As a generalization of Problem 3.24, about groups with two conjugacy classes, Andrew noticed that a similar result is true for finite groups with three conjugacy classes: if G is finite and has three conjugacy classes, then G is Z₃ or S₃. To see this, let m and n be the sizes of the two conjugacy classes besides {1_G}. Then |G| = 1 + m + n, and, since both m and n divide |G| (since the size of a conjugacy class equals the index of the centralizer of one of its elements), m divides 1 + n and n divides 1 + m. We may assume m is less than or equal to n. Since n divides 1 + m, either n=m=1 or n = 1 + m. In the latter case, since m divides 1 + n = 2 + m, m equals 1 or 2. Then we are left with three possibilities: (m,n) = (1,1), (1,2), or (2,3). In the first case |G|=3 so G = Z₃. In the second case, |G| = 4, and, since there are only up to isomorphism two groups of order four, both of which are abelian, G has four conjugacy classes, a contradiction. Finally, if (m,n) = (2,3), then |G|=6. There is up to isomorphism only one nonabelian group of order 6, and it is S₃ and, indeed, S₃ has three conjugacy classes, namely {1}, {(123),(132)}, and {(12), (13), (23)}.