Are the orders of the generators of a group and the product of pairs of thereof enough for this group to be isomorphic to a Coxeter group?

Question

Let's say we have $n$ generators $x_1, x_2, \cdots, x_n$ along with the following facts concerning their orders: \begin{eqnarray*} ord(x_i) &=& 2 \text{ for } i = 1, 2, \cdots, n \\ ord(x_i x_j) &=& m_{i,j}. \end{eqnarray*} Are those orders enough to prove that the (finite) group generated by the $x_i$'s is isomorphic to the Coxeter group $$\langle r_1, r_2, \cdots, r_n \mid r_i^2 = 1, (r_i r_j)^{m_{i,j}} = 1\rangle?$$

Edit: What extra conditions do we need to ensure that the two groups above are isomorphic?

Answer 1

I think not in general ( at least when $n >2$), if I am reading the question correctly. Any finite group generated by its involutions has such a generating set-which may be refined to an irredundant generating set of involutions if desired- (in particular, any non-Abelian finite simple group has such a generating set) and these are not usually Coxeter groups.

Edit following Jim Humphreys' answer: I was not quite assertive enough above. There definitely are finite simple groups which are not Coxeter groups, since finite Coxeter groups are known to be reflection groups. For example, it is easy to check from the character table that ${\rm PSL}(2,11)$ has no faithful complex irreducible representation in which any involution has an eigenspace of codimension $1$, which means it is not a reflection group in any complex linear representation (irreducible or not). As I said, every finite simple group has an irredundant generating set of involutions as in the question, so such groups are not all Coxeter groups.

Answer 2

Here is a concrete counterexample.

Start with $G$ generated by $x,y,z$, with defining relators saying each of $x,y,z$ has order $2$ and each product $xy$, $yz$, $zx$ has order $3$. This is an example of an infinite Coxeter group. It is isomorphic to the group of isometries of the Euclidean plane generated by reflections in the three sides of an equilateral triangle.

This group $G$ has a finite index normal subgroup $N$ which does not contain $x$, $y$, $z$ (each of which is a reflection of the plane), or the first or second powers of $xy$, $yz$, and $zx$ (each of which is a rotation of the plane). This subgroup $N$ is simply the set of elements of $G$ which are translational isometries of the Euclidean plane.

It follows that the finite group $G/N$ has generators $x,y,z$ each of order $2$ such that each product $xy$, $yz$, $zx$ has order $3$. But it is not a Coxeter group.

Answer 3

As Geoff points out, the answer is likely to be no, though it takes some work to apply the known finite group theory here. It should be emphasized in the question that the group considered is finite as indicated by the tag 'finite-groups', so I edited the question and added another tag. (The question would also make sense in the more open-ended setting of infinite groups, where there are lots of Coxeter groups whose products of involutive generators have finite order---but no clearcut classification.)

It's also essential here to realize that the property of being a "Coxeter group" depends on specifying particular generators: an abstract group may be a Coxeter group in more than one way. Also, it's important to distinguish between "irreducible" Coxeter groups and direct products of smaller ones. Finite irreducible Coxeter groups have a nice classification: they are either crystallographic (Weyl groups of simple Lie algebras) or among the remaining dihedral groups along with two extra reflection groups called $H_3$ and $H_4$. In all these cases one has good information about the internal structure (Sylow theory, etc.) and the character tables.

To exhibit a very small example of a group $G$ satisfying your kind of conditions but not a Coxeter group, probably the best you could do is to start with the Weyl group $W$ of type $D_4$ and factor out its center (of order 2) to get $G$. Here $G$ has order 96 and like $W$ has subgroups isomorphic to $S_4$ of order 24. So by the classification, $G$ couldn't be one of the irreducible Coxeter groups. Since $G$ is solvable (though not nilpotent), it has plenty of normal subgroups. To rule out a decomposition of $G$ as direct product of two proper normal subgroups, you'd have to do more work using the known characters of $W$ or perhaps the Sylow structure of $W$. Though I haven't done all the work at this point, I agree with Geoff that there should be counterexamples.

Finally, I'd suggest that the only reasonable necessary-and-sufficient condition for a finite group generated by involutions to be a Coxeter group is that the other relations involving orders of products of generators be the sole relations one has to add to define the group. Short of this, I'm not sure whether there are any useful necessary-or-sufficient conditions.

Answer 4

Maybe I misunderstand the question. I interpret it as follows: you have a finite group $G$ and some (minimal) set $S$ of generators for $G$ that satisfies the relations defining a Coxeter group $W$ with the same number of generators. It follows that $G$ is a quotient of $W$, and follows in turn that $G$ and $W$ are isomorphic if and only if they have the same order.

I expect that you can use the classification of finite Coxeter groups to determine which of these have proper quotients in which all of the original simple reflections remain nontrivial and form a minimal generating set, if this is the kind of answer you are after.