Let G be a finite group with exactly two conjugacy classes P

Let G be a finite group with exactly two conjugacy classes. Prove that G must be cyclic of order 2.

Solution

If Z(G) G, then there exists an element x G \\ Z(G), and the conjugacy class CG(x) of x contains more than one element. On the other hand, the identity element 1 is in the center Z(G), and forms its own conjugacy class. Since G has exactly two conjugacy classes, these are all the conjugacy classes of G. Therefore, there is exactly one element in Z(G).

By the Theorem, we have |G| = |Z(G)| + |G : CG(x)|

and hence |G : CG(x)| = |G| |Z(G)| = |G| 1. Therefore, |G| 1 is the index of the subgroup CG(x), and so must divide |G|. Since |G| is a positive integer, this can only happen when |G| = 2. We conclude that any finite group with exactly two conjugacy classes is isomorphic to the cyclic group of order 2.

Hence proved