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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!

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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.)

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