What are the Weyl group of type $E_8$, $F_4$,$G_2$? This problem is as titled.
The textbook states that the order of the Weyl group of type $E_8$, $F_4$ are $2^{14}3^55^27$ and 1152 respectively, but I am wondering how are these groups like, namely, how can they be decomposed into simpler groups, or what kind of subgroup or ideals do they have.
Thanks for any attention and help~
 A: According to GAP, $F_4$ is
(((((C2 x D8) : C2) : C3) : C3) : C2) : C2

Here C2 and so on denote cyclic groups, D8 denotes the dihedral group of order $8$, x denotes direct product, and : denotes semidirect product.
That is not the most enlightening description of the group, but shows that it is made up from rather simple pieces.
You should download GAP and play with it a lit to explore the subgroups and what not in the groups. It's fun.
A: Every Weyl group is a Coxeter group with Coxeter diagram given more or less by the Dynkin diagram of the corresponding Lie algebra (up to some minor alterations). A good reference is Humphreys' Reflection groups and Coxeter groups.
A: The Weyl group of type $E_8$ is the group $O(8,\mathbb{F}_2)^+$ of order $696729600$.  It is a stem extension by the cyclic group $C_2$ of an extension of $C_2$ by a group $G$, namely where $G$ is the unique simple group of order $174182400$, known as $PSΩ_8^+(\mathbb{F}_2)$. For a discussion see also this MO question (there is a discussion on the notation $O(8,\mathbb{F}_2)^+$).
The Weyl group of $F_4$ is a soluble group of order $1152$,
for detailed references see here. One of the presentations of it is
$$
< x, y \mid x^2 = y^6 = (xy)^6 = (xy^2)^4 = (xyxyxy^{-2})^2 = 1 >.
$$
