– YCorCommented May 10, 2016 at 20:38
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No. There are Tarski monsters in which all proper subgroups are conjugated. I don't have a reference for this, but I read that Mann said that Rips said so.
answered Feb 16, 2016 at 20:04
Jan-Christoph Schlage-PuchtaJan-Christoph Schlage-Puchta
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1$\begingroup$ Jan-Christoph I think he meant that $B(m,n)$ is the free group of exponent $n$ on the $m$ generators. So to give a negative answer you need to find a group of exponent $n$ with $m$ generators which has infinite class number or a sequence of such groups which have unbounded class numbers. $\endgroup$ Commented Feb 16, 2016 at 20:11
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$\begingroup$ Oh, I read the question as "Suppose I can prove that $B(n,m)$ has finitely many conjugacy classes, can I deduce that $B(n,m)$ is finite?". A Tarski monster with finitely many conjugacy classes shows that this conclusion is not immediate. $\endgroup$ Commented Feb 16, 2016 at 20:15
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$\begingroup$ By a Burnside group $B(m,n)$, I mean the group $\langle x_1,\dotsc,x_n: w^n=1,\, w\text{ is a word in $x_i^{\pm1}$}\rangle$, not the quotient of this group. $\endgroup$ Commented Feb 17, 2016 at 8:08
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$\begingroup$ @Jan-ChristophSchlage-Puchta: Your understanding of problem is correct. But I think a Tarski monsters are quotients of Burnside groups, aren't? $\endgroup$ Commented Feb 17, 2016 at 8:24