4
$\begingroup$

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?

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    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.2011-12-04
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    ... 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 :-(2011-12-04
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    Jyrki: It should be $D_4$ instead of $B_4$, isn't it? (I may be wrong; but, what I know is that long roots of $F_4$ give root system of $D_4$; I don't know if there is $B_4$.)2017-01-28
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    @Beginner: It is true that the long roots in $F_4$ form a root system of type $D_4$. But there are also copies of $B_4$ in $F_4$. They consist of all the long roots and additionally 8 of the short roots, as Jyrki wrote. Actually, $F_4$ can be realised as the vectors $\pm e_i$ ($1 \le i \le 4$, gives 8 short vectors), $\pm e_i \pm e_j$ ($1\le i < j \le 4$, gives 24 long vectors) and $\frac{1}{2}(e_1 \pm e_2 \pm e_3 \pm e_4)$ (gives 16 more short vectors) in $\mathbb{R}^4$. The first 30 vectors there are the standard realisation of $B_4$.2017-02-03

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