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Zbl 0924.20015
Reid, Michael
The number of conjugacy classes. (English)
[J] Am. Math. Mon. 105, No.4, 359-361 (1998). ISSN 0002-9890

Let $\cal G$ stand in general for a finite group; $p$ for a prime number. Let $m\in\bbfZ_{\geq 1}$. Consider $${\cal G}_m=\{{\cal G};\ p\mid|{\cal G}|\Rightarrow p\equiv 1\pmod m\}.$$ 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 $\cal G$.\par 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 $B(1)=1$.\par 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 {\it 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).\par 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.
[R.W.van der Waall (Amsterdam)]

MSC 2000:: *20D60 Arithmetic and combinatorial problems on finite groups
11A07 Congruences, etc.

Keywords: conjugacy classes of finite groups; numbers of conjugacy classes

Citations: Zbl 0828.11002

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