Class Equation for Conjugation Actions in A_4, D_8, and D_{10}

[mathusers]

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 :)

 

[mathusers]

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

my working:
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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> = <$xy$> = <$x^2y$> = <$x^3y$> = {$y,xy,x^2y,x^3y$}
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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 :)