# Classes of group

1. Jan 8, 2014

### LagrangeEuler

1. The problem statement, all variables and given/known data
$e = \begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$,
$a =\frac{1}{2} \begin{bmatrix} 1 & -\sqrt{3} \\[0.3em] -\sqrt{3} & -1 \\[0.3em] \end{bmatrix}$.
$b =\frac{1}{2} \begin{bmatrix} 1 & \sqrt{3} \\[0.3em] \sqrt{3} & -1 \\[0.3em] \end{bmatrix}$
$c= \begin{bmatrix} -1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$
$d=\frac{1}{2} \begin{bmatrix} -1 & \sqrt{3} \\[0.3em] -\sqrt{3} & -1 \\[0.3em] \end{bmatrix}$
$f=\frac{1}{2} \begin{bmatrix} -1 & -\sqrt{3} \\[0.3em] \sqrt{3} & -1 \\[0.3em] \end{bmatrix}$
Find classes of group.

2. Relevant equations

3. The attempt at a solution
Classes are $\{e\}$,$\{a,b,c\},\{d,f\}$.
My problem is what is easiest way to find them? I think by watching characters. Am I right? $\{e\}$ is only element with character $2$. Elements $\{a,b,c\}$ have character $0$, and elements $\{d,f\}$ have character $-1$. What is the other way to search this classes?

Last edited: Jan 8, 2014
2. Jan 8, 2014

### Dick

Classes means conjugacy classes, right? And what you are calling a 'character' I would call a 'trace', but ok. But, yes, that's a good clue. If two matrices don't have the same trace then they can't be conjugate. But they can be not conjugates and still have the same trace. So it's best to either check directly or rely on some other argument besides trace. And I think your matrix c has a factor of 1/2 that doesn't belong there.

3. Jan 8, 2014

### LagrangeEuler

Yes classes are conjugacy classes. In $c$ is mistake. Yes. You're right. But how you see that so fast that in $c$ $\frac{1}{2}$ is problem? Characters are traces of elements. Can you tell me some other way to find conjugacy classes?

4. Jan 8, 2014

### Dick

Use the definition of conjugate. a is conjugate to b if there is an element of the group g such that gag^(-1)=b. If you try to calculate c^2 it's pretty obvious the (1/2) doesn't belong there.