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 $m_{i_r,n}=3$.