TITLE: Finite Groups with exactly $n$ conjugacy classes $(n=2,3,...)$
QUESTION [61 upvotes]: I am looking to classify (up to isomorphism) those finite groups $G$ with exactly 2 conjugacy classes.
If $G$ is abelian, then each element forms its own conjugacy class, so only the cyclic group of order 2 works here.
If $G$ is not abelian, I am less sure what is going on.  The center $Z(G)$ is trivial since each of it's elements also form their own conjugacy class.  Now assume $G-\{1_G\}$ is the other conjugacy class.
The Class Equation says $|G|=|Z(G)|+\sum [G:C_G(x)]$ where the sum is taken over representatives from the conjugacy classes (not counting the singleton ones from the center).  (Here $C_G(x)=\{g\in G~|~gx=xg\}$ is the centralizer of $x$ in $G$.)
For us this simplifies to $|G|-1=[G:C_G(x)]$ for any $x\in G-\{1_G\}$.  Therefore $|C_G(x)|=\frac{|G|}{|G|-1}$ is an integer.  But this can only happen when $|G|=2$ which we have already covered.  So does this mean up to isomorphism there is only one group with 2 conjugacy classes?
If so, how would the argument change if we allowed 3 conjugacy classes?

REPLY [10 votes]: Please see $\text{Bounding size by Number of Conjugacy Classes}$ in the below Keith Conrad's article.

http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/conjclass.pdf