Centralizers of reflections in parabolic subgroups of Coxeter groups Let us consider a (not necessarily finite) Coxeter group $W$ generated by a finite set of involutions $S=\{s_1,...,s_n\}$ subject (as usual) to the relations $(s_is_j)^{m_{i,j}}$ with $m_{i,j}=m_{j,i}$ and  $m_{i,j}=1$ if and only if $i=j$ (if necessary you may also assume that $m_{i,j}<\infty$ for all $i,j$). Let $P\leq W$ be a subgroup generated by all but one of the $s_i$, say wlog $P=\langle s_1,...,s_{n-1}\rangle$.
I am interested in the centralizer of $s_n$ in $P$. In particular I would like to know if $C_P(s_n)=C_W(s_n) \cap P=\langle s_i~|~ 1\leq i\leq n-1, m_{i,n}=2\rangle=:Z$ always holds.
Obviously this is true if $n=2$ and I believe (though I have not written it down rigorously) I can prove it for Coxeter groups of type $A_n$ by using the standard isomorphism to $S_n$. On the other hand the centralizer of $s_n$ in $W$ is not necessarily a standard parabolic subgroup (look at the dihedral group of order $8$ for example).
There are some results on centralizers of reflections in Coxeter groups and on normalizers/centralizers of parabolic subgroups (which is the same in this special case) to be found in the literature but most deal with the centralizer in $W$. In principal it should be possible to obtain the centralizer in $P$ from these results by simply taking the intersection but the results I found so far are not explicit/ simple enough for this to be a feasible solution.
Edit: Some thoughts so far: I can show that elements of $C_P(s_n)$ of length $1$ or $2$ already lie in $Z$ (the case $1$ being trivial) and that elements of $C_P(s_n)$ of length $3$ where all three occurring simple reflections are pairwise distinct already belong to $Z$. 
On the other hand look at $s_1s_2s_1 \in C_P(s_n)$ which centralizes $s_n$ if and only if $s_2$ centralizes $s_1s_ns_1$. I don't see any reason why this should not be the case so I tried constructing a counterexample consisting of $s_1,s_2$ and $s_3$ such that $s_1,s_2$ do not commute and $s_1s_3$ do not commute but $s_2$ and $s_1s_3s_1$ do. Any ideas on how to do that? 
 A: I posted the question over at MathOverflow and just posted a proof. So to not have the question without an answer on this site, here it goes: 
Surprisingly the proof only uses standard facts on Coxeter-groups (exchange condition, solving the word problem via braid-moves,...).
Let me first make the notation a bit easier:
Claim: Let $P\leq W$ be a special subgroup of $W$ generated by some subset $S'\subsetneq S$ of $S$ and $s \in S-S'$. Then the centralizer of $s$ in $P$ is generated by those involutions in $S'$ which commute with $s$, $C_P(s)=\langle s' \in S'~|~ s's=ss' \rangle$.
Proof: Let $w \in C_P(s)$ and $w=s_1...s_r$ be a reduced expression. By induction it is enough to prove that $s_rs=ss_r$ since the elements of length $1$ in the centralizer are precisely the simple involutions commuting with $s$.
We have $\ell(ws)=\ell(w)+1$ and since $\ell(wsw^{-1})=\ell(s)=1$ we conclude that $\ell(wss_r)<\ell(ws)$, so $\ell(wss_r)=\ell(w)$. By the exchange condition there is a reduced expression for $ws$ ending in $s_r$ and since $s_1...s_rs$ is already a reduced expression for $ws$ there exists a finite series of braid-moves connecting these two expressions. 
The expression $s_1...s_rs$ contains $s$ only once and no simple involution that does not commute with $s$ shows up to the right of $s$. Consider now any braid-move in this situation. If $s$ is not involved in the move the two conditions obviously still hold afterwards. If $s$ is involved the other simple involution involved must commute with $s$ since any braid-move involving $s$ and a non-commuting $s'$ requires either at least two occurrence of $s$ (two the left and two the right of $s'$) or an occurrence of $s'$ to the right of $s$ neither of which there is. Hence any braid-move fixes our two conditions and after finitely many braid moves there is still no simple involution to the right of $s$ which does not commute with $s$. 
On the other hand there is, as noted above, a finite series of braid-moves after which the expression ends in $s_r$ so $s_r$ has to commute with $s$ as asserted.
