Number of conjugacy classes in a finite non-abelian simple group

Question

Can we say that every finite non-abelian simple group has at least 4 non-identity conjugacy classes?

Answer 1

Here's a way to approach this question.

Let's find all the finite groups with exactly 3 conjugacy classes. Let such a group have classes of sizes $1$, $a$, and $b$, with $1\le a\le b$. The size of a conjugacy class must divide the order of the group (since it equals the index of a centralizer), from which it follows that $a$ divides $b+1$, and $b$ divides $a+1$. Think about that for a while, and you find that the only possibilities for $(a,b,c)$ are $(1,1,1)$, $(1,1,2)$, and $(1,2,3)$. The first describes the 3-element group, the second can't happen (since every group of order 4 is abelian), and the third describes the symmetric group on 3 letters. So, no simple nonabelian groups here.

Now the same approach should reduce the exactly-4-conjugacy-class problem to a finite number of possibilities, which you can then compare to the list of known simple nonabelian groups, and see whether there are any.