# Coxeter groups generated by one finite conjugacy class

Let $$(W,S)$$ be an arbitrary Coxeter system. We consider the following scenario:

Let $$\mathcal{O}$$ be a conjugacy class of an element $$w$$ in $$W$$ which is finite and which generates the whole group $$W$$.

The only example of the above scenario I know arises in simply laced irreducible finite Coxeter groups (i.e. all Coxeter integers are $$\leq 3$$) and for elements $$w=s\in S$$.

My question is: Are there any other examples, especially examples which occur in non simply laced Coxeter groups (i.e. with a Coxeter integer $$\geq 4$$), or examples which occur if $$W$$ is infinite?

This question is related. I found it by searching online. I impose finiteness of $$\mathcal{O}$$ on purpose to treat the case where $$W$$ is infinite, and this question tells us that a finite $$\mathcal{O}$$ can exist in this case.

EDIT. As I noticed meanwhile, I missed one example of the above scenario: The dihedral group $$\mathbb{D}_m$$ with $$m\geq 5$$ odd ($$m$$ is the unique Coxeter integer) and for elements $$w=s\in S$$ as in the first example. It seems I still had a false intuition concerning "non simply laced". Even if this second example is "non simply laced" according to the terminology, unlike Weyl groups, it only has one orbit on the root system / reflections. Sorry for the confusion. Let us focus on infinite $$W$$ from now on. We ask the

QUESTION. Does the above scenario occur for infinite $$W$$?

I think it can't. Proof (attempts), comments, thoughts, references are all welcome!

Let me summarize some inputs of Nadeau and Speyer (thanks to both) and let me add clarification. It holds that: $$W$$ is generated by one conjugacy class if and only if all elements of $$S$$ lie in one conjugacy class if and only if the subgraph of the Coxeter graph with odd labels and same vertices $$S$$ is connected.

This completely solves the question for finite Coxeter groups in the same sense as my text above. In the finite case, there are the simply laced Weyl groups (type $$\mathsf{ADE}$$) and $$\mathbb{D}_m$$ with $$m$$ odd - and only those.

As being said before the input, my question evolves around the infinite case where the above condition is necessary but not sufficient as the conjugacy classes of elements in $$S$$ are infinite by a result of Speyer linked above.

• S is included in a single conjugacy class if the subgraph of the Coxeter diagram given by edges with odd m_{st} is connected (the converse also holds), so yes your scenario happens for many infinite W. Apr 7 '19 at 15:38
• Please read the linked question. The conjugacy class of a $w=s\in S$ is always infinite if $W$ is infinite. Thanks anyway. Your description of "$S$ in one conjugacy class" makes some ideas more precise - and covers some cases where $\mathcal{O}$ generates for $w=s\in S$. But it is not finite - as linked. Apr 7 '19 at 15:44
• My intuition behind the linked answer of Speyer is somehow: The root system $R$ is finite if and only if $R^{\mathrm{re}}$=(real roots) is finite. At least this is precise for symmetrizable Kac-Moody algebras. Apr 7 '19 at 15:48
• The FC-center of any Coxeter group with no (locally finite)/affine component is trivial (i.e., all elements $\neq 1$ have infinite conjugacy classes), see arxiv.org/abs/1211.5635. In the affine case, it's not an FC-group so is not generated by a finite conjugacy class. Remains the cases $A_\infty,A'_\infty, B_\infty,D_\infty$ which probably have a trivial FC-center too.
– YCor
Apr 7 '19 at 20:09
• Thanks, @YCor. Your answer and reference is nice and interesting. It probably also solves other questions of mine in a similar direction as the post - although I have to study some terminology to fully grasp its meaning. For the moment, I think I will post an elementary solution below which I found in the meanwhile. Apr 8 '19 at 11:22

Besides the deeper comment of YCor above which points out further solutions/directions, I add a very simple consideration as answer to "close the case".

Let $$(W,S)$$ be infinite and $$\mathcal{O}_w$$ a finite conjugacy class of some $$w\in W$$. Let $$G$$ be the group generated by $$\mathcal{O}_w$$.

First note that $$\mathcal{O}_{w^{-1}}$$ is finite too. Indeed, it is in bijection to $$\mathcal{O}_w$$ by taking inverse.

Let $$G$$ be the group generated by $$\mathcal{O}_w$$. Let $$v\in G$$. Then, $$v$$ may be written as $$g_1w^{\epsilon_1}g_1^{-1}\cdots g_nw^{\epsilon_n}g_n^{-1}$$ with $$g_1,\ldots,g_n\in W$$ and some signs $$\epsilon_1,\ldots,\epsilon_n$$. But such an expression is "stable" under conjugation, and contained in the pointwise product $$\mathcal{O}_{w^{\epsilon_1}}\cdots\mathcal{O}_{w^{\epsilon_n}}$$ which is finite by the above. Thus, the conjugacy class of any element of $$G$$ is finite. In other words, $$G$$ is covered by finite conjugacy classes.

I want to adress and dissipate some doubts expressed and resolved in the comments by YCor. I therefore edit/rewrite the final part of the proof.

Recall that $$W$$ is finitely generated if and only if $$S$$ is finite.

If we now assume that $$W=G$$, then $$W$$ is generated by the finite conjugacy class $$\mathcal{O}_w$$, which means that $$S$$ is finite. In that case, there exists a infinite component $$(W',S')$$ of $$(W,S)$$ where $$S'\subseteq S$$ is finite. By the exact result of Speyer, we have that any $$s'\in S'$$ gives rise to an infinite conjugacy class which contradicts the fact that $$W$$ is a "FC-group" (a group which has only finite conjugacy classes) by assumption $$W=G$$.

This proves:

Let $$(W,S)$$ be an infinite Coxeter system. Then there exists no finite conjugacy class which generates $$W$$.

I actually expect the following uniform statement to be true, which is, as the further question below, more or less reduced to an analysis of the affine case by what was said in the comments above (maybe I will figure it out alone later on):

Let $$(W,S)$$ be an arbitrary Coxeter system. Then any finite conjugacy class generates a finite group.

• What you're proving is that in a group $G$, the set $FC(G)$ of elements with finite conjugacy class is a subgroup. This is classical, and immediate once one observes that $g\in FC(G)$ $\Leftrightarrow$ the centralizer $C_G(g)$ of $g$ has finite index in $G$.
• As immediate as it is, it answers my very initial question. What you call here "spherical type", seems to be just a finite component. In any case, according to my understanding, the exception you are referring to occurs only in the reducible case, and even then, only if you have infinitely many finite irreducible components (recall that $W$ is supposed to be finite), in which case also $S$ is infinite. OK, that is an exception. Apr 8 '19 at 12:46
• ERRATUM FOR THE LAST COMMENT: (recall that $W$ is supposed to be infinite) Apr 8 '19 at 13:07