By Theorem 2 $\Gamma$ is a tree and thus (i) is given in [Bou1968] (Lemma 1, Chapter V, section 6). For a graph $\Gamma$ with just one vertex labelled with the integer $p$ we see from [STo1954] that $d_1=p$ and thus (ii) and (iii) are obvious. Now if $\Gamma$ satisfies $p_i=2,$ $1\le i\le \ell ,$ so $W\left(\Gamma \right)$ is a Coxeter group, (ii) and (iii) are given by a general argument in [Bou1968] (Proposition 3, Chapter V, section 6). If $\Gamma \in {𝒞}_2^{+},$ say $\Gamma$ is $p_1\left[q\right]p_2,$ Coxeter observes [Cox1962] that $d_1=2h/q$ and $d_2=h\text{.}$ He also obtains $\gamma$ and ${\gamma }^{1+h-\frac{2h}{q}}$ as the eigenvalues of $S_1^t{S}_2^t$ and thus we have ${\gamma }^{\frac{2h}{q}-1}$ and ${\gamma }^{h-1}$ as the eigenvalues of $R_1{R}_2\text{.}$

We now consult List 1 of Theorem 2 to determine which graphs still remain and we will verify (ii) and (iii) for these groups in a case by case manner.

Suppose $\Gamma$ is $3\left[3\right]3\left[3\right]3\text{.}$ By comparing $\left[3\right]$ with $\left[2\right]$ we see that $d_1=6,$ $d_2=9,$ $d_3=12\text{.}$ Now the matrix for $R_1{R}_2{R}_3$ in the basis $\{v_1,v_2,v_3\}$ is $\left(\begin{array}{ccc}0& 0& -\omega \\ \alpha & 0& {\omega }^2\alpha \\ 0& \alpha & {\omega }^2\end{array}\right)$ where $\omega =e^{2\pi i/3}$ and $\alpha =\frac{1-{\omega }^2}{\sqrt{3}}\text{.}$

The characteristic equation is thus ${\lambda }^3-{\omega }^2{\lambda }^2+\omega \lambda -1=0$ which has roots ${\gamma }^5,$ ${\gamma }^8,$ ${\gamma }^{11}$ where $\gamma =e^{2\pi i/12}\text{.}$

For the graph $\Gamma$ $3\left[3\right]3\left[4\right]2$ we find from $\left[3\right]$ and $\left[2\right]$ that the degrees are $d_1=6,$ $d_2=12,$ $d_3=18\text{.}$ We compute that the matrix for $R_1{R}_2{R}_3$ in the basis $\{v_1,v_2,v_3\}$ is $\left(\begin{array}{ccc}0& \alpha & B{\omega }^2\\ \alpha & 1& -\alpha B\\ 0& B& -1\end{array}\right)$ where $\alpha =\frac{1-{\omega }^2}{\sqrt{3}}$ and $B=3^{\frac{1}{4}}\text{.}$ Thus the characteristic equation is ${\lambda }^3+\omega =0$ and the eigenvalues of $R_1{R}_2{R}_3$ are ${\gamma }^5,$ ${\gamma }^{11},$ ${\gamma }^{17}$ where $\gamma =e^{2\pi i/18}\text{.}$

If $\Gamma$ is $3\left[3\right]3\left[3\right]3\left[3\right]3$ we find again from $\left[3\right]$ and $\left[2\right]$ that $d_1=12,$ $d_2=18,$ $d_3=24,$ $d_4=30\text{.}$ This time the matrix for $R_1{R}_2{R}_3{R}_4$ in the basis $\{v_1,v_2,v_3,v_4\}$ is $\left(\begin{array}{cccc}0& 0& 0& -\omega \alpha \\ \alpha & 0& 0& -\omega \\ 0& \alpha & 0& {\omega }^2\alpha \\ 0& 0& \alpha & {\omega }^2\end{array}\right)$ where $\omega =e^{2\pi i/3}$ and $\alpha =\frac{1-{\omega }^2}{\sqrt{3}}\text{.}$ The characteristic equation is thus ${\lambda }^4-{\omega }^2{\lambda }^3+\omega {\lambda }^2-\lambda +{\omega }^2=0$ which has roots ${\gamma }^{11},$ ${\gamma }^{17},$ ${\gamma }^{23},$ ${\gamma }^{29}$ where $\gamma =e^{2\pi i/30}\text{.}$

