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Can you show me a reasonably simple (using only elementary group-theoretic tools) example of infinite group with just 2 conjugacy classes ?

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Okay so as Arturo points out my answer was crap so I deleted it. But we should know that the example we're looking for should have a trivial center. – Patrick Da Silva Dec 6 '11 at 19:25
Probably not - the usual way of doing this is ascending HNN extensions. – user641 Dec 6 '11 at 19:32
up vote 13 down vote accepted

I doubt anyone can give you an elementary construction. There are such groups, but the only constructions I know involve things like HNN extensions and free products with amalgamations; hardly "elementary group-theoretic tools". And the only presentations I know for such groups are infinite presentations.

Here's one construction:

Recall that an HNN extension is a construction that achieves the following: let $H$ and $K$ be subgroups of a group $G$, and let $\varphi\colon H\to K$ be a group isomorphism. Then there is an overgroup $\mathfrak{G}$ of $G$ and an element $t\in\mathfrak{G}$ such that $tht^{-1} = \varphi(h)$ for every $h\in H$; that is, $H$ and $K$ are conjugate in $\mathfrak{G}$, and the conjugation "realizes" the isomorphism $\varphi$.

The construction is as follows: let $t$ generate an infinite cyclic group; then consider the free product $G*\langle t\rangle$; let $N$ be the smallest normal subgroup of $G*\langle t\rangle$ containing all elements of the form $tht^{-1}\varphi(h)^{-1}$. Then $\mathfrak{G}=G*_{\varphi} = (G*\langle t\rangle)/N$ is the desired group. Showing that it contains a copy of $G$ is the nontrivial part of the construction (that is, showing that $N$ does not contain any elements of $G$). The element $t$ is called a "stable letter".

Taking for granted that this can be done, then from any torsion-free group $G$ we can construct a new torsion-free group $\mathfrak{G}$ that contains $G$, and in which every nonidentity element of $G$ is conjugate (in $\mathfrak{G}$). Just perform a (possibly infinite) series of HNN-extensions that makes pairs of non-identity non-conjugate elements of $G$ conjugate (the subgroups they generate are isomorphic). It is not hard to verify that if $G$ is torsion free then so is $\mathfrak{G}$.

So, start with your favorite torsion-free nontrivial group; e.g., $G_1 = \mathbb{Z}$. Let $G_2$ be a torsion-free group in which every nonidentity element of $G_1$ is conjugate to one-another. Then let $G_3$ be the same construction for $G_2$. Inductively, we obtain an ascending chain of groups $$G_1\subseteq G_2\subseteq G_3\subseteq\cdots\subseteq G_n\subseteq \cdots$$ such that for all $i$, if $x,y\in G_i$ are both different from the identity, then there exists $z\in G_{i+1}$ such that $zxz^{-1}=y$.

Now verify that $\cup G_n$ is a group in which any two nontrivial elements are conjugate.

This is Exercise 11.78 in Rotman's An Introduction to the Theory of Groups, 4th edition.

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