Only Weyl group of rank two root system can be dihedral 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.
 A: 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 $\langle s, t \mid s^2 = t^2 = 1\rangle$).
A: Thanks to @Moishe Kohan comment, I was able to find a proof, with the following steps:

*

*If $\Phi$ is reducible, then it is fairly easy to show that the Weyl Group is a product of the corresponding Weyl groups. The only dihedral group which is a product is $D_4\cong\mathbb{Z}_2\times\mathbb{Z}_2$, and the root system is $A_1\times A_1$, which is of rank 2.
So we can assume onwards that $\Phi$ is irreducible.

*The Weyl group $W$ acts on $V$ (the real vector space which $\Phi$ span). Since $\Phi$ is irreducible, $V$ is irreducible representation of $W$. (This claim is present in Humphreys, but can be shown quite easily).

*Now, the scalar extension $V_\mathbb{C}$ is (complex) irreducible representation of $W$. Thanks to this answer, the only non-elementary ingredient in the proof is the scalar extension itself.

*every complex irreducible representation of a dihedral group is 1- or 2-dimensional: the group is generated by reflection $s$ and rotation $t$. Choose an eigenvector of $t$, so because $V$ is irreducible it is the span of $v$ and $s\cdot v$.
Since the dimension of $V$ is equal to the complex dimension of $V_\mathbb{C}$, the result follows.