We finally consider the infinite family of graphs $B_{\ell }^p\text{.}$ Here, comparing $\left[3\right]$ and $\left[2\right]$ yields that the degrees are given by $d_k=kp$ $(1\le k\le \ell )\text{.}$ Letting $\alpha ={\{2\text{sin}\hspace{0.17em}\pi /p\}}^{-\frac{1}{2}}$ we define a new basis $\{x_1,\dots ,x_{\ell }\}$ by $x_i=\alpha v_1+v_2+v_3+\dots +v_i\text{.}$ Then $$\begin{array}{c}{S}_1\left(x_1\right)=\epsilon x_1(\epsilon =e^{2\pi i/p})\\ S_1\left(x_j\right)=x_j\text{for}\hspace{0.17em}j\ne i\text{.}\end{array}$$ Further if $2\le k\le \ell ,$ $S_k$ interchanges $x_{k-1}$ and $x_k$ and fixes every other $x_j\text{.}$ Hence the matrix for $R_1{R}_2·\dots ·R_{\ell }\text{.}$ in this basis is $$\left(\begin{array}{ccccccc}0& & & & & & {\epsilon }^{-1}\\ 1& 0& & & & \\ & 1& 0\\ & & 1\\ & & & & & 0\\ & & & & & 1& 0\end{array}\right)\text{.}$$ The characteristic equation is thus ${\lambda }^{\ell }-{\epsilon }^{-1}=0\text{.}$ Hence $h=\ell p$ and letting $\gamma =e^{2\pi i/h}$ the roots are ${\gamma }^{kp-1}$ $1\le k\le \ell \text{.}$

This completes the verification.

$\square$

We may assume the $p_k$ are not all $2\text{.}$ If $\ell =1$ the statement is obvious. If $\ell =2,$ $\Gamma$ is $p_1\left[q\right]p_2\text{.}$ From [Cox1962] we have that $m_1=\frac{2h}{q}-1,$ $m_2=h-1,$ and $h=\frac{2p_1{p}_{2}q}{(p_1+p_2)q-p_1{p}_2(q-2)}\text{.}$ Thus we have $$\begin{array}{ccc}\frac{1}{h}(m_1+m_2)& =& \frac{2}{q}-\frac{1}{h}+1-\frac{1}{h}=\frac{2}{q}+1-\frac{2}{h}\\ & =& \frac{2}{q}+1-\frac{p_1+p_2}{p_1{p}_2}+1-\frac{2}{q}\\ & =& (1-\frac{1}{p_1})+1-\frac{1}{p_2}\\ & =& \frac{p_1-1}{p_1}+\frac{p_2-1}{p_2}\text{.}\end{array}$$ If $\ell \ge 3$ we have just computed the numbers $m_1,\dots ,m_{\ell }$ and $h$ in the verification of Theorem 6. It thus becomes a trivial arithmetic task to verify the desired result for the remaining cases.

$\square$

Since $m_i=d_i-1,$ and $\sum _{i=1}^{\ell }(d_i-1)$ is the number of reflections in $\theta \left(W\left(\Gamma \right)\right)$ this follows immediately from Theorem 7.

$\square$

If $W$ is a Coxeter group, $m_1=1$ and thus (10) occurs in [Sol1966, p.378] where Solomon credits it to Witt.

If $\Gamma$ has just one vertex labelled with the integer $p,$ we have $\left|W\right|=p$ and $m_1=p-1$ so (10) is trivial.

