# Math Help - Class Equation

1. ## Class Equation

hi, any hints to give me a good idea on how to solve this question would be greatly appreciated:

Question:
let G be a group acting on itself by conjugation, $g . x = gxg^{-1}$

Describe the set-theorem and numerical forms of the Class Equation for these actions explicitly when

(i) $G = A_4$
(ii) $G = D_8$
(iii) $G = D_{10}$

thanx

2. Originally Posted by joanne_q
hi, any hints to give me a good idea on how to solve this question would be greatly appreciated:

Question:
let G be a group acting on itself by conjugation, $g . x = gxg^{-1}$

Describe the set-theorem and numerical forms of the Class Equation for these actions explicitly when

(i) $G = A_4$
(ii) $G = D_8$
(iii) $G = D_{10}$
These problems are very computational. Post what you did so far. Do you know what to do here? Pick an element in G and conjugate it with other elements that will give you its conjugacy class. Now take a another element from G which is not in that conjugacy class and conjugate it with other elements of G that will give you a new conjugacy class. Keep on going until you exhaust all of G. Note, that elements in Z(G) will simply be conjugate to themselves only.

3. ok let me try part (ii) $G = D_8$

my working:
------------------------------------------------------------------------
Firstly $D_8 = [1,x,x^2,x^3,y,xy,x^2y,x^3y]$
$x^4 = 1, y^2 = 1, yx = x^3y$

If we calculate the orbits then we have:
orbit of <1> = {1}

<x>:
$(1)x(1^{-1}) = x$,
$(x)x(x)^{-1} = x$,
$(x^2)x(x^2)^{-1} = x$,
$(x^3)x(x^3)^{-1} = x$,
$(y)x(y)^{-1} = x^2yy=x^2$,
$(xy)x(xy)^{-1} = xyxx^3y=xx^2yx^3y=x^6=x^2$,
$(x^2y)x(x^2y)^{-1} = x^2yxx^2y=x^2x^2yx^2y=x^4yx^2y=x^6yy=x^6y^2=x^6=x^ 2$,
$(x^3y)x(x^3y)^{-1} = x^3yxxy=x^3x^2yxy=x^3x^2x^2yy=x^7y^2=x^7=x^3$,
so orbit of <x> = <x^2> = <x^3> = { $x,x^2,x^3$}

similarly,
<y>:
$(1)y(1^{-1}) = y$,
$(x)y(x)^{-1} = x^3y$,
$(x^2)y(x^2)^{-1} = x^2y$,
$(x^3)y(x^3)^{-1} = xy$,
$(y)y(y)^{-1} = y$,
$(xy)y(xy)^{-1} = x^3y$,
$(x^2y)y(x^2y)^{-1} = x^2y$,
$(x^3y)y(x^3y)^{-1} =xy$,
so orbit of <y> = <xy> <x^2y> = <x^3y> = { $y,xy,x^2y,x^3y$}

concluding, the if $D_8$ cuts on itself by conjugation, the orbits are:
<1> = {1}
<x> = < $x^2$> = < $x^3$> = { $x,x^2,x^3$}
<y> = < $xy$> = < $x^2y$> = < $x^3y$> = { $y,xy,x^2y,x^3y$}
------------------------------------------------------------------------

i believe i've done the hard part: so from here how would i descibe explicitly, the set theorem and numerical forms of the class equation for D_8 ? thanks

4. Originally Posted by joanne_q
ok let me try part (ii) $G = D_8$
When you wrote $\text{D}_8$ I assumed you meant the dihedral group on $8$ vertices which has $16$ elements. But some people rather use $\text{D}_{2n}$ (like you) rather than $\text{D}_n$ (like me). In that case this group is $G = \{ x^ay^b| 0\leq a\leq 3 , \ 0\leq b\leq 1,\ yx = x^3 y\}$.

To help us with computations I will use a simple theorem, maybe you learned it or not.

Theorem: Let $G$ be a finite group and $a\in G$ then the number of elements in the conjugacy class of $a$ is $[G:\text{C}(a)]$ (the centralizer).

The first step that I would do is compute the center $\text{Z}(G)$, those are precisely the elements in their own conjugacy class. Of course $1$ is one such element. I leave it as a simple excerise to prove that $yx^k = x^{-k} y$. Thus, for $x^k$ to compute with $y$ we require that $x^k = x^{-k}$ where $0\leq k \leq 3$ thus $k=2$. This means that $x^2\in \text{Z}(G)$. Thus, $\text{Z}(G) = \{ 1,x^2\}$, and that gives rise to two conjugacy classes $\{1\}$ and $\{x^2\}$ (again because elements in the center are only conjugate to themselves).

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.

The fourth step that I would do is choose an element from from the ones chosen, say $xy$. No theorem this time, we reason differently. Since two more elements are left it cannot be that $xy$ is in its own conjugacy class for that would imply $xy\in \text{Z}(G)$ which is not true thus the conjugacy class of $xy$ contains $2$ elements. Which we know have to be (without computations) $\{xy, x^3y\}$.
That takes a long time.

That means the conjugacy class equation is: $8 = 2 + 2 + 2 + 2$. Note it is not $8 = 1+1+2+2+2$ because the center elements are grouped (a pun ) together.