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Related to your question 1: I don't know if that is what you are looking for, but if you let $G$ act on $M_n(\mathbb{C})$ and consider one orbit, say $\mathcal{O}_x=G\cdot x$ with $x\in M_n(\mathbb{C})$, then you get an action of $B$ on $G/Z_x$ where $Z_x$ is the centralizer of $x$, and there is a $1$ to $1$ correspondence between $B$-orbits on $G/Z_x$ and $Z_x$-orbits on the flag variety $G/B$ (preserving inclusions of orbit closures). This is at least one situation where "flags" come into the picture when considering the action of $G$ on $M_n(\mathbb{C})$...
Do you have any information about the center of your group ? Since the abelianization is $\mathbb{Z}/2$, if it is a Coxeter group it must be irreducible, hence have trivial center since it is infinite. So if you find a nontrivial central element, you're done...
I guess that you do not want just an embedding as abstract groups at the level of Artin groups, but you at least want compatibility with the quotient maps to the Coxeter group ? For instance, the free group on two generators (which is an Artin group of rank two) embeds into the three strand braid group, while the universal Coxeter group of rank two is infinite, hence cannot embed into $S_3$.
Thank you very much for your very detailed answer ! I am not convinced yet that every step which you describe can be turned into an algorithm (for instance, checking that there is a point on a real line segment that has squared length equal to zero), but I'll think about it. Anyway this is very helpful !
Yes, sure ! I also think that $\mathbb{Q}$-linear combination of roots is OK and that's why I specified... But maybe further restrictions are necessary. For arbitrary $v$ in $V$ I guess that one indeed has incoding problems and the answer is probably "no". :)
Thanks for the misprint, I'll edit. About your question : let $s_1 s_2\cdots s_k$ be a reduced expr. of $t_2'$. As $t_2'\in W$ and $W$ is a stand. parabolic of $\widetilde{W}$, then all the $s_i$'s lie in $W$. It follows that $s' s_1 s_2 \cdots s_k$ is reduced. Now if $s'$ and $t_2'$ commute, then $s' t_2'=s' s_1 s_2 \cdots s_k$ has a reduced expression ending by $s'$. It is not possible, as at least one of the $s_i$'s must be in $S\backslash \{ s\}$, hence has $m_{s' s_i}=\infty$ : you cannot move $s'$ to the right of the first occurrence of $s_i$ in $s_1 s_2 \cdots s_k$ by a braid move.
