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Let $(W,S)$ be a finite and irreducible Coxeter Group. For $J \subseteq S$, let $W_J = \langle s | s \in J \rangle$, a parabolic subgroup. For which $J$ is the action (group multiplication on the left) of $W$ permuting the (left) cosets of $W_J$ faithful exactly?

Is there a good reference to find this result in the literature?

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    $\begingroup$ Shouldn't this happen for any proper $J\subset S$? If $g\in W$ fixes every coset of $W_J$ then $g$ is in every conjugate of $W_J$, and any intersection of parabolic subgroups is parabolic (i.e., conjugate to a standard parabolic), so the intersection of all conjugates of $W_J$ is a proper normal parabolic subgroup. But $(W,S)$ is finite and irreducible, so the only proper parabolic subgroup that's normal is $\{1\}$, hence $g=1$. $\endgroup$
    – Matt Zaremsky
    Commented Jun 7, 2021 at 16:40

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Stumbled on this again, I should make my comment an official answer.

This is true for any proper $J\subseteq S$. Suppose $g\in W$ fixes every coset of $W_J$, so $g$ lies in the intersection of all conjugates of $W_J$. Any intersection of parabolic subgroups (meaning conjugates of standard parabolic subgroups) is a parabolic subgroup, so the intersection of all conjugates of $W_J$ is a proper normal parabolic subgroup of $W$. But $(W,S)$ is finite and irreducible, so the only proper normal parabolic subgroup of $W$ is $\{1\}$. Hence $g=1$, i.e., the action is faithful.

(As a remark, I don't think this used finiteness of $W$ (??), just irreducibility of $(W,S)$. The key is just that there are no proper non-trivial normal parabolic subgroups.)

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    $\begingroup$ If $W$ is allowed to be infinite, probably you have to write "Any finite intersection of parabolic... is parabolic". Then you probably use (in an arbitrary Coxeter group) that if $K$ is a subset of $S$ and is not a union of components, then for $w\in S-K$ adjacent to $K$, $w$ doesn't normalize $W_K$. $\endgroup$
    – YCor
    Commented Jun 18, 2021 at 12:08
  • $\begingroup$ Ah yes, good point, the infinite case is a little more complicated. $\endgroup$
    – Matt Zaremsky
    Commented Jun 18, 2021 at 12:49

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