Suppose next that $\Gamma$ has two vertices; say $\Gamma$ is $p_1\left[q\right]p_2\text{.}$ Letting $h$ denote the order of $S_1{S}_2$ we know from Theorem 6 that $h=m_2+1$ and thus from Theorem 7 that $1-\frac{1}{p_1}-\frac{1}{p_2}=\frac{m_1-1}{m_2+1}\text{.}$ Using this together with the previously mentioned fact that $\left|W\right|=(m_1+1)(m_2+1)$ we compute $$\begin{array}{ccc}\sum _{J\subset I}{(-1)}^J|W:W\left(J\right)|& =& \left|W\right|(1-\frac{1}{p_1}-\frac{1}{p_2})+1\\ & =& (m_1+1)(m_2+1)\frac{(m_1-1)}{m_2+1}+1\\ & =& m_1^2\text{.}\end{array}$$

Now ignoring the infinite family $B_{\ell }^p$ for a moment and using the table in [Cox1967] in conjunction with that in [STo1954] the verification of (10) for the other three non-Coxeter groups is a trivial arithmetic task.

Finally suppose $\Gamma$ is the graph $$B_{\ell }^p\begin{array}{c} p 4 ⋯ \end{array}$$ Consulting [STo1954] and [Cox1967] we see that $\left|W\right|=p^{\ell }\ell !$ and $m_1=p-1\text{.}$ We number the vertices from left to right and denote the $i^{\text{th}}$ vertex by $a_i\text{.}$ We must show that $$\sum _{J\subset I}{(-1)}^J{p}^{\ell }\ell !/\left|W\left(J\right)\right|={(p-1)}^{\ell }\text{.}$$ Let $S$ be the set of all subsets of the vertex set $I=\{a_1,\dots ,a_{\ell }\}\text{.}$ Write $$S=\bigcup _{k=0}^{\ell }{S}_k$$ a disjoint union where we put a subset $J\subset I$ in $S_k$ if $\{a_1,\dots ,a_k\}\subset J$ and $a_{k+1}\notin J\text{.}$ Let $I_k=\{a_{k+2},\dots ,a_{\ell }\}$ for $0\le k\le \ell -2$ and put $I_{\ell -1}=I_{\ell }=\varphi \text{.}$ If $J\in S_k$ then $J=\{a_1,\dots ,a_k\}\cup H$ where $H\subseteq I_k$ and thus $$\left|W\left(J\right)\right|=p^{k}k!\left|W\left(H\right)\right|$$ (Here of course $\left|W\left(\varphi \right)\right|=1\text{).}$

Hence $$\sum _{J\subseteq I}{(-1)}^J{\left|W\left(J\right)\right|}^{-1}=\sum _{k=0}^{\ell }\frac{{(-1)}^k}{p^{k}k!}\sum _{H\subseteq I_k}{(-1)}^H{\left|W\left(H\right)\right|}^{-1}$$ But $W\left(I_k\right)$ is the symmetric group on $\ell -k$ letters, a Coxeter group, so (10) implies $$\sum _{H\subseteq I_k}{(-1)}^H{\left|W\left(H\right)\right|}^{-1}=\frac{1}{(\ell -k)!}$$ Thus $$\begin{array}{ccc}\sum _{J\subseteq I}{(-1)}^J|W:W\left(J\right)|& =& \sum _{k=0}^{\ell }{(-1)}^k\frac{\ell !}{k!(\ell -k)!}{p}^{\ell -k}\\ & =& {(p-1)}^{\ell }\text{.}\end{array}$$

$\square$

For both (a) and (b) it is clear that the right side is included in the left. So suppose $\bar{x}\in {ℂ}_{\bar{G}}\left(\bar{g}\right)\text{.}$ Hence $g^x=\lambda g$ some $\lambda \in Z\text{.}$ Taking traces on both sides of this equation we have $$\text{tr}\left(g\right)=\lambda \text{tr}\left(g\right),$$ so $\text{tr}\left(g\right)\ne 0$ implies $\lambda =1\text{.}$ Thus $g^x=g$ and $\bar{x}\in \overline{{ℂ}_G\left(g\right)},$ yielding (a). In the situation of (b) we take determinants on both sides of the equation $g^x=\lambda g$ to obtain $$\text{Det}\left(g\right)={\lambda }^2\text{Det}\left(g\right)\text{.}$$ Thus ${\lambda }^2=1,$ so $g^x=\pm g$ and $\bar{x}\in \overline{{ℂ}_G^{\prime }\left(g\right)}\text{.}$

