Last update: 10 January 2014

Let $\Gamma \in {\mathcal{C}}_{\ell}$ and put $R=\{{r}_{1},\dots ,{r}_{\ell}\}$ the set of generators for $W=W\left(\Gamma \right)\text{.}$ We define an equivalence relation $\text{"}\sim \text{"}$ on $R$ as follows. We say ${r}_{i}\sim {r}_{i}$ and if $i\ne j$ we say ${r}_{i}\sim {r}_{j}$ if and only if there is a sequence of vertices of $\Gamma ,$ ${a}_{i}={a}_{{i}_{1}},{a}_{{i}_{2}},\dots ,{a}_{{i}_{m}}={a}_{j},$ such that ${q}_{{i}_{k}{i}_{k+1}}$ is odd for all $1\le k\le m-1\text{.}$ We then write $R={R}^{1}\cup {R}^{2}\cup \dots \cup {R}^{n}$ as the disjoint union of equivalence classes. Since ${r}_{i}$ is conjugate to ${r}_{j}$ if ${q}_{ij}$ is odd all the elements of an equivalence class ${R}^{k}$ are conjugate and we denote by ${p}_{{i}_{k}}$ the integer which is the common label on all those vertices in $\Gamma $ corresponding to the generators in ${R}^{k}\text{.}$

**Lemma 3.** Let $\Gamma \in {\mathcal{C}}_{\ell}$ and let
$W=W\left(\Gamma \right)\text{.}$ Let
$W\prime $ denote the commutator subgroup of $W\text{.}$ Then using the notation
just introduced above we have
$$|W/W\prime |=\prod _{k=1}^{n}{p}_{{i}_{k}}\text{.}$$

Proof. | |

We denote the natural map $W\to W/W\prime $ by $x\to \stackrel{\u203e}{x}\text{.}$ Then since $W$ is generated by ${r}_{1},\dots ,{r}_{\ell},$ $\stackrel{\u203e}{W}$ is generated by ${\stackrel{\u203e}{r}}_{1},\dots ,{\stackrel{\u203e}{r}}_{\ell}\text{.}$ But in the map $W\to W/W\prime $ all the elements of an equivalence class ${R}^{k}$ are identified. Thus $W/W\prime $ is generated by $n$ elements (perhaps not all different) of orders at most ${p}_{{i}_{1}},\dots ,{p}_{{i}_{n}}\text{.}$ Hence $\left|\stackrel{\u203e}{W}\right|\le \prod _{k=1}^{n}{p}_{{i}_{k}}\text{.}$ Now let $A=\u27e8{\epsilon}_{{i}_{1}}\u27e9\times \dots \times \u27e8{\epsilon}_{{i}_{n}}\u27e9$ the direct product of $n$ cyclic groups of orders ${p}_{{i}_{1}},\dots ,{p}_{{i}_{n}}\text{.}$ We define $\gamma :R\to A$ by $$\gamma \left({r}_{j}\right)=(1,\dots ,1,\underset{\underset{{k}^{\text{th}}\hspace{0.17em}\text{position}}{\u23df}}{{\epsilon}_{{i}_{k}}},1,\dots ,1)$$ if ${r}_{j}\in {R}^{k}\text{.}$ We claim that $\gamma $ can be extended to a homomorphism of $W$ onto $A\text{.}$ For clearly we have ${\left(\gamma \left({r}_{j}\right)\right)}^{{p}_{j}}=1\text{.}$ Further if the ${j}^{\text{th}}$ and ${j\prime}^{\text{th}}$ vertices of $\Gamma $ are not joined by an edge, then $\gamma \left({r}_{j}\right)$ and $\gamma \left({r}_{j\prime}\right)$ commute as $A$ is abelian. Now if ${q}_{j,j\prime}$ is odd, then ${r}_{j}$ and ${r}_{j\prime}$ are in the same equivalence class and hence $\gamma \left({r}_{j}\right)=\gamma \left({r}_{j\prime}\right)$ so the relation $\gamma \left({r}_{j}\right)\gamma \left({r}_{j\prime}\right)\gamma \left({r}_{j}\right)\dots =\gamma \left({r}_{j\prime}\right)\gamma \left({r}_{j}\right)\gamma \left({r}_{j\prime}\right)\dots $ with ${q}_{jj\prime}$ factors on each side is obviously satisfied. Finally if ${q}_{jj\prime}$ is even the relation $\gamma \left({r}_{j}\right)\gamma \left({r}_{j\prime}\right)\gamma \left({r}_{j}\right)\dots =\gamma \left({r}_{j\prime}\right)\gamma \left({r}_{j}\right)\gamma \left({r}_{j\prime}\right)\dots $ with ${q}_{jj\prime}$ factors on each side becomes $${\left(\gamma \left({r}_{j}\right)\gamma \left({r}_{j\prime}\right)\right)}^{\frac{{q}_{jj\prime}}{2}}={\left(\gamma \left({r}_{j\prime}\right)\gamma \left({r}_{j}\right)\right)}^{\frac{{q}_{jj\prime}}{2}}$$ which is obvious as $A$ is abelian. So we have our claim and thus $$\left|\stackrel{\u203e}{W}\right|\ge \left|A\right|=\prod _{k=1}^{n}{p}_{{i}_{k}}\text{.}$$ $\square $ |

