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Is there an infinite group with exactly two conjugacy classes?

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    $\begingroup$ @Gerry, I do not think this is a homework. An example can be easily constructed using iterated HNN-extensions. A finitely generated example also exists but this is a highly non-trivial result of D. Osin. $\endgroup$
    – Anton Klyachko
    Nov 3, 2013 at 10:41
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    $\begingroup$ If one additionally requires the group to be finitely generated, the question is Problem 9.10 in the Kourovka Notebook, and has been answered in: D. V. Osin, Small Cancellations over Relatively Hyperbolic Groups and Embedding Theorems, Ann Math. 172, no. 1 (2010), 1-39, annals.math.princeton.edu/2010/172-1/p01, ArXiV link: arxiv.org/abs/math/0411039. $\endgroup$
    – Stefan Kohl ♦
    Nov 3, 2013 at 11:01
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    $\begingroup$ A nice exercise in a first course in group theory, is to show that a finite group with 2 conjugacy classes, is the 2-element group... $\endgroup$
    – Alain Valette
    Nov 3, 2013 at 11:12
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    $\begingroup$ It seems to be a duplicate from MSE: see math.stackexchange.com/questions/88980/… $\endgroup$
    – Francesco Polizzi
    Nov 3, 2013 at 13:55
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    $\begingroup$ Boy, I'm relieved this (probably) wasn't homework, because I was not finding this easy. $\endgroup$
    – Todd Trimble ♦
    Nov 3, 2013 at 14:09

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A detailed construction of a non-finitely generated version can be found at https://math.stackexchange.com/questions/88980/infinite-group-with-only-two-conjugacy-classes as pointed out in the comments.

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As a psychological curiosity, Per Enflo writes in his Autobiography that the existence of groups with two conjugacy classes was a key insight behind his many solutions to outstanding problems in Functional Analysis.

"I made important progress in mathematics in 1966, but it was more on the level of new insights, than actual results. When thinking about topological groups in the spirit of Hilbert's fifth problem (I had gradually modified the Hibert problem to some very general program: To decide whether different classes of topological groups shared properties with Lie groups) I was wondering whether there exist "very non-commutative" groups i.e. groups, where all elements except e, are conjugate to each other. I constructed such groups, by finding the right finite phenomenon and then make an induction. I understood, that this is a very general construction scheme (or "philosophy"), that probably could be applied to various infinite or infinite-dimensional problems. And actually – this philosophy is behind several of my best papers - the solution of the basis and approximation problem, the solution of the invariant subspace problem for Banach spaces, the solution of Smirnov's problem on uniform embeddings into Hilbert space and more."

( http://perenflo.com/sida9.html )

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    $\begingroup$ Wow, that is incredibly interesting! $\endgroup$
    – Todd Trimble ♦
    Nov 4, 2013 at 19:58
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The answer to the question is yes, even if one additionally requires the group to be finitely generated. In this case, the question is Problem 9.10 in the Kourovka Notebook, and has been answered in:

D. V. Osin, Small Cancellations over Relatively Hyperbolic Groups and Embedding Theorems, Ann Math. 172, no. 1 (2010), 1-39, http://annals.math.princeton.edu/2010/172-1/p01, arXiv link.

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  • $\begingroup$ Is there an example of a finitely presented group with two generators? $\endgroup$
    – Thomas
    Nov 12, 2013 at 1:56
  • $\begingroup$ @Thomas: I don't know. $\endgroup$
    – Stefan Kohl ♦
    Nov 16, 2013 at 10:14
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According to Peter Cameron (in his "Notes On Classical Groups", available on his webpage) "Paul Cohn constructed an example of a division ring $F$ such that all elements of $F\backslash\{0,1\}$ are conjugate in the multiplicative group of F." This in particular implies that the group can be chosen to be the multiplicative group of a division ring.

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I again thank Yves Cornulier for providing the reference to Cohn's paper. The credit shoud be given to him.\

An easy but interesting fact: Let $G$ be a group and consider the {\it diagonal action} of $G\times G$ on the set $X=G$ by

$$(g,h)\cdot x=g^{-1}xh.$$ It can be shown that this action is $2$-transitive if and only if all non-identity elements of $G$ are conjugate. (This is an exercise in {\it P. J. Cameron, Permutation groups, LMS, 1999.})

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    $\begingroup$ This is Corollary 2 of Theorem 6.3 in: P. M. Cohn, The embedding of firs in skew fields, Proc. London Math. Soc. (3) 23 (1971), 193-213. $\endgroup$
    – YCor
    Nov 3, 2013 at 15:43
  • $\begingroup$ Yves Cornulier, thank for the exact reference. $\endgroup$
    – Name
    Nov 3, 2013 at 15:52

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