# Conjugacy class with two elements in G implies that G is not simple

1. Jan 7, 2010

### 3029298

1. The problem statement, all variables and given/known data
If some conjugacy class of an element in a group G contains precisely two elements, show that G cannot be a simple group.

3. The attempt at a solution
This question was longer, with two questions before this one which I could answer and which probably lead to the answer on this question.

I showed that for an element x in G the elements of G which commute with x, form a subgroup C(x) of G, called the centralizer. I also proved that the size of the conjugacy class of x is equal to the number of left cosets of C(x) in G. But now for the last question, I need a hint.

2. Jan 7, 2010

### rochfor1

What can you say about a subgroup of index 2?

3. Jan 7, 2010

### 3029298

A subgroup H<G of index two is normal, since if you take x in G, H and xH partition G and H and Hx partition G, which gives Hx=xH, and H is normal. If I have a normal subgroup of index 2, that means that the order of this subgroup is half of the order of G, which cannot be {e} or G, unless G={e}. This is not the case, since a conjugacy class of G contains two distinct elements. Therefore H is a normal subgroup which is not {e} or all of G, hence G is not simple.

Is this correct? Thanks for helping me out in your free time!

4. Jan 7, 2010

### rochfor1

But how do you know you have a subgroup of index two? Otherwise you are correct.

5. Jan 7, 2010

### 3029298

Because I proved earlier that the index of the centralizer C(x) is the same as the order of the conjugacy class of x, and C(x) is a subgroup.
Thanks thanks :)

6. Jan 7, 2010

### rochfor1

Exactly, I figured you knew why, you just hadn't explicitly stated it. Don't thank me too much; I just gave a nudge in the right direction.