**Corollary 6.** Let $\Gamma \in \mathcal{C}$ and let
$W=W\left(\Gamma \right)\text{.}$ If
${r}_{i}$ and ${r}_{j}$ are in distinct equivalence classes under the equivalence relation
$\text{"}\sim \text{"},$ then no non-identity element of
$\u27e8{r}_{i}\u27e9$ is conjugate in $W$
to any element of $\u27e8{r}_{j}\u27e9\text{.}$

Let $V$ be a finite dimensional complex vector space and let $G$ be a finite subgroup of $GL\left(V\right)\text{.}$ We say $R\in G$ is a reflection if $R\ne 1$ and if there is a hyperplane $U\subset V$ fixed by $R\text{.}$ $U$ is called the reflecting hyperplane for the reflection $R\text{.}$ We let $\mathcal{U}=\mathcal{U}\left(G\right)$ be the collection of hyperplanes $U$ in $V$ such that $U$ is the reflecting hyperplane for some reflection $R\in G\text{.}$ Note that if $U\in \mathcal{U}$ is the reflecting hyperplane for $R\in G$ and if $T\in G,$ then $T\left(U\right)\in \mathcal{U}$ as it is the reflecting hyperplane for the reflection $TR{T}^{-1}\text{.}$ Thus $G$ acts on $\mathcal{U}\text{.}$ For $U\in \mathcal{U}$ we let $C\left(U\right)$ be the subgroup of $G$ consisting of all elements of $G$ which are the identity on $U\text{.}$ Since $G$ is finite each $C\left(U\right),$ for $U\in \mathcal{U},$ is isomorphic to a finite subgroup of $\u2102\backslash \left\{0\right\}$ and hence is a cyclic group. We denote its order by $e\left(U\right)\text{.}$ Then if $\Delta $ is an orbit of $G$ on $\mathcal{U}$ and if $U,U\prime \in \Delta $ we have $C\left(U\right)$ and $C\left(U\prime \right)$ are conjugate, so $e\left(U\right)=e\left(U\prime \right)\text{.}$ We thus denote by $e\left(\Delta \right)$ the common value of $e\left(U\right)$ for $U\in \Delta \text{.}$

The following result appears in [Spr1974].

