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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.

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  • $\begingroup$ I edited the header a bit to reflect what seems to be the stated question. (Maybe it's better to add the word "irreducible" too.) Aside from that, this kind of question looks interesting but the thread of answers and comments got a little out of control and could use some clean-up. $\endgroup$ Dec 8, 2011 at 20:36

4 Answers 4

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The conjugacy class of a reflection in an infinite irreducible Coxeter group is always infinite. This follows from a result of mine, which was earlier proved by Kleiner and Pelley in the case of a symmetrizable integer Cartan matrix:

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.

Now, recall the following criterion for a word to be reduced: Let $x_1$, $x_2$, ..., $x_N$, be any sequence of elements of $S$.

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.

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.


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.

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Assume that your group is a product of two Coxeter groups. Finite and infinite one. Then generators of the finite subgroup have finite conjugacy classes.

Otherwise, you may assume that W is minimal infinite. Then W is either affine or hyperbolic. In both cases you could find, for a given reflection t another reflection s, such that the mirrors of t and s (hyperplanes in Eucliden or Lobačevskiĭ space) do not intersect. Therefore they span infinete dyhedral group with infinite conjugacy class of t.

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  • $\begingroup$ Hi Swiatoslaw, welcome to MO! $\endgroup$ Dec 7, 2011 at 22:17
  • $\begingroup$ Yeah, welcome, but for the future - you should edit your answer, not post a new one (or is it that new users can't edit their answers?) $\endgroup$ Dec 8, 2011 at 1:54
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Actually, I was too fast. It may happen that removing any generator different from t the group is finite, but after removing t the group (call it W') is still infinite. (Think of triangular group with exponents p, q, and ∞.) If W' is affine than hyperbolic intuition might be right (one has to argue that W contains hyperbolic elements).

In the general case one needs a better argument:

Let t∈T⊂S be such, that WT is infinite, and for any s∉t WT-{s} is finite. Assume that for T'=T-{t} WT' is infinite. Ad that t does not commute with WT. This means that there exists s∈T such that t and s do not commute.

I look at my paper On normal subgroups of Coxeter groups generated by standard parabolic subgroups http://dx.doi.org/10.1007/s10711-005-7890-1 where I seem to claim (Cor. 3.3.) that t commutes with w∈WT' if and only if any reduced form of w does not contain a generator not commuting with t. (I hope I understand what I wrote correctly and this is true).

In fact the set of generators that occur in a reduced form of w is always the same (this I do understand).

Now, take infinite order element w∈WT'. It exists because WT' is an infinite Coxeter group (those are virtually torsion free) and contains all generators (as a reduced word) as WT' is minimal infinite. Therefore t cannot commute with w. By the same argument it cannot commute with any power of w (those are still infinite order).

Therefore all conjugates of t by powers of w are different.

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  • $\begingroup$ Nice argument! I believe this. $\endgroup$ Dec 8, 2011 at 16:45
  • $\begingroup$ Riding bike home I understood Cor. 3.3. Let R&sub;T' denote subset of geerators commuting with t. Let v be unique shortest representative of W<sub>R</sub>wW<sub>R</sub>. Ie. w=uvu' for some u,u'&isin;W<sub>R</sub> and v does not start nor end with a generator commuting with t. Then twt=w is equivalent to tvt=v. And no Tits move is possible with tvt unles v=e. $\endgroup$
    – Swiat Gal
    Dec 8, 2011 at 19:20
  • $\begingroup$ It is still incomplete (i.e. T as at the beggining might not exist). This T' generating minimal infinite subgroup can be far from s. That is one needs a whole chain t_0=s, ..., t_{n+1}\in T' such that t_i and t_{i+1} do not commute. then the powers of w'=t_1t_2...t_nwt_n...t_1 (where w as above is of infinite order in W_{T'} would conjugate s to different elements. $\endgroup$
    – Swiat Gal
    Dec 9, 2011 at 10:41
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Here is a counterexample. Take $W = < a,b,c >$, where the only relations are $ab = ba, bc = cb, a^{2} = b^{2} = c^{2} = 1$. Take $S = \{ a, b, c \}$. Then the conjugacy class of $b$ only has one element, but $W$ is infinite.

Edit: the question has now been edited to include the conditions that we want to consider only conjugacy classes of non-central elements of $S$. Here is another counterexample for the new question:

Take $W = < a, b, c, d>$ where the only relations are $ab = ba, cb = bc, ad = da, cd = dc, (bd)^{3} = a^{2} = b^{2} = c^{2} = d^{2} = 1$. Take $S = \{ a, b, c, d\}$. Then $b$ is noncentral and the conjugacy class of $b$ consists of the three elements $b, dbd, bdbdb$.

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    $\begingroup$ Or more generally: Any Coxeter group of the form $W_1\times W_2$ weith $W_1$ finite gives a counter example. The natural modification of the question would be "Let (W,S) be an irreducible Coxeter-system..." What about that question? T think it'll still be the same answer and I suspect that $\tilde{A}_2$ is an example. But I haven't checked that. $\endgroup$ Dec 7, 2011 at 18:46
  • $\begingroup$ I've added a second example, which is an irreducible Coxeter system. $\endgroup$
    – Sam Nolen
    Dec 7, 2011 at 18:51
  • $\begingroup$ Your second example is not irreducible. It decomposes as $W_1 \times W_2$ where $W_1 = \langle a,c \rangle$ and $W_2 = \langle b,d \rangle$. $\endgroup$ Dec 7, 2011 at 20:05
  • $\begingroup$ Ah, you're right. But I think you can make the following small change to get an irreducible counterexample. Add an involution $e$ which commutes with, say, $a, b$, and $d$, but not with $c$. $\endgroup$
    – Sam Nolen
    Dec 7, 2011 at 20:24
  • $\begingroup$ Nope. That's $\langle b,d \rangle \times \langle a,c,e \rangle$. $\endgroup$ Dec 7, 2011 at 20:29

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