TITLE: Centralizers of reflections in special subgroups of Coxeter groups
QUESTION [6 upvotes]: 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 $i,j$ or even that $W$ is an affine reflection group). 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 reflection 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 principle 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.
Here are some thougts so far: I can show that elements of $C_P(s_n)$ of length $1$ or $2$ already lie in $Z$ (the case of length $1$ 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 $s_1s_2s_1 \in P$ 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 $s_1,s_3$ do not commute but $s_2$ and $s_1s_3s_1$ do. Any ideas on how to do that? 
Edit: I should note that I already posted this question to math.StackExchange (https://math.stackexchange.com/questions/1193740) but did not get any helpful feedback.
Edit 2: Regarding the question whether $s_1s_2s_1$ can centralize $s_3$ (all reflections pairwise distinct; $s_1$ neither centralizing $s_2$ nor $s_3$) in the case of $W$ being an affine reflection group I did a case by case check on the possible Dynkin-diagrams. The cases to consider are $A_3,B_3,\tilde{A_2},\tilde{B_2}$ and $\tilde{G_2}$ and using the standard representation on the root space I found that $s_1s_2s_1$ never centralizes $s_3$.
Edit 3: Here is a proof in the case that $m_{i,n} \neq 3$ for all $1 \leq i \leq n-1$:
Assume $C_P(s_n) \neq Z$ and take an element $w \in C_P(s_n) - Z$ of minimal length. Then each reduced expression for $w$ neither starts nor ends with one of the simple reflections in $Z$.
Let $w=s_{i_1}...s_{i_r}$ be such a reduced expression. Since 
$s_{i_1}...s_{i_r}s_ns_{i_r}...s_{i_1}=s_n$
there is a sequence of braid- and nil-moves that reduces $s_{i_1}...s_{i_r}s_ns_{i_r}...s_{i_1}$ to $s_n$. Since $m_{i_r,n} >3$ we cannot start with a braid- or nil-move involving $s_n$ and since we chose a reduced expression for $w$ there are certainly no nil-moves possible at all. Hence all we can start with is a braid-move in the reduced expression for $w$ (or $w^{-1}$). But after finitely many of such braid-moves the expression we get for $w$ still ends with a simple reflection $s_{i_k}$ which does not commute with $s_n$ (since this would yield an element of shorter length in $C_P(s_n) - Z$). Furthermore $m_{i_k,n} \neq 3$ so we still are unable to perform a braid-move involving $s_n$ and we still have a reduced expression for $w$ so there are no possible nil-moves.
In conclusion: After finitely many steps we will never have performed any nil-moves and hence we cannot reduce $ws_nw^{-1}$ to $s_n$ which is a contradiction and hence such a $w$ does not exist.
I hope one can use an analogous argument in the case $m_{i_r,n}=3$.

REPLY [2 votes]: I believe I finally found a proof. Surprisingly it does only use 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 occurrences of $s$ (to the left and to the right of $s'$) or an occurrence of $s'$ to the right of $s$ neither of which happens. 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.