**Proposition 9.** Let $\Gamma \in {\mathcal{C}}^{+}$ and let
$G=\theta \left(W\left(\Gamma \right)\right)\text{.}$
Let $G\prime $ denote the commutator subgroup of $G\text{.}$ Then
$G/G\prime $ is the direct product of cyclic groups of orders $e\left(\Delta \right)$
as $\Delta $ runs through the orbits of $G$ on
$\mathcal{U}\left(G\right)\text{.}$ In particular
$|G/G\prime |=\prod _{\Delta}e\left(\Delta \right)\text{.}$

Since for $\Gamma \in {\mathcal{C}}^{+},$ $\theta $ is a faithful representation of $W\left(\Gamma \right),$ Proposition 9 gives another way of computing the order of $W/W\prime \text{.}$ 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 $\Gamma \in {\mathcal{C}}_{\ell}^{+}$ and
let $G=\theta \left(W\left(\Gamma \right)\right)\text{.}$
Then every reflection in $G$ is conjugate in $G$ to a power of one of the generating reflections,
${S}_{1},\dots ,{S}_{\ell}\text{.}$

Proof. | |||||||

We may assume $\Gamma $ is connected. Thus $\Gamma $ occurs in List 1. Now if $\Gamma $ has just one vertex there is nothing to prove. So we consider the case where $\Gamma $ has two vertices. Say $\Gamma $ is ${p}_{1}\left[q\right]{p}_{2}\text{.}$ For $i=1,2,$ we put ${U}_{i}=$ the reflecting hyperplane for ${S}_{i},$ ${\Delta}_{i}=$ the orbit containing ${U}_{i},$ and ${e}_{i}=e\left({\Delta}_{i}\right)\text{.}$ Since ${S}_{i}\in C\left({U}_{i}\right)$ we have ${p}_{i}|{e}_{i}$ so in particular ${p}_{i}\le {e}_{i}\text{.}$ Now if $q$ is odd we apply Lemma 3 to obtain $|G/G\prime |={p}_{1}\text{.}$ Now clearly Proposition 9 implies $|G/G\prime |\ge {e}_{1}\text{.}$ We thus obtain the sequence $|G/G\prime |\ge {e}_{1}\ge {p}_{1}=|G/G\prime |\text{.}$ Hence ${e}_{1}={p}_{1}$ and ${\Delta}_{1}$ is the only orbit of $G$ on $\mathcal{U}\left(G\right)\text{.}$ Thus every reflection in $G$ is conjugate in $G$ to a power of ${S}_{1}\text{.}$ If $q$ is even we apply Lemma 3 to obtain $|G/G\prime |={p}_{1}{p}_{2}\text{.}$ Now if ${\Delta}_{1}\ne {\Delta}_{2}$ we can apply Proposition 9 to get $|G/G\prime |\ge {e}_{1}{e}_{2}\text{.}$ Thus if ${\Delta}_{1}\ne {\Delta}_{2}$ we have the sequence $|G/G\prime |\ge {e}_{1}{e}_{2}\ge {p}_{1}{p}_{2}=|G/G\prime |$ which forces ${p}_{i}={e}_{i}$ implying the desired result. Thus we must show ${\Delta}_{1}\ne {\Delta}_{2}\text{.}$ Assume to the contrary that ${\Delta}_{1}={\Delta}_{2}\text{.}$ Then $C\left({U}_{1}\right)$ is conjugate in $G$ to $C\left({U}_{2}\right)\text{.}$ Let $d=({p}_{1},{p}_{2})\text{.}$ If $d\ne 1,$ $\u27e8{S}_{i}\u27e9$ has a subgroup ${D}_{i}$ of order $d>1$ $(i=1,2)\text{.}$ Now ${D}_{1}\subset C\left({U}_{1}\right)$ and ${D}_{2}\subset C\left({U}_{2}\right)$ and thus ${D}_{1}$ is conjugate in $G$ to ${D}_{2}$ since $C\left({U}_{1}\right)$ and $C\left({U}_{2}\right)$ are cyclic and conjugate. Thus $\u27e8{r}_{1}\u27e9$ has a non trivial subgroup which is conjugate in $W\left(\Gamma \right)$ to a subgroup of $\u27e8{r}_{2}\u27e9\text{.