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Is there an example of a finitely presented infinite group in which every element has finite order? Or, is it known that every finitely presented infinite group has an element of infinite order?

I asked this question of math.stackexchange, thinking it might be trivial (for finitely generated groups there are numerous counterexamples...), but it seems the question is wide open. So more specifically, could someone point me to partial results, or give a good reason why this problem won't be solved any time soon?

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This is a famous open problem (which is still open as far as I know). See front.math.ucdavis.edu/0208.5237 for an example of a finitely presented monster, and for a discussion why they are hard to come by. –  Igor Belegradek Oct 18 '11 at 2:38
    
So the von Neumann problem talks about non-amenable groups without non-abelian free subgroups, but I'm just looking for the existence of a copy of $\mathbb{Z}$. Wouldn't this be easier? –  JeremyKun Oct 18 '11 at 3:01
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I do not have expertise to comment on what is easier but I suggest you read the discussion on page 3 of Sapir's survey arxiv.org/pdf/0704.2899.pdf, where he reviews what is known and also says "...I am reasonably sure that a finitely presented infinite bounded torsion group exists (note that unbounded torsion infinite finitely presented groups are not known also)...". –  Igor Belegradek Oct 18 '11 at 11:59

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up vote 19 down vote accepted

This is just an expanded version of Igor's comment. Indeed this is an open problem. The common opinion (I believe) is that such groups do exist, but the best result in this direction so far is the Olshanskii-Sapir group, which is finitely presented and (infinite torsion)-by-cyclic.

There is a general idea, commonly attributed to Rips, which shows that such groups should exist. The idea is the following. Take the free Burnside group $B(m,n)$ on $m\ge 2$ generators and of exponent $n>>1$, so that $B(m,n)$ is infinite. Embed it into a finitely presented group $G$ by using a version of the Higman embedding theorem. Let $G=\langle x_1, \ldots , x_n\rangle$. Then take the quotient group $Q$ of $G$ by the relations

(*) $x_1=b_1, \ldots , x_n=b_n$, where $b_1, \ldots , b_n\in B(m,n)$.

Clearly $Q$ is torsion being a quotient of $B(m,n)$ and is finitely presented. And it is believable that if the embedding of $B(m,n)$ in $G$ is "reasonably hyperbolic" and elements $b_1, \ldots , b_n$ are chosen randomly, then $Q$ is infinite with probability close to $1$. This idea was essentially implemented by Olshanskii and Sapir, but they managed to impose all relations of type (*) but one. (This is in fact a very rough interpretation of what they did, so Mark may correct me here.) There are even very particular choices of $b_1, \ldots , b_n$ (e.g., long aperiodic words with small cancellation), for which the group $Q$ should be infinite. So there are even particular finite presentations which should represent infinite torsion groups, but nobody knows how to prove that.

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