So, the question is: Determine all finite groups that have at most three conjugacy classes I'm a little confused by how to start. Right now, we can say for sure that cyclic groups of order 1, 2, and 3 satisfy this criterion. Also, with Lagrange's Theorem and the counting formula(I'm using this from Artin's book, which states that the product of the order of a conjugacy class with the order of its stabilizer is equal to the order of the group) , we know that the orders of the conjugacy classes divide the order of the group. Also, the identity commutes with every element in the group, and so would be in its own conjugacy class. This means that if we have only one conjugacy class, it would be the group of only the identity. If we have two conjugacy classes, we let x be the order of the nontrivial conjugacy class. But then, [tex] x+1=|G|[/tex], and, if [tex]x[/tex] divides [tex]|G|[/tex], then [tex]x=1[/tex]. If we have three, then we let the two nontrivial conjugacy classes have order [tex]x[/tex] and [tex]y[/tex]. Then, [tex]|G|=1+x+y[/tex]. But, if [tex]x[/tex] and [tex]y[/tex] divide [tex]|G|[/tex], then [tex]x[/tex] divides [tex]y+1[/tex], and [tex]y[/tex] divides [tex]x+1[/tex]. If [tex]x=y[/tex], then [tex]x=y[/tex], and this is a cyclic group of order 3. If not, then let's assume that [tex]x>y[/tex]. Then, [tex]y+1=x[/tex]. This is only true if [tex]y=1[/tex] or [tex]y=2[/tex]. So the class equation in this case would be [tex]3=1+1+1[/tex] or [tex]4=1+1+2[/tex]. However, I'm not sure, here, how to describe all finite groups with the latter class equation. I know that all groups of order [tex]n[/tex] are isomorphic to a subgroup of [tex]S_n[/tex]. Should I just do casework from there?