Abstract Nonsense

Crushing one theorem at a time

The Number of Conjugacy Classes of a Finite Group of Odd Order is Equivalent to The Order of The Group Modulo Sixten


Point of Post: In this post we prove a remarkable fact about finite groups of odd order which isn’t easily proven without using representation theory. Namely, we prove that if G is a finite group of odd order and k is the number of conjugacy classes of G then |G|\equiv k\text{ mod }16.

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Motivation

So far we’ve seen a few theorems in pure finite group theory which are very difficult (if not impossible) to prove without using representation theory– of course Burnside’s Theorem is the quintessential example. In this post we prove another such theorem, namely that for a finite group G of odd order that the number of conjugacy classes is equivalent to the order of G mod sixteen. This is really a remarkable and unexpected theorem. And at first it may seem quite useless, but think about its implications for finite group theory. Namely, suppose you had a group G of odd order which suspected is abelian. Of course all you need to do is show that the number of conjugacy classes is |G|. Thus, one is reduced to disproving the other possible number of conjugacy classes, but our little theorem will erase (in the case of most finite groups of small order) all but one or two possible cases. As a less impressive application it clearly implies that all groups of odd order less than or equal to sixteen are abelian (i.e. 1,3,5,7,11,13,15 the only remotely non-trivial one of these is 15 and since 5\not\equiv 1\text{ mod }3 this has been previously established). Probably the most fantastic part of this theorem is how incredibly easy it is to prove using representation theory. Namely, it almost falls out from our previous theorem that every non-trivial irrep of a group of finite order is complex.

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The Theorem

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Without further ado we prove our very interesting theorem:

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Theorem: Let G be a finite group of odd order and k the number of conjugacy classes of G. Then, |G|\equiv k\text{ mod }16.

Proof: Since |G| is odd we know that every non-trivial irrep of G is complex. Thus, for every \alpha in \widehat{G} one has that \overline{\alpha}\ne\alpha and d_\alpha=d_{\overline{\alpha}}. But, since |G| is odd and we know that d_\alpha\mid |G| we may conclude that d_\alpha=2k_\alpha+1 for some k_\alpha\in\mathbb{N}. Thus, since we know that the cardinality of \widehat{G} is k=2r+1 (where r\in\mathbb{N}) we enumerate the elements of \widehat{G} as \alpha_{\text{triv}},\alpha_1,\overline{\alpha_1},\cdots,\alpha_r,\overline{\alpha_r}. Thus, we have by prior theorem that

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\displaystyle \begin{aligned} |G| &= 1+\sum_{j=1}^{r}2d_{\alpha_j}^2\\ &= 1+\sum_{j=1}^{r}2(2k_{\alpha_j}+1)^2\\ &= 1+\sum_{j=2}^{k}(8k_{\alpha_j}^2+8k_{\alpha_j}+2)\\ &= 2r+1+8\sum_{j=1}^{r}k_{\alpha_j}(k_{\alpha_j}+1)\\ &\equiv 2r+1\text{ mod }16\end{aligned}

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from where the conclusion follows. \blacksquare

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.

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April 23, 2011 - Posted by | Algebra, Group Theory, Representation Theory | , , , ,

2 Comments »

  1. Very interesting. I sent you an email about citation (I’d like to cite it in my work) – do you have this published as part of some preprint? Also, could you please use a short URL? It’s a bit tricky to give the URL when it’s as long as the current one.

    Comment by mcmurthy | April 8, 2012 | Reply

    • Dear McMurthy,

      While I technically thought up of this proof “on my own” it was only with help of being a problem in a book on the correct subject, in the correct chapter. Moreover, this is a fairly famous problem and solution which goes back very far (perhaps to Schur). I don’t think there is a need to cite anyone for this result, and it would be downright inappropriate to cite me since I was FAR from the first person figure this out/talk about it.

      I hope that helps!

      Best,
      Alex

      Comment by Alex Youcis | April 9, 2012 | Reply


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