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In "Embedding Theorems for Groups", Higman, Neumann, Neumann state that "the only group containing elements of finite order, with only two classes of conjugacte elements is the cyclic groups of order two".

In the finite case this is easy. In the infinite case I was just able to prove that such a group must be a $2$-group. Of course it is also simple. What I miss?

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    $\begingroup$ You might be interested in this old question. It constructs an infinite group with precisely two conjugacy classes. $\endgroup$
    – user1729
    Apr 29 '14 at 10:05
  • $\begingroup$ Thanks but I already know that construction (since it is in the same paper of Higman, Neumann, Neumann) ;) $\endgroup$
    – W4cc0
    May 8 '14 at 23:00
  • $\begingroup$ Ah, sorry, I haven't looked at that paper in ages! $\endgroup$
    – user1729
    May 9 '14 at 8:07
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As you noted, any such group is a 2-group. Since all non-identity elements are conjugate, they must all have order 2, and so the group is abelian.

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  • $\begingroup$ It was easy! Thanks! $\endgroup$
    – W4cc0
    Apr 28 '14 at 23:01

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