Is it possible to prove (in a simple way, maybe non-constructive) the existence of an infinite group with finitely many conjugacy classes, or to build a simple example?
I am aware of this question, but it doesn't help me. This one provides an example, but it is a bit complicated in my opinion; moreover the question asked for an infinite group with exactly two many conjugacy classes, which is more restrictive than my question.
Thank you!
It's easy (and quite explicit) with HNN extensions.
1) For every tf (torsion-free) group $G$ and $g,h\in G\smallsetminus\{1\}$, there exists a tf overgroup of $G$ in which $g$ and $h$ are conjugate; if $G$ is countable then $H$ can be chosen countable. (Namely, take $H$ to be the HNN extension of $G$ over the isomorphism between $\langle g\rangle \to \langle h\rangle$ mapping $g$ to $h$.)
2) For every tf group $G$ there exists a tf overgroup $H$ of $G$ in which any two nontrivial elements of $G$ are conjugate (again, countable if $G$ is countable): enumerate the nontrivial elements of $G$ as $g_0,\dots$, define $G_0=G$, and for $i\ge 1$ define $G_i$ as an overgroup of $G$ in which $g_0$ and $g_i$ are conjugate; finally define $H=\bigcup G_i$.
3) Every tf group $G$ embeds into a group $K$ in which all nontrivial elements are conjugate (countable if $G$ is countable): define $K_0=G$ and $K_{i+1}$ as a tf group in which all elements of $K_i$ are conjugate (by (2)). Then $K=\bigcup K_i$ works.
(Note: this is explicit in principle [e.g., this proves that the resulting group can be chosen to be computable] but however it relies on enumeration of the HNN extension at each step, which makes this not so explicit in practice.)
