# word problem in free Burnside groups (and other torsion groups)

Question 1. Is it known that for some free Burnside groups the word problem is undecidable?

Provided that the answer is negative, what about the following easier question.

Question 2. Is there a known example of a finitely generated (and preferably finitely presented) group $G$ and an integer $k$ such that all elements of $G$ have order at most $k$ and the word problem in $G$ is undecidable?

• I am far from being an expert, but are there any examples of infinite finitely presented torsion groups? Commented Jun 2, 2011 at 11:38
• @Yiftach: I've just googled that this was an open question in 2002 (paragraph 3, page 3 in the article "non-amenable finitely presented torsion-by cyclic groups" by Ol'shanskii and Sapir), so perhaps such examples are still not known. Commented Jun 2, 2011 at 11:50
• @Yiftach: there's also an article of M. Sapir from 2007 "Some group theory problems", where he mentions existence of finitely presented infinite torsion groups as an open problem. Commented Jun 2, 2011 at 13:22
• I asked Martin Bridson in January 2010, at which point the question was still open. I have no reason to think this has changed. Commented Jun 20, 2011 at 12:39

Concerning question 1: for free Burnside groups of odd exponent $n\geq 665$ the decidability was shown by S.I.Adian. Lysionok proved the same in the case of even exponent $n=16k\geq 8000$. The corresponding deciding procedure is just Dehn's algorithm.
It will be interesting to know how to solve the word problem for groups $B(m,n)$ when $n=2k$, where $k\geqslant 665$ and odd. These groups are infinite (it follows from the cited result of Adian), but it seems that the decidability of the word problem for these groups is an open question.
The pure existence of non-finitely presented group with undecidable w.p. easily follows from the result of S.I.Adian who proved that there are continuum non-isomorphic periodic groups with fixed number of generators $m\geqslant 2$ satisfying the periodic law $X^n=1$ if $n$ is odd and $n \geq 665$. One should mention, that the number of "additional relations" (besides all periodic) is not necessarily finite.