Last update: 10 January 2014
Let and put the set of generators for We define an equivalence relation on as follows. We say and if we say if and only if there is a sequence of vertices of such that is odd for all We then write as the disjoint union of equivalence classes. Since is conjugate to if is odd all the elements of an equivalence class are conjugate and we denote by the integer which is the common label on all those vertices in corresponding to the generators in
Lemma 3. Let and let Let denote the commutator subgroup of Then using the notation just introduced above we have
We denote the natural map by Then since is generated by is generated by But in the map all the elements of an equivalence class are identified. Thus is generated by elements (perhaps not all different) of orders at most Hence
Now let the direct product of cyclic groups of orders We define by if We claim that can be extended to a homomorphism of onto For clearly we have Further if the and vertices of are not joined by an edge, then and commute as is abelian. Now if is odd, then and are in the same equivalence class and hence so the relation with factors on each side is obviously satisfied. Finally if is even the relation with factors on each side becomes which is obvious as is abelian. So we have our claim and thus
Corollary 6. Let and let If and are in distinct equivalence classes under the equivalence relation then no non-identity element of is conjugate in to any element of
Let be a finite dimensional complex vector space and let be a finite subgroup of We say is a reflection if and if there is a hyperplane fixed by is called the reflecting hyperplane for the reflection We let be the collection of hyperplanes in such that is the reflecting hyperplane for some reflection Note that if is the reflecting hyperplane for and if then as it is the reflecting hyperplane for the reflection Thus acts on For we let be the subgroup of consisting of all elements of which are the identity on Since is finite each for is isomorphic to a finite subgroup of and hence is a cyclic group. We denote its order by Then if is an orbit of on and if we have and are conjugate, so We thus denote by the common value of for
The following result appears in [Spr1974].
Proposition 9. Let and let Let denote the commutator subgroup of Then is the direct product of cyclic groups of orders as runs through the orbits of on In particular
Since for is a faithful representation of Proposition 9 gives another way of computing the order of By using both these methods and comparing results we can give a case by case verification of the following theorem which is a known result for Coxeter groups [Bou1968, p.74].
Theorem 5. Let and let Then every reflection in is conjugate in to a power of one of the generating reflections,
We may assume is connected. Thus occurs in List 1. Now if has just one vertex there is nothing to prove. So we consider the case where has two vertices. Say is For we put the reflecting hyperplane for the orbit containing and Since we have so in particular
Now if is odd we apply Lemma 3 to obtain Now clearly Proposition 9 implies We thus obtain the sequence Hence and is the only orbit of on Thus every reflection in is conjugate in to a power of
If is even we apply Lemma 3 to obtain Now if we can apply Proposition 9 to get Thus if we have the sequence which forces implying the desired result. Thus we must show Assume to the contrary that Then is conjugate in to Let If has a subgroup of order Now and and thus is conjugate in to since and are cyclic and conjugate. Thus has a non trivial subgroup which is conjugate in to a subgroup of This is a contradiction to Corollary 6. Hence we are forced to assume Thus and we have the existence of a reflection of order Now consists of scalar matrices as is irreducible. Thus also has order when viewed as an element in But is isomorphic to the alternating group on four letters, the symmetric group on four letters, or the alternating group on five letters [STo1954, p.279]. Now none of these groups contain an element of order higher than five. But forces So we have the theorem for graphs with two vertices.
If has three vertices but is not in any of the infinite families in List 1, we see it is one of
For (1) we apply Lemma 3 to get Proposition 9 certainly implies that We thus obtain the sequence Hence, and is the only orbit.
For (2) the argument is the same as for (1).
For (3) we apply Lemma 3 to obtain Thus if we can apply Proposition 9 as we have before to obtain the result. If there is a reflection of order So the eigenvalues of R are where is a primitive sixth root of unity. We obtain a contradiction as follows:
acts naturally on a two dimensional vector space over Viewing the vector space as an abelian group of row vectors we can form the semi-direct product can be viewed as the matrix group We note that We define a homomorphism denoted by by putting It is easy to see that Further and in we compute a central element of order six. Now comparing, with we see that Hence the kernel of the homomorphism defined above is Hence, since is a reflection, Now it is easy to see that there are exactly two conjugacy classes of elements of order six in -- one determined by and the other by its inverse. Thus, in must be conjugate to an element of or of Now the eigenvalues of are and those of are In particular for either or the eigenvalues are distinct. Hence the same is true for any scalar multiple of either or Thus the element with eigenvalues cannot be conjugate to such a scalar multiple.
If has four vertices but is not in one of the infinite families in List 1 it is either or The first two are Coxeter groups for which the result is known. For the last graph we number the vertices from left to right and use the familiar notation We apply Lemma 3 to obtain the equality Now Proposition 9 gives us the inequality Putting them together with the fact that we have the sequence Thus and is the only orbit.
Now all of are graphs of Coxeter groups for which the result is known. Or, just as easily for all these groups, an application of Lemma 3 gives and thus applying Proposition 9 yields the result. The only remaining case is Applying Lemma 3 gives So numbering from the left as usual we see that if an application of Proposition 9 will again yield the result. Now if is even we can argue just as we did for the graph with even and to show that So we assume is odd. Then forces the existence of a reflection in of order In fact, by Proposition 9, in this situation we must have and is the only orbit. But in the basis defined by the matrices for the generators are the standard generators for the full monomial group of Namely for If interchanges and and fixes the other basis vectors. In this basis it becomes clear that a cyclic group of order Thus,
This is a typed version of David W. Koster's thesis Complex Reflection Groups.
This thesis was submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy (Mathematics) at the University of Wisconsin - Madison, 1975.