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I have read that for any group $G$ of order $2m+1$ (odd) with $n$ conjugacy classes, it is always the case that $16$ divides the value $(2m+1)-n = |G|-n$.

This seems to me like an astonishing result: what on earth would $16$ have to do with every single odd group and its conjugacy classes? At any rate, I am wondering, assuming it is true, how would you go about proving it? I would like to show it in a reasonably simple way, but nothing whatsoever comes to mind, could anyone help? Many thanks - M.

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Yes, I am. As you said, it's a past paper: I thought the homework tag would be the closest thing. I didn't use the representation tag out of curiosity on the offchance there was a non-character-theory approach, but I will add the rep theory tag if that's the only way to do it. I can see why that would be odd, yes: but I'm afraid I can't seem to make enough use of that to obtain the result. (Something to do with the class equation en.wikipedia.org/wiki/… perhaps?) –  Spyam May 12 '11 at 14:38
    
Even if there were a non-character-theoretic approach, it's likely to 1) be longer and 2) not be the intended solution anyway. –  Qiaochu Yuan May 12 '11 at 14:46

2 Answers 2

up vote 21 down vote accepted

The number of conjugacy classes is the same as the number of irreducible representations; if $d_i$ is the set of dimensions of the irreducible representations, we know that $d_i | |G|$, hence $d_i$ is odd, and $|G| = \sum d_i^2$. Since $d_i$ is odd, we compute that $d_i^2 \equiv 1, 9 \bmod 16$, so it already follows that $|G| \equiv n \bmod 8$. To get the result $\bmod 16$ it suffices to show that the representations such that $d_i^2 \equiv 9 \bmod 16$ occur an even number of times.

In fact, we claim that no non-trivial irrep is self-dual, from which the above follows (since we can take the dual of an irrep with $d_i^2 \equiv 9 \bmod 16$ to get another one). This is equivalent to the claim that no non-identity element is conjugate to its inverse, which follows from the fact that no element of $G$ can act by a permutation of order two.

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Very beautiful proof. –  Joel Cohen May 12 '11 at 14:59
    
Ah, that's brilliant, thankyou. (Incidentally, while the non-character theoretic approach may have been longer, I had hoped it might have been a little more interesting too!) –  Spyam May 12 '11 at 15:14
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Note $\ $ This well-known classic proof is due to Burnside (1911) - see my answer. –  Bill Dubuque May 12 '11 at 16:13

This classic theorem of Burnside is often presentated as an example of a theorem that is difficult to prove without using representation theory. However, one can in fact provide elementary proofs for results of this sort. For example, see Reid: The number of conjugacy classes, AMM, 1998, 359-361, where, generalizing results in an earlier 1995 AMM paper by Poonen), if $G$ is a finite group such that every prime divisor $p$ of its order satisfies $p \equiv 1\ (mod\ m)\:,\:$ he proves the strongest possible congruence between $|G|$ and $n$. Below is the Zbl review by R. W. van der Waall (Amsterdam) followed by Burnside's original proof, from S.222, p.294 of Theory of groups of finite order, 1911.


Let $\cal G$ stand in general for a finite group; $p$ for a prime number. Let $m\in\mathbb Z_{\geq 1}$. Consider $${\cal G}_m=\{{\cal G}\ :\ p\mid|{\cal G}|\ \Rightarrow\ p\equiv 1\ (mod\ m)\}.$$ Let $B(m)$ be the greatest common divisor of all numbers $|{\cal G}|-s$, where $\cal G$ runs through ${\cal G}_m$. Here $s$ stands for the number of conjugacy classes of $\cal G$.

In this paper the following is proved. Theorem. If $m >2$, then $B(m)$ is the least common multiple of $48$ and $2m^2$. $\:$ Also $B(2)=16$ and $B(1)=1$.

The proof is elementary, without using representation theory. It uses the facts that each of $3, 16$ and $2m^2$ divides $B(m)$. The case $3\mid B(m)$ is proved here by elementary means; the case $2m^2\mid B(m)$ likewise as done by B. Poonen [in Am. Math. Mon. 102, No. 5, 440-442 (1995; Zbl 0828.11002)]. The fact $16\mid B(m)$ follows for $m >2$ from $B(2)\mid B(m)$, whereas Burnside proved that $16\mid B(2)$ (before 1911).

Reviewer's remark: Using representation theory, Burnside proved that if $|\cal G|$ is odd, then $|\cal G|\equiv s\pmod{16}$. Does there exist an elementary proof, i.e. without using representation theory and without using the Feit-Thompson theorem on the solvability of finite groups of odd order? A result by K. A. Hirsch in that direction seems to be unjustified.


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