$\square$

Suppose $p_i\ne 2\text{.}$ Since $G$ is irreducible $𝒵\left(G\right)$ consists of scalar matrices. Put $\bar{G}=G/𝒵\left(G\right)\text{.}$ Since $S_i$ is a reflection ${\bar{S}}_i$ has order $p_i\text{.}$ From [STo1954] we know that $\bar{G}$ is the alternating group on four letters, the symmetric group on four letters, or the alternating group on five letters, and any non identity element of order different from two in any of these groups is self centralizing. Hence ${ℂ}_{\bar{G}}\left({\bar{S}}_i\right)=⟨{\bar{S}}_i⟩\text{.}$ Now $p_i\ne 2$ further forces $\text{tr}\left(S_i\right)\ne 0,$ and applying Lemma 4(a) we have $\overline{{ℂ}_G\left(S_i\right)}=⟨{\bar{S}}_i⟩\text{.}$ Thus ${ℂ}_G\left(S_i\right)=⟨S_i⟩\times 𝒵\left(G\right)\text{.}$

Now suppose $p_i=2\text{.}$ Consulting List 1 of Theorem 2 we see $\Gamma$ must be one of $3\left[6\right]2,$ $4\left[6\right]2,$ $3\left[8\right]2,$ $5\left[6\right]2,$ $3\left[10\right]2\text{.}$ From [STo1954] we have $2|\left|𝒵\left(G\right)\right|$ so that $-I\in 𝒵\left(G\right)\text{.}$ Hence $-S_2$ is a reflection in $G\text{.}$ It follows immediately from Theorem 5 that $-S_2$ is conjugate to $S_2$ for all of the above graphs with the possible exception of $4\left[6\right]2$ where we must rule out the possibility that $-S_2$ is conjugate to $S_1^2\text{.}$

In the group $4\left[6\right]2,$ $S_1{S}_2$ has order 24 and $𝒵\left(G\right)$ is generated by ${\left(S_1{S}_2\right)}^3,$ an element of order 8. Hence $-I={\left(S_1{S}_2\right)}^{12}\text{.}$ If $\lambda$ is a linear character of $G,$ $\lambda \left(S_1\right)=i^{\alpha }$ and $\lambda \left(S_2\right)={(-1)}^{\beta }$ and thus $\lambda {\left(S_1{S}_2\right)}^{12}=1\text{.}$ Thus $-I={\left(S_1{S}_2\right)}^{12}\in G\prime ,$ the commutator subgroup of $G\text{.}$ So, $-S_2$ and $S_2$ have the same images in $G/G\prime \text{;}$ but a glance at the proof of Lemma 3 reveals that $S_2$ and $S_1^2$ have distinct images in $G/G\prime \text{.}$ Hence, $-S_2$ is not conjugate to $S_1^2\text{.}$

So using the notation of Lemma 4 we have $${ℂ}_G^{\prime }\left(S_2\right)>{ℂ}_G\left(S_2\right)\text{.}$$ In fact, we have $$\mid {ℂ}_G^{\prime }\left(S_2\right):{ℂ}_G\left(S_2\right)\mid =2\text{.}$$ In light of Lemma 4(b) we see that to show ${ℂ}_G\left(S_2\right)=⟨S_2⟩\times 𝒵\left(G\right)$ it suffices to show $$\begin{array}{cc}\text{(12)}& \left|{ℂ}_{\bar{G}}\left({\bar{S}}_2\right)\right|=4\text{.}\end{array}$$ If $\bar{G}$ is the alternating group on tour or five letters (12) is obvaous. So we assume $\bar{G}$ is the symmetric group on four letters. Consulting [STo1954] we see that $\Gamma$ is $3\left[8\right]2$ or $4\left[6\right]2\text{.}$ If ${\bar{S}}_2$ is a transposition (12) is again obvious.

