Let $\Phi$ be a (reduced, crystallographic) root system, and $W$ its Weyl group.
Is it possible to prove that if we know $W$ is dihedral, then the rank of $\Phi$ is two, Without using the classification of root system, but only elementary facts? (In Humphreys' Lie Algebras book, this correspond to ch. 9-10)
(It can be assumed that $\Phi$ is irreducible)
Remark: My failed attempts so far:
- In dihedral group there are at most two elements of order two that commute with each other (Only reflections of two perpendicular roots commute) – However, not every root system of rank greater than two has three perpendicular roots.
- Dihedral groups are generated just by two elements of order two – It turns out that also the symmetric group can be generated by two elements of order two.
- I thought, in similar vein, to show that if dihedral group is generated by three elements of order two then it can be generated by two of these elements – This is simply wrong, since for example the group $D_{2\cdot 30}$ with reflection $s$ and rotation $t$ of order 30 is generated by the three elements $s,st^2,st^5$ but no two of them generate the group.
Edit: If a group can be generated by two non-commuting elements of order two the it is dihedral as shown for example here. (The symmetric group can be generated by three elements of order two.)
So to show the claim it suffices to show that if the rank of $\Phi$ is greater than two, then $W$ cannot be generated by two elements of order two. For reflections it can be deduced easily from the transitivity of Weyl chambers (since a reflection fixes codimension 1 subspace) but it is not clear how to show the claim for general elements.
Best Answer
Your problem is that it is not true that the symmetric group can be generated by two elements of order 2 (except in the case $n = 3$, which is dihedral). "Can be generated by two elements of order 2" is a classification of dihedral groups (if you include the infinite dihedral group $\langle s, t \mid s^2 = t^2 = 1\rangle$).