Is There a Faster Way to Find Conjugacy Classes in Group Theory?

[catcherintherye]

I was wondering if anyone knows a more efficient method of finding conjugacy classes than the one i am currently using.
tex/ Example D_6* =<x,y| x^3=1, y^4=1, yx=x^2y>

now to find the conjugacy classes of this group i would first write out
the orbit of x <x> ={ 1x1, xxx^2, x^2xx, yxy^3,...x^2y^2xy^2x^2,...etc}

then i would use the set relation yx=(x^2)y to work out each of these 12 conjugates individually. Once this is done i continue with <x^2>, <y> etc...

..the only short cut i have found is the theorem that says <x>intersection<y> = empty set or <x>=<y>. But even with this surely there is a quicker way??

 

[masnevets]

Hello,

In terms of finding the conjugacy classes, there doesn't look to be a faster way. I mean, if you actually wanted to know what they are, how would you get them without doing the calculations? If you don't want to do them by hand, I recommend GAP: http://www-gap.mcs.st-and.ac.uk/
It will compute conjugacy classes for you.

There are some other theorems that might save you some time. For example, if G has odd order g, and if h is the number of conjugacy classes of G, then g = h (mod 16). Once you have computed most of the classes, this will probably tell you if the rest of the elements form a single conjugacy class or not. There is another variant of this theorem which says that if all of the primes dividing g are congruent to 1 (mod m), then g = h (mod 2m^2).
http://links.jstor.org/sici?sici=0002-9890(199505)102%3A5<440%3ACRTOOA>2.0.CO%3B2-%23