# [Algebra] Conjugacy classes of Finite Groups

1. Nov 13, 2007

### daveed

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, $$x+1=|G|$$, and, if $$x$$ divides $$|G|$$, then $$x=1$$.

If we have three, then we let the two nontrivial conjugacy classes have order $$x$$ and $$y$$. Then,
$$|G|=1+x+y$$. But, if $$x$$ and $$y$$ divide $$|G|$$, then $$x$$ divides $$y+1$$, and $$y$$ divides $$x+1$$.

If $$x=y$$, then $$x=y$$, and this is a cyclic group of order 3.

If not, then let's assume that $$x>y$$. Then, $$y+1=x$$. This is only true if $$y=1$$ or $$y=2$$.
So the class equation in this case would be $$3=1+1+1$$ or $$4=1+1+2$$.

However, I'm not sure, here, how to describe all finite groups with the latter class equation. I know that all groups of order $$n$$ are isomorphic to a subgroup of $$S_n$$. Should I just do casework from there?

Last edited: Nov 13, 2007
2. Nov 13, 2007

### matt grime

Let's just think about the case of 2 conjugacy classes.

One must be that of the identity. Let the other one have A elements. Then what can we say about things? A must divide the order of the group. What is the order of the group?

There, that's sorted the first one out, as I think you did. Incidentally, it would help to say what you're about to do, what you're doing, and what you just did when writing a proof so people can read it easily.

Now, what about the 3 classes. I can't believe anything more than hard work and the orbit-stabilizer theorem is required. So just play around with them.

Last edited: Nov 13, 2007
3. Nov 13, 2007

### morphism

Also, is it just me or is the TeX not rendering properly?

For example, I'm seeing the sentence "Then, x." but when I clicked to see the TeX for this, I saw it was supposed to be "Then, |G|=1+x+y." Regardless, I don't see any reason why things like "|G|=1+x+y" or "x>y" should be written in TeX when normal text would more than suffice.

4. Nov 13, 2007

### Kreizhn

Yeah, I see what you mean. That's quite confusing.

As far as writting things script unecessarily, I guess it's just for consistency. I know that when actually writting a LaTeX document, you're supposed to put everything mathematical (even if it's just a mention of the variable x) in math font. So this could be a habit carried over from there.

5. Nov 14, 2007

### matt grime

The reason it might not render correctly is because the OP used tex tags and not itex tags for inline mathematics. This may create some formatting issues. Even with itex it sometimes renders badly in my browser. Hitting reload cures this.