Word processing of Coxeter group $E_6$

Question

Exercise 5.4.11 of Hartshorne book "Algebraic geometry" gives a definition of the relationship between Weyl groups, as follows: Given any diagram consisting of points and line segments joining some of them, we define an abstract group, given by generators and relations, as follows: each point represents a generator $x_i$. The relations are $x_i^2=1$ for each $i$, $(x_ix_j)^2=1$ if $i$ and $j$ are not joined by a line segment, and $(x_ix_j)^3=1$ if $i$ and $j$ are joined by a line segment.

The Weyl group $E_6=\langle x_1,x_2,x_3,x_4,x_5,y\rangle$ is defined using the following diagram: $x_i$ and $x_{i+1}$ are joined by a line segment $(1\leq i\leq 4)$, and there is an additional node $y$ which is only joined by a line segment to $x_3$.

The last question of this exercise is to calculate the order of $E_6$ and finally prove that it is isomorphic to the automorphism group of 27 lines on a cubic surface (the mapping has been constructed in the second question). I can prove isomorphism, but Hartshorne seems to mean to use elementary group theory to calculate the number of elements(which means relying only on the above definition). I consulted some literature on Lie algebra, and the methods they used are not elementary. And I know the number of elements of $E_6$ is $51840$.

I discussed this problem with my classmates and some people who are willing to help me in mathematics. One person said that this is a word processing of the Coxeter group, and no papers in this field have been searched. So how to calculate and whether my understanding is wrong. Does Hartshorne mean this?

Answer 1

Similarly to this proof of part (a), one can calculate right coset representatives. The trick is doing it twice: once for the inclusion $H=\langle x_1,x_2,x_3,y\rangle \subset \langle x_1,x_2,x_3,y,x_4 \rangle = F$ and once for $F=\langle x_1,x_2,x_3,y,x_4 \rangle\subset \langle x_1,x_2,x_3,y,x_4,x_5 \rangle= E_6$. As $\langle x_1,x_2,x_3,y \rangle \cong S_5$ from part (a), the first group is of order 120, so if we can show that there are at most 16 right coset representatives of $H$ in $F$ and at most 27 right coset representatives of $F$ in $E_6$, then we can bound from above the order of the group given by generators and relations by $120\cdot 16\cdot 27 = 51840$.

Here is a graph of the right coset representatives for the first inclusion, where each right coset is a node, and a labeled edge between two nodes corresponds to right multiplication by the labeled element taking you between the two nodes. If a node corresponding to the right coset $Hz$ has no edge labeled by $x_i$ or $y$, then $Hzx_i=Hz$ or $Hzy=Hz$, respectively. (This is actually the meat of the claim: proving this shows that there are at most this many right coset representatives. I will skip recording this in the post, it is somewhat tedious.)

Here is the same graph with the same rules for the right coset representatives of the second inclusion (quiver link):

Answer 2

The procedure I use to count the order of a Weyl group is to use orbit-stabilizer. We know that the root system has $72$ roots, all the same length, so they form a single orbit of the Weyl group. The highest root is the only one in the closure of the dominant Weyl chamber, and it is also the fundamental dominant weight related to the simple root that "sticks out" in the Dynkin diagram. Hence the stabilizer that root is a Coxeter group of type $A_5$. That is, isomorphic to $S_6$.

Therefore the order of this Weyl group is $$|\Phi|\cdot |S_6|=72\cdot720=51840.$$