Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a systematic way (other than trial-and-error) of finding the conjugacy classes, as well as the number of these and representatives of these classes, for a given finite group?

share|cite|improve this question
For symmetric groups, you'll find this answer of mine helpful. For alternating groups, a similar argument thread can be woven. – user21436 Mar 10 '12 at 19:03

Many computer algebra packages do this in a way which must be considered reasonably systematic. If you want to do it by hand, then even for "well-known" groups, the answer is quiet difficult and complicated. The theory of the rational canonical form describes how to determine the conjugacy classes in ${\rm GL}(n,F)$ when $F$ is a finite field. For other finite "classical" groups, the description of the conjugacy classes becomes more difficult, and has received attention from some very strong mathematicians. As for exceptional groups of Lie type such as $E_{6}(q),E_{7}(q)$ and $E_{8}(q),$ the situation is yet more complex. In principle, (say when dealing with a reasonably small group), if you want to determine the representative for the conjgacy classes, the strategy is clear: pick an element $x \in G,$ determine $C_{G}(x),$ and exhibit the $[G:C_{G}(x)]$ conjugates of $x.$ Then choose an element $y \in G$ not already listed, determine $C_{G}(y),$ etc., until eventually all elements of $G$ are accounted for. A priori, there may not be any significant shortcut available for a general finite group, unless you have further information.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.