Let $(W,S)$ be a Coxeter system, where $S$ is finite. Assume that $W$ has an infinite number of elements.


Is it true that conjugacy classes of elements of non-central elements of $S$ have always an infinite number of elements? 


Or maybe it is better to ask if there there exists examples of infinite Coxeter groups $(W,S)$ in which the conjugacy class of some non-central element $s\in S$ is finite.


EDIT: Sam Nolen gave me an example in which $(W,S)$ is reducible. So the natural addendum to the original question would be to assume that $(W,S)$ is irreducible. 



Infinite Coxeter groups with a non-trivial finite conjugacy class?
The conjugacy class of a reflection in an infinite irreducible Coxeter group is always infinite. This follows from a result of [mine](http://arxiv.org/abs/0710.3188), which was earlier proved by [Kleiner and Pelley](http://arxiv.org/abs/math.RT/0608612) in the case of a symmetrizable integer Cartan matrix:



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> Let $W$ be an infinite irreducible Coxeter group, with generating set $S = \{s\_1, s\_2, \cdots, s\_n\}$. Then the word $s\_1 s\_2 \cdots s\_n s\_1 s\_2 \cdots s\_n \cdots s\_1 s\_2 \cdots s\_n$, consisting of any number of repetitions of $s\_1 s\_2 \cdots s\_n$, is reduced.
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Now, recall the following criterion for a word to be reduced: Let $x\_1$, $x\_2$, ..., $x\_N$, be any sequence of elements of $S$. 



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> The word $x\_1 x\_2 \cdots x\_N$ is reduced if and only if the $N$ group elements $x\_1$, $x\_1 x\_2 x\_1$, $x\_1 x\_2 x\_3 x\_2 x\_1$, ... $x\_1 x\_2 \cdots x\_N \cdots x\_2 x\_1$ are all distinct.
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Write $c$ for $s\_1 s\_2 \cdots s\_n$. Then, combining these two results, we see that $s\_1$, $c s\_1 c^{-1} = s\_1 s\_2 \cdots s\_n s\_1 s\_n \cdots s\_2 s\_1$, $c^2 s\_1 c^{-2}$, $c^3 s\_1 c^{-3}$, etcetera are all distinct. So the conjugacy class of $s\_1$ is infinite. Since the ordering of $S$ was arbitrary, every element of $S$ has infinite conjugacy class.




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I had earlier read the question as asking whether every conjugacy class in an infinite irreducible Coxeter group was infinite. My answer to that remains below. Thanks to Swiat Gal for pointing out the misreading.


This is not true; the class of a translation in an affine group provides a counterexample.


The simplest such counterexample is the group generated by $a$ and $b$ subject to $a^2=b^2=1$. The conjugacy class of $ab$ is $\{ ab, ba \}$, as you can verify by seeing that $a (ab) a^{-1} = b (ab) b^{-1} = ba$ and $a (ba) a^{-1} = b (ba) b^{-1} = ab$. Geometrically, one can think of this as the group of maps $\mathbb{R} \to \mathbb{R}$ of the form $x \mapsto \pm x + k$, for $k \in \mathbb{Z}$. The generators $a$ and $b$ are $x \mapsto -x$ and $x \mapsto -x+1$.


More generally, every affine Coxeter group is of the form $W\_0 \ltimes \mathbb{Z}^r$ for some finite group $W\_0$. If you take a group element of the form $(e, v)$ with $v$ a nonzero element of $\mathbb{Z}^r$, its conjugacy class will be $\{ (e, gv): g \in W\_0 \}$, which is finite. Geometrically, you should think of this as the group of maps $\mathbb{R}^r \to \mathbb{R}^r$ of the form $x \mapsto gx+v$ where $v$ ranges through $\mathbb{Z}^r$ and $g$ ranges through a finite subgroup $W\_0$ of $GL\_r$. Then the conjugacy class of a translation $x \mapsto x+v$ is finite.


