3
$\begingroup$

Let $W$ be a finite Coxeter group. Let $$ N_W=\operatorname{max}_{g\in W}\operatorname{ord}(g) $$ where $\operatorname{ord}(g)$ denotes the order of an element $g$. By Fermat's little theorem, we know that $N_W$ divides the order of $W$. What else can we say about $N_W$? Is $N_W$ related to the Coxeter number of $W$?

$\endgroup$
8
  • 6
    $\begingroup$ For the symmetric group see oeis.org/A000793. $\endgroup$ Oct 4, 2019 at 2:58
  • 2
    $\begingroup$ Let it be $f(n)$ for $S_n$. So in type $A_n$ it's $f(n+1)$. In type $B_n=C_n$ (perm. wreath product $C_2\wr S_n$) for $n\ge 2$, say we get $h(n)$. Then $f(n)\le h(n)\le 2f(n)$. If $f(n)$ is even we get $2f(n)$ by an easy argument. The OEIS list suggests the only exception is $n=8$ for which $f(n)=15$. But $S_8$ has an element of order $12$ and then $h(8)=24$. So if indeed $f(n)$ is even for all $n\ge 9$, then $h(n)=2f(n)$ for all $8\neq n\ge 2$. $\endgroup$
    – YCor
    Oct 4, 2019 at 9:23
  • 4
    $\begingroup$ The Coxeter number of $S_m$ is neither $m!$ nor $\frac{m(m-1)}{2}$. It is $m$. (Every Coxeter element of $S_m$ is an $m$-cycle, so the order of a Coxeter element is $m$.) The numbers $m!$ and $\frac{m(m-1)}{2}$ are, respectively, the order of $S_m$ and the number of reflections (or positive roots) in $S_m$. $\endgroup$ Oct 4, 2019 at 11:50
  • 1
    $\begingroup$ @NathanReading Ah yes, I got my notation mixed up. Landau proved that for $S_n$ we have $H_W \sim O(n \ln n)$, and since the Coxeter number is $n$, we have evidence that there isn't an obvious relation. $\endgroup$ Oct 4, 2019 at 17:16
  • 4
    $\begingroup$ I haven't verified for $S_m$ that $N$ is the LCM of $(1,\ldots,m)$, but if so, you should think of it as the LCM of $(2,\ldots,m)$. That suggests a general question: For a finite Coxeter group $W$, is the smallest positive integer $N$ such that $w^N=1$ for all $w\in W$ equal to the LCM of the degrees of $W$ (or equivalently, the LCM of $(e_i+1:i=1,\ldots,\mathrm{rank}(W))$, where the $e_i$ are the exponents of $W$)? $\endgroup$ Oct 8, 2019 at 19:52

0

You must log in to answer this question.

Browse other questions tagged .