Finite Groups with exactly nn conjugacy classes (n=2,3,…)(n=2,3,…)

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{1G} is the other conjugacy class.

The Class Equation says |G|=|Z(G)|+[G:CG(x)] where the sum is taken over representatives from the conjugacy classes (not counting the singleton ones from the center). (Here CG(x)={gG | gx=xg} is the centralizer of x in G.)

For us this simplifies to |G|1=[G:CG(x)] for any xG{1G}. Therefore |CG(x)|=|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?


Nice question! The n=3 case is fun and I think small values of n are going to make very good exercises so I encourage you to work on them yourself, but if you really want to know a solution….

If H1,H2 denote the stabilizers of the non-identity conjugacy classes with |H1||H2|, then the class equation reads |G|=1+|G||H1|+|G||H2|, or


The reason this is useful is that if either |G| or |H1| gets too big, then the terms on the RHS become too small to sum to 1. Since we know that |G|3, it follows that we must have |H1|3; otherwise, 1|H1|+1|H2|12 and the terms can’t sum to 1.

Now, if |H1|=2 then |H2|3, hence |G|6. Since every group of order 4 is abelian we can only have |G|=6,|H2|=3. The unique nonabelian group of order 6 is S3, which indeed has 3 conjugacy classes as desired.

If |H1|=3, then |G|3 implies |H2|3, hence |H2|=|G|=3 and G=C3.

Source : Link , Question Author : RHP , Answer Author : Qiaochu Yuan

Leave a Comment