}$ This is a contradiction to Corollary 6. Hence we are forced to assume $d=1\text{.}$ Thus ${p}_{1}{p}_{2}|{e}_{1}$ and we have the existence of a reflection $R\in C\left({U}_{1}\right)$ of order ${p}_{1}{p}_{2}\text{.}$ Now $Z\left(G\right)$ consists of scalar matrices as $G$ is irreducible. Thus $R$ also has order ${p}_{1}{p}_{2}$ when viewed as an element in $G/Z\left(G\right)\text{.}$ But $G/Z\left(G\right)$ 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 $({p}_{1},{p}_{2})=1$ forces ${p}_{1}{p}_{2}\ge 6\text{.}$ So we have the theorem for graphs with two vertices. If $\Gamma $ 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 $|G/G\prime |=2\text{.}$ Proposition 9 certainly implies that $|G/G\prime |\ge {e}_{1}\text{.}$ We thus obtain the sequence $$|G/G\prime |\ge {e}_{1}\ge {p}_{1}=2=|G/G\prime |\text{.}$$ Hence, ${e}_{1}={p}_{1}$ and ${\Delta}_{1}$ is the only orbit. For (2) the argument is the same as for (1). For (3) we apply Lemma 3 to obtain $|G/G\prime |=6\text{.}$ Thus if ${\Delta}_{1}\ne {\Delta}_{3}$ we can apply Proposition 9 as we have before to obtain the result. If ${\Delta}_{1}={\Delta}_{3},$ there is a reflection $R\in G$ of order $6\text{.}$ So the eigenvalues of R $$are $1,1,\delta $ where $\delta $ is a primitive sixth root of unity. We obtain a contradiction as follows: $SL(2,3)$ acts naturally on a two dimensional vector space $P$ over $GF\left(3\right)\text{.}$ Viewing the vector space as an abelian group of row vectors we can form the semi-direct product $K=P\u22caSL(2,3)\text{.}$ $K$ can be viewed as the matrix group $$\left\{\left(\begin{array}{cc}A& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}x& y\end{array}& 1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}A\in SL(2,3)\hspace{0.17em}x,y\in GF\left(3\right)\}\text{.}$$ We note that $\left|K\right|=216\text{.}$ We define a homomorphism $G\to K,$ denoted by $x\to \stackrel{\u203e}{x},$ by putting $$\begin{array}{c}{\stackrel{\u203e}{S}}_{1}=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),\\ {\stackrel{\u203e}{S}}_{2}=\left(\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right),\phantom{\rule{2em}{0ex}}{\stackrel{\u203e}{S}}_{3}=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 1& 0\end{array}\right)\text{.}\end{array}$$ It is easy to see that $\stackrel{\u203e}{G}=K\text{.}$ Further ${\left({\stackrel{\u203e}{S}}_{1}{\stackrel{\u203e}{S}}_{2}{\stackrel{\u203e}{S}}_{3}\right)}^{3}={1}_{K}$ and in $G$ we compute ${\left({S}_{1}{S}_{2}{S}_{3}\right)}^{3}=-{\epsilon}_{1}^{2}{I}_{3}$ a central element of order six. Now comparing, $\left[3\right]$ with $\left[2\right]$ we see that $\left|G\right|=1296,$ $\left|Z\left(G\right)\right|=6\text{.}$ Hence the kernel of the homomorphism defined above is $Z\left(G\right)\text{.}$ Hence, since $R$ is a reflection, $\left|R\right|=\left|\stackrel{\u203e}{R}\right|\text{.