# Class number of Burnside groups

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?

• 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.
– YCor
Commented May 10, 2016 at 20:38

• 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. Commented Feb 16, 2016 at 20:11
• 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. Commented Feb 16, 2016 at 20:15
• 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. Commented Feb 17, 2016 at 8:08