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 $m\_{i\_r,n}=3$.



Centralizers of reflections in special subgroups of Coxeter groups
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.


