Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Weyl Group of $F_4$ is of order $1152=2^{7} \cdot 3^{2}$. By Burnside's theorem the group is solvable.

Is there a way to see solvability from the root system? Is it possible to see the order of the group there?

share|improve this question
The root system has a copy of $B_4$ as a subset. The Weyl group $H$ of the latter is of order $384=2^7\cdot3$. Therefore the bigger Weyl group $G$ acts on the three cosets of $H$ giving rise to a homomorphism $f: G\rightarrow S_3$. That gets us started. The homomorphism is actually surjective. This follows from the fact that the 24 short roots of $F_4$ fall into two orbits of $H$ of unequal size: the 8 short roots of $B_4$ and the other 16 (Look up e.g. Humphreys' book for a description of the root systems to see all this). Thus $H$ cannot be normal. –  Jyrki Lahtonen Dec 4 '11 at 20:04
... and the group $H$ is a semidirect product of $C_2^4$ being acted on by $S_4$, so $\ker f$ is an index two subgroup of it. I'm afraid this is probably not quite what you wanted, so leaving it as a comment/food for thought :-( –  Jyrki Lahtonen Dec 4 '11 at 20:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.