Let $G$ a connected semisimple simply connected group over $\mathbb{C}$ and $W$ his Weyl group.

What can be said about $W'$, the subgroup of $W$ generated by the Coxeter elements of $W$?


Suppose the group $G$ is simple; the general case should be similar. Let $\Gamma$ be the subgroup of $W$ generated by all Coxeter elements and suppose $W$ is generated by the simple reflections $s_i$, where $i\in I$ and $|I|=l$. For distinct $i,j\in I$ consider two Coxeter elements $c_1:=s_is_j\prod_{k\ne i,j}s_k$ and $c_2=s_js_i\prod_{k\ne i,j}s_k$. Then $c_1,c_2\in \Gamma$ as all Coxeter elements are conjugate in $W$, implying $s_is_js_is_j=c_1c_{2}^{-1}\in \Gamma$. If the roots $\alpha_i$ and $\alpha_j$ have the same length and are linked on the Dynkin graph of $W$ then $(s_is_j)^3=1$. By interchanging the roles of $i$ and $j$ we then deduce that $s_is_j\in\Gamma$ for all such $i$ and $j$.

If all roots of $G$ have the same length then it is immediate from the above that $s_is_j\in \Gamma$ for all $i,j\in I$. But then the derived subgroup $[W,W]$ is contained in $\Gamma$. As it has index $2$ in $W$ we deduce that $\Gamma=[W,W]$ if $l$ is even and $\Gamma=W$ if $l$ is odd.

When $G$ has roots of different lengths one has to use some case-by-case arguments. This is especially easy for types $B_{2n}$, $C_{2n}$ and $G_2$ where we still have that $\Gamma=[W,W]$.

  • $\begingroup$ For F4, I also get Gamma = the unique normal subgroup of index 4 in W = [W,W] $\endgroup$ – ya-tayr May 6 '13 at 16:23
  • $\begingroup$ @Sasha: I don't see a role here for the crystallographic root system, since long and short roots don't enter into the Weyl group structure. Here types $B, C$ coincide. $\endgroup$ – Jim Humphreys May 6 '13 at 17:12
  • $\begingroup$ @Jim: Of course, I used type $B$ throughout and only mentioned type $C$ for completeness. By the way, if the questioner is interested in the group generated by all Coxeter elements associated with a {\it fixed} generating set of $W$ then this group may not be normal, in general. $\endgroup$ – Alexander Premet May 6 '13 at 18:21

In a finite irreducible reflection group all Coxeter elements are conjugate because the Coxeter graph contains no circuits. In particular, such elements generate a large (clearly noncentral unless $|W|=2$) normal subgroup. Beyond this you probably need to look at them case-by-case, using the standard realizations. When the group isn't irreducible (connected Coxeter graph), it just decomposes as a direct product of irreducible ones.

I'm not sure what motivates the question, but it doesn't really concern algebraic groups or their Weyl groups; instead it's a question about arbitrary finite Coxeter groups, including all dihedral groups. So a tag `coxeter-groups' could be substituted.

[Helpful hint: taking $W$ to consist of orthogonal matrices, a Coxeter element has determinant 1 precisely when the rank is even.]

ADDED: In more detail, each finite (irreducible) Coxeter group $W$ has a "rotation" subgroup $W^+$ of index 2, the kernel of the sign (or determinant) map. Say $W_c$ is the subgroup generated by all Coxeter elements; these are all conjugate, so $W_c$ is normal. If $W$ has rank $n$, a Coxeter element has length $n$ and thus $W_c \subset W^+$ iff $n$ is even.

My suggestion is to rely as much as possible on the known normal subgroup structure of $W$. Though case-by-case work seems necessary, the actual results look suspiciously uniform; so there might be a more conceptual approach. Here are a few easy examples.

By Coxeter's classification, there are familiies $A_n, B_n (= C_n), D_n$ and the dihredral groups $I_2(n)$, along with isolated groups $E_n (n = 6,7,8), F_4, G_2, H_3, H_4$ (with $A_2, B_2, G_2$ dihedral). In the dihedral case, the Coxeter number is $n$ and thus $W_c = W^+$. Most groups $W$ in the three classical families involve a large simple group (alternating). It's easy to check for type $A_n$ that $W_c = W$ for $n$ odd and $W^+$ for $n$ even. At the other extreme, $W$ of type $E_8$ has a center $\{\pm 1\}$ of order 2 in $W^+$ and in $W_c$ (each Coxeter element has order 30, with 15th power equal to $-1$). As in Bourbaki, $W^+$ modulo the center is the simple group $O_8^+(2)$. Thus $W_c = W^+$.

  • $\begingroup$ How do we know that they generate a normal subgroup? And for $C_{2n+1}$ or $B_{2n+1}$ do we still obtain W? $\endgroup$ – prochet May 6 '13 at 7:09
  • $\begingroup$ @prochet: I hope I answered your first question, since the subgroup generated by a conjugacy class is normal. For the second question, I don't understand all of your notation, but I expect one has $W_c=W$ for types $B_n, D_n$ with odd $n$. (P.S. The question only concerns finite irreducible Coxeter groups, so you should replace the 'algebraic-groups' tag.) $\endgroup$ – Jim Humphreys May 6 '13 at 15:35
  • $\begingroup$ In type $B_{2n+1}$ we get neither $W$ nor $[W,W]$. What we get is a subgroup of index $2$ which can be explicitly described by identifying $W$ with the semidirect product of $S:=\mathfrak{S}_{2n+1}$ and a normal subgroup $A:=\mathbb{Z}_2^{2n+1}$. Then $\Gamma$ is the semidirect product of $[S,S]$ and $A$. The hardest case is actually $F_4$, but it's just too boring to look into. $\endgroup$ – Alexander Premet May 6 '13 at 16:26
  • $\begingroup$ @Sasha: That looks right. I didn't get quite so far in looking at the cases. In a way it's reassuring that the pattern is not totally uniform. I'm also uncertain about the non-Weyl groups $H_3, H_4$. $\endgroup$ – Jim Humphreys May 6 '13 at 17:10

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