# Thread: Conjugacy Class Equation

1. ## Conjugacy Class Equation

Hello there,

Can someone please explain to me the quickest method of finding out the Class equation of Dihedral groups, D_8, D_10, D_12 etc.

I am sure there is a quicker way than doing all the calculations, if someone can please tell me in ENGLISH what it is [since I can't understand the mathematical way]

I had initially thought there was a pattern, but it didnt work for D_8, since I thought if you have D_2n, then the class equation is
{1}, {x^k,x^m}, {x^w, x^v},....,{xy, ...., x^n-1 y}

where k+m = n, and w+v = n

This however was not the case for D_8, since {xy, ...., x^n-1 y} was split into 2, so if someone can please explain.

Thanks

2. Look here, the last post.

3. I dont understand this part:

The second step that I would do is choose an element not from the ones chosen, say $x$. Now we will use the theorem which will make computations a little easier. The number of elements conjugate with $x$ has to be $[G:\text{C}(x)]$. Now $1,x,x^2,x^3 \in \text{C}(x)$ thus it has at least four elements, by Lagrange's theorem it can therefore have $4\mbox{ or } 8$ elements it cannot be $8$ since $x$ is not in the center thus $\text{C}(x) = \{1,x,x^2,x^3\}$ thus $[G:\text{C}(x)] = 2$ which means the number of elements conjugate with $x$ is just two. Now $yxy^{-1} = x^{3}yy^{-1} = x^3$ thus $x^3$ is conjugate to $x$, we can stop here by the theorem thus $\{x,x^3\}$ is another conjugacy class.

The third step that I would do is choose an element not from the ones chosen, say $y$. Now we will use the theorem which will make computations a little easier. The number of elements conjugate with $y$ has to be $[G:\text{C}(y)]$. Now $1,x^2,y\text{C}(y)$ thus it has at least three elements it cannot have all eight thus by Lagrange's theorem $|\text{C}(y)| = 4$ which implies that $[G:\text{C}(y)] = 2$. Now $xyx^{-1} = x^2y$ thus $x^2y$ is conjugate to $y$, we can stop here by the theorem thus $\{y,x^2y\}$ is another conjugacy class.
Can you please simplify the above if possible?

4. Originally Posted by dcclanuk
Can you please simplify the above if possible?
The centralizer of an element $x$, denoted by $x$, is the subgroup [tex]\{ a\in G|ax = xa\}. Think of it as a set of all elements which commute with $x$, we denote it by $\text{C}(x)$.

The theorem (for finite groups) says that the number of elements in the same conjugacy class as $x$ is equal to $[G:\text{C}(x)]$.

Remember what $[G:\text{C}(x)]$ means. It means the number of cosets of $\text{C}(x)$. It is in fact equal to $|G|/|\text{C}(x)|$.

5. Originally Posted by ThePerfectHacker
The centralizer of an element $x$, denoted by $x$, is the subgroup [tex]\{ a\in G|ax = xa\}. Think of it as a set of all elements which commute with $x$, we denote it by $\text{C}(x)$.

The theorem (for finite groups) says that the number of elements in the same conjugacy class as $x$ is equal to $[G:\text{C}(x)]$.

Remember what $[G:\text{C}(x)]$ means. It means the number of cosets of $\text{C}(x)$. It is in fact equal to $|G|/|\text{C}(x)|$.
Sorry to sound dumb, but I don't understand how u still carried out the above calculations