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

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.

share|cite|improve this question
up vote 22 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.

share|cite|improve this answer
    
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
2  
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.


enter image description here enter image description here

share|cite|improve this answer

Your Answer

 
discard

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.