}$ Now it is easy to see that there are exactly two conjugacy classes of elements of order six in $K$ -- one determined by $\stackrel{\u203e}{{S}_{1}{S}_{3}}$ and the other by its inverse. Thus, in $G,$ $R$ must be conjugate to an element of $Z\left(G\right){S}_{1}{S}_{3}$ or of $Z\left(G\right){\left({S}_{1}{S}_{3}\right)}^{-1}\text{.}$ Now the eigenvalues of ${S}_{1}{S}_{3}$ are $1,$ $-1,$ ${\epsilon}_{1}$ and those of ${\left({S}_{1}{S}_{3}\right)}^{-1}$ are $1,$ $-1,$ ${\epsilon}_{1}^{2}\text{.}$ In particular for either ${S}_{1}{S}_{3}$ or ${\left({S}_{1}{S}_{3}\right)}^{-1}$ the eigenvalues are distinct. Hence the same is true for any scalar multiple of either ${S}_{1}{S}_{3}$ or ${\left({S}_{1}{S}_{3}\right)}^{-1}\text{.}$ Thus the element $R,$ with eigenvalues $1,1,\delta ,$ cannot be conjugate to such a scalar multiple. If $\Gamma $ has four vertices but is not in one of the infinite families in List 1 it is either ${H}_{4},$ ${F}_{4},$ or $3\left[3\right]3\left[3\right]3\left[3\right]3\text{.}$ 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 ${U}_{i},$ ${\Delta}_{i},$ ${e}_{i}\text{.}$ We apply Lemma 3 to obtain the equality $|G/G\prime |=3\text{.}$ Now Proposition 9 gives us the inequality $|G/G\prime |\ge {e}_{1}\text{.}$ Putting them together with the fact that ${e}_{1}\ge {p}_{1}$ we have the sequence $$|G/G\prime |\ge {e}_{1}\ge {p}_{1}=3=|G/G\prime |\text{.}$$ Thus ${e}_{1}={p}_{1}$ and ${\Delta}_{1}$ is the only orbit. Now all of ${E}_{6},$ ${E}_{7},$ ${E}_{8},$ ${A}_{\ell},$ ${D}_{\ell}$ 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 $|G/G\prime |=2$ and thus applying Proposition 9 yields the result. The only remaining case is ${B}_{\ell}^{p}\text{.}$ Applying Lemma 3 gives $|G/G\prime |=2p\text{.}$ So numbering from the left as usual we see that if ${\Delta}_{1}\ne {\Delta}_{2}$ an application of Proposition 9 will again yield the result. Now if $p$ is even we can argue just as we did for the graph ${p}_{1}\left[q\right]{p}_{2}$ with $q$ even and $({p}_{1},{p}_{2})\ne 1$ to show that ${\Delta}_{1}\ne {\Delta}_{2}\text{.}$ So we assume $p$ is odd. Then ${\Delta}_{1}={\Delta}_{2}$ forces the existence of a reflection in $C\left({U}_{1}\right)$ of order $2p\text{.}$ In fact, by Proposition 9, in this situation we must have $\left|C\left({U}_{1}\right)\right|=2p$ and ${\Delta}_{1}$ is the only orbit. But in the basis $\{{x}_{1},\dots ,{x}_{\ell}\}$ defined by ${x}_{i}=\frac{1}{\alpha}{v}_{1}+{v}_{2}+\cdots +{v}_{i}$ $(\alpha ={(2\text{sin}\hspace{0.17em}\pi /p)}^{\frac{1}{2}})$ the matrices for the generators ${S}_{k}$ are the standard generators for the full monomial group of ${p}^{\ell}\ell !$ Namely ${S}_{1}\left({x}_{1}\right)={\epsilon}_{1}{x}_{1},$ ${S}_{1}\left({x}_{j}\right)={x}_{j}$ for $j\ne 1\text{.}$ If $k\ne 1,$ ${S}_{k}$ interchanges ${x}_{k}$ and ${x}_{k-1}$ and fixes the other basis vectors. In this basis it becomes clear that $C\left({U}_{1}\right)=\u27e8{S}_{1}\u27e9$ a cyclic group of order $p\text{.}$ Thus, ${\Delta}_{1}\ne {\Delta}_{2}\text{.}$ $\square $ |

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.