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My knowlege in group theory is very limited and this question is out of pure curiousity: Given a random finite group, how many conjugacy classes will it probably have? I can try to make this question more precise:

Define $a_n$ to be the average number of conjugacy classes in finite groups of order less or equal $n$.

What is the asymptotic behaviour of the sequence $a_n$?

Jan Weidner
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    This model of "random" is almost hopeless to work with. It's a difficult problem even for abelian groups! I suspect that a reasonable answer might exist for narrower classes of groups, such as $p$-groups or simple groups. – Victor Protsak Aug 03 '10 at 08:45
  • Could you please indicate the context which brought you to this question? Does it give equal probabilities for each isomorphism class? – user2035 Aug 03 '10 at 09:46
  • My model of randomness was just a try to make the question more precise, if there are answers with more clever models, I would be equally happy. I would really like to know the answer for the class of all groups, however if there is no hope restriction to simple groups would also be okay.

    @ a-fortiori Yes, equal probability to each isomorphism class, but since the model is useless, it doesn't matter anyway. This question is completetly unrelatet to what I normally do, I just asked myself how "nonabelian" is the generic group.

     – Jan Weidner Aug 03 '10 at 10:26
  • For some class of finite groups you can give an upper bound for the number of conjugacy classes, see for instance the paper of Liebeck and Pyber "Upper Bounds for the Number of Conjugacy Classes of a Finite Group", J. Algebra 198. But probably this is not exactly what you are looking for. – Francesco Polizzi Aug 03 '10 at 11:38
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    By most measures, the generic finite group is a three-step nilpotent 2-group, and I think it has exponent 8. You should expect your sequence to jump at powers of 2, and approach relative smoothness between such powers. – S. Carnahan Aug 03 '10 at 12:36
  • @Francesco Polizzi Yes, I am not really interested in a bound, but in the generic behaviour. But thanks for the comment anyway :) @Scott:Thanks Scott, this comment is really helpful. That the generic finite group is a 2-group sounds very plausible to me as well as the behaviour for the sequence you predict. However I would not have expected that the generic 2 group is three step nilpotent. That sounds surprising to me, I would have guessed that the number of steps "is infinity" for the generic 2 group. Do you have a reference for the three step nilpotence? – Jan Weidner Aug 03 '10 at 13:22
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    Look at http://arxiv.org/abs/math/0608491 and the references therein for the 3-step nilpotent results. – Noah Snyder Aug 03 '10 at 16:02
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    If the goal is to determine "how nonabelian" a typical group in a family of groups of varying orders is, wouldn't the ratio $|C|/|G|$ of the number of the conjugacy classes to the group order be a better invariant to track than $|C|$ itself? – Victor Protsak Aug 03 '10 at 18:10
  • @ Noah: Thanks for the reference! @ Victor: Yes, you might be right. – Jan Weidner Aug 04 '10 at 08:19

1 Answers1

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In elementary terms, you have to analyze the following class equation $ n = 1 + h_2 + ... + h_r $ where

  • n is the order of the group G
  • $h_k$ denotes the number of elements in the k-th conjugacy class, and $ n = c_k.h_k$.

Dividing by n, you get $1 = \frac{1}{n} + \frac{1}{c_2} + ... + \frac{1}{c_r} $ which has a finite number of solutions.

Christine Ayoub in her paper On the number of conjugate classes in a group (Proc. Internat. Conf Theory of Groups Canberra 1967) has worked out this analysis for p-groups and there are probably more recent papers on this aspect, which Scott and others allude to in the comments. See for example

  1. MR2557143 Keller, Thomas Michael . Lower bounds for the number of conjugacy classes of finite groups. Math. Proc. Cambridge Philos. Soc. 147 (2009), no. 3, 567--577.

Another way of looking at your question is to see that the number of conjugacy classes is the same as the number of irreducible representations. The character table is always square. Therefore, one could ask "what are the number of irreducible characters Irr(G) in a finite group of order n?". The number of linear characters are [G:G'] where G'=commutator subgroup but the nonlinear ones are tougher and there are papers establishing various bounds for these.

  1. MR2526321 (2010d:20010) Aziziheris, Kamal ; Lewis, Mark L. Counting the number of nonlinear irreducible characters of a finite group. Comm. Algebra 37 (2009), no. 5, 1572--1578.
  2. MR0689258 (84d:20014) Wada, Tomoyuki . On the number of irreducible characters in a finite group. Hokkaido Math. J. 12 (1983), no. 1, 74--82.
  3. MR0798751 (87a:20006) Wada, Tomoyuki . On the number of irreducible characters in a finite group. II. Hokkaido Math. J. 14 (1985), no. 2, 149--154.
SandeepJ
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  • Thanks for providing this interesting perspective of counting irreps/irreducible characters. I will check out the papers you mentioned. – Jan Weidner Aug 04 '10 at 08:23