Infinite torsion group with finitely many conjugacy classes Do there exist infinite torsion groups with finitely many conjugacy classes? One can easily see that there are no such groups with only two conjugacy classes. Note also that one can construct torsion-free groups with finitely many conjugacy classes via HNN extensions.
 A: Yes there are groups like this. In Geometry of Defining Relations in Groups by Ol'shanskii this is described in the last chapter, and as a particular example he gives infinite $p$ groups with exactly $p$ conjugacy classes (Thm 41.2), and I think he give credit to Ivanov for the basic approach.
He uses the same small cancellation technology used to construct Tarski monsters (fg infinite p-groups).
A: The first finitely generated examples were first constructed by Ivanov (I found this fact in Osin's paper, below).
Note that Denis Osin constructed the first examples of finitely generated groups with exactly two conjugacy classes. He points out on page 2 of his paper that Ivanov's ideas cannot be extended to prove Osin's result. (Osin's paper appeared in the Annals of Mathematics and is the culmination of a large piece of work on small cancellation theory for relatively hyperbolic groups.)
Ivanov's construction is as a limit of hyperbolic groups, so $G$ is such that there exist normal subgroups $N_1\lhd N_2\lhd\cdots \lhd F$ of a free group $F$ such that $F/N_i$ is hyperbolic for all $i$ and $G=F/N$ where $N=\cup N_i$. Suppose that $G$ has two conjugacy classes. Then there exist elements $g, t$ such that $t^{-1}gt=g^2$, and hence there exists some $i$ such that this identity holds in $F/N_i$. However, this identity never holds in hyperbolic groups (Osin cites here the old texts of hyperbolic groups, but I think this is due to Gersten and Short and is slightly later).
This identity is also the reason why a group with two conjugacy classes must be torsion-free (just in case anyone is interested!). Suppose that $G$ has two conjugacy classes and contains torsion. Then every non-trivial element has prime order $p>2$. Note that there exists $g, t$ such that $t^{-1}gt=g^2$, and so we have $g^{2p-1}=t^{-p}gt^pg^{-1}=gg^{-1}=1$, and hence $2p-1=0\mod p$. But of course, by Fermat's little theorem we have that $p-1=0\mod p$, so $2p-2=0\mod p$, a contradiction.
