We denote the natural map $W\to W/W\prime$ by $x\to \bar{x}\text{.}$ Then since $W$ is generated by $r_1,\dots ,r_{\ell },$ $\bar{W}$ is generated by ${\bar{r}}_1,\dots ,{\bar{r}}_{\ell }\text{.}$ 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 $\left|\bar{W}\right|\le \prod _{k=1}^n{p}_{i_k}\text{.}$

Now let $A=⟨{\epsilon }_{i_1}⟩\times \dots \times ⟨{\epsilon }_{i_n}⟩$ the direct product of $n$ cyclic groups of orders $p_{i_1},\dots ,p_{i_n}\text{.}$ We define $\gamma :R\to A$ by $$\gamma \left(r_j\right)=(1,\dots ,1,\underset{\underset{k^{\text{th}}\hspace{0.17em}\text{position}}{⏟}}{{\epsilon }_{i_k}},1,\dots ,1)$$ 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 $$\left|\bar{W}\right|\ge \left|A\right|=\prod _{k=1}^n{p}_{i_k}\text{.}$$

$\square$

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 $𝒰\left(G\right)\text{.}$ 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 $d\ne 1,$ $⟨S_i⟩$ 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 $⟨r_1⟩$ has a non trivial subgroup which is conjugate in $W\left(\Gamma \right)$ to a subgroup of $⟨r_2⟩\text{.}$ 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 $K=P⋊SL(2,3)\text{.}$ $K$ 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 $x\to \bar{x},$ by putting $$\begin{array}{c}{\bar{S}}_1=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),\\ {\bar{S}}_2=\left(\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right),{\bar{S}}_3=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 1& 0\end{array}\right)\text{.}\end{array}$$ It is easy to see that $\bar{G}=K\text{.}$ Further ${\left({\bar{S}}_1{\bar{S}}_2{\bar{S}}_3\right)}^3=1_K$ 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, $\left|R\right|=\left|\bar{R}\right|\text{.}$ Now it is easy to see that there are exactly two conjugacy classes of elements of order six in $K$ -- one determined by $\overline{S_1{S}_3}$ 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 $C\left(U_1\right)=⟨S_1⟩$ a cyclic group of order $p\text{.}$ Thus, ${\Delta }_1\ne {\Delta }_2\text{.}$

$\square$