Periodic group with bound on order of finite subgroups

I have asked the same question previously on stackexchange without any answer (https://math.stackexchange.com/questions/923638/periodic-group-with-bounded-subgroups):

I am looking for infinite periodic groups $G$ (by periodic I mean that every element has finite order), whose finite subgroups are not arbitrarily large. So there is a constant $M$ such that for any finite subgroup $H$ of $G$ one has $|H| \leq M$.

A class of groups that satisfy these requirements are the Tarski monster groups. Here one can take $M = p$ for some prime $p$.

Are there any other notable examples or classes of examples?

• en.wikipedia.org/wiki/Burnside%27s_problem gives information. Yes there are such groups. None has an easily understood structure. Take any even integer $n=512 k \gt 2^{48}$ and $m \ge 2$ generators $x_1,\cdots,x_m$ with the relations $w^m=1$ for every possible finite word. This does define some group and it is infinite. Tarski Monster groups are know to exist for any prime $p \gt 10^{75}.$ – Aaron Meyerowitz Sep 16 '14 at 17:36
• It is known that finitely generated periodic finitely generated dimensional complex linear groups are finite, which is one reason to expect that the groups asked about will not be straightforward. – Geoff Robinson Sep 16 '14 at 17:51
• For the Burnside groups of even exponent, there are arbitrarily large finite subgroups: sciencedirect.com/science/article/pii/S0021869396969410 – Ian Agol Sep 16 '14 at 18:27
• I believe for odd exponent free Burnside groups, sufficiently large, all finite subgroups are cyclic. – Benjamin Steinberg Sep 16 '14 at 19:42
• @BenjaminSteinberg: At least in the mentioned paper by Ian Agol they additionally assume that the finite subgroup is also abelian (first paragraph of the introduction). – scalar Sep 17 '14 at 6:11