Prove: If a coxeter system $(W,S)$ is reducible, then it is the product of parabolic subgroups.

Reducible system means the coxeter diagram is disconnected.

Parabolic subgroup: let $S$ be the set of generators and $I \subset S$. Define $R$ to be the subgroup generated by $I$. Then the parabolic subgroups are defined as $rRr^{-1}$ for each $r \in R$.

First and foremost: I don't know much about coxeter groups and can't find any literature online to allay my deficiency and wikipedia is not that helpful.

Would anyone mind giving some definitions or examples and direction to start this proof? I'd like to have a collaborative effort in this.

• If would help if you made excplicit your definitions of reducibility and of parabolic subgroups. – Mariano Suárez-Álvarez Apr 3 '14 at 19:12

Nodes in a Coxeter diagram are not adjacent iff the corresponding generators commute. So, if there are two connected components $C_1$ and $C_2$, all the generators corresponding to nodes in $C_1$ commute with all the generators corresponding to nodes in $C_2$. Say the nodes in $C_1$ are $I \subset S$, so the nodes in $C_2$ are $S \setminus I$; then it's not hard to show, based on the commuting generators, that each element of the parabolic subgroup $R_1 = \langle I \rangle$ commutes with each element of the parabolic subgroup $R_2 = \langle S \setminus I \rangle$. Hence, they normalize eachother, so the product $R_1 R_2$ is a subgroup of $S$. Moreover, $R_1 R_2$ contains all the generators in $S$, so $S = R_1 R_2$.
If there are more than two connected components, just let $C_1$ be one of them, and $C_2$ be the union of all the rest.