A group with only two conjugacy classes?

Question

I'm struggling with another problem from Rotman, namely:

If a group G has exactly two conjugacy classes, and at least one element of finite order, prove that it is isomorphic to the cyclic group of order 2.

Here's what I've got.

A. Let a be an element of order n. Then any other non-identity element is a conjugate of a, and hence has order dividing n.

B. By a very similar argument, all non-identity elements have the same order.

C. Obviously, that order must be a prime.

D. If that prime equals 2, the result is easy to prove as follows:

For any non-identity a and b, ab has order 2 (or 1). So abab = 1. So bab = a.

We just proved that all conjugates of a equal a. Since every non-identity element of G is by assumption a conjugate of a, we're done.

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So what am I missing?

Lobosito,

That's a little bit too advanced of a proof for me. I'm just working my way through my old textbook (Rotman), and am rusty on this stuff because I last studied it in 1975.

CAM

Answer 1

by A,B,C, you obtain that the order of G is finite and G is a p-group.( its order is a power of p).

Now every p-group has a non-trivial center Z(G).(that is it has a non-identity element that commutes with all G).

That is a classical exercise proved using the conjugacy class equation.

If a is non-identity in Z(G), results that a equal its conjgates, so equals all non-identity elements in G.