So assume ${\bar{S}}_2$ is a product of two disjoint transpositions. Thus ${\bar{S}}_2$ is an element of the alternating group on four letters. But if $\Gamma$ is $3\left[8\right]2$ then ${\bar{S}}_1,$ an element of order three, is also in the alternating group and hence so is $\bar{G}=⟨{\bar{S}}_1,{\bar{S}}_2⟩\text{;}$ a contradiction. If $\Gamma$ is $4\left[6\right]2$ then as mentioned before ${\bar{S}}_1{\bar{S}}_2$ has order three and hence is in the alternating group. This again is impossible since ${\bar{S}}_1$ has order four and does not lie in the alternating group.

Using Corollary 9 we can now proceed to verify our conjecture (11).

If $w\in W,$ $w\ne 1,$ and $\theta \left(w\right)$ is not a reflection, then $w$ is not conjugate to any element of either $⟨r_1⟩$ or $⟨r_2⟩$ and $\mu \left(w\right)=0$ so $k\left(w\right)=2\text{.}$ Hence $$\Psi \left(w\right)=1={(-1)}^{k\left(w\right)}{m}_1^{\mu \left(w\right)}\text{.}$$

If $\theta \left(w\right)$ is a reflection, then by Theorem 5, $w$ is conjugate to an element of $⟨r_1⟩$ or an element of $⟨r_2⟩\text{.}$ By looking at the matrices one easily sees that if $x\in ⟨r_i⟩,$ $x\ne 1,$ then ${ℂ}_w\left(x\right)={ℂ}_w\left(r_i\right)$ and thus by Corollary 9, ${ℂ}_w\left(x\right)=⟨r_i⟩\times 𝒵\left(W\right)\text{.}$ Further from [Cox1974] we know that $\left|𝒵\left(W\right)\right|\frac{(q,2)h}{q}$ and that $m_1=\frac{2h}{q}-1\text{.}$ We next separate two cases.

Case (i) $q$ is even. Without loss of generality we assume $w$ is to an element of $⟨r_1⟩\text{.}$ By Corollary 6 $w$ is not conjugate to an element of $⟨r_2⟩\text{.}$ Hence $1_{⟨r_2⟩}^W\left(w\right)=0\text{.}$ Since $S_1,$ $S_1^2,$ $\dots ,$ $S_1^{p_1-1}$ have distinct non identity eigenvalues the conjugacy classes determined by $r_1,$ $r_1^2,$ $\dots ,$ $r_1^{p_1-1}$ are all distinct. Hence $$\begin{array}{ccc}{1}_{⟨r_1⟩}^W\left(w\right)& =& \frac{1}{p_1}\left|{ℂ}_W\left(w\right)\right|\\ & =& \left|𝒵\left(W\right)\right|\\ & =& \frac{2h}{q}\end{array}$$ So $\Psi \left(w\right)=1-\frac{2h}{q}=-m_1={(-1)}^{k\left(w\right)}{m}_1^{\mu \left(w\right)}\text{.}$

Case (ii) $q$ is odd. Thus, $p_1=p_2=p,$ say and $r_1$ is conjugate to $r_2\text{;}$ so $w$ is conjugate to an element of $⟨r_1⟩$ and to an element of $⟨r_2⟩\text{.}$ Hence $$1_{⟨r_1⟩}^W\left(w\right)=1_{⟨r_2⟩}^W\left(w\right)=\frac{1}{p}\left|{ℂ}_W\left(w\right)\right|=\frac{h}{q},$$ So again, $\Psi \left(w\right)=1-\frac{2h}{q}=-m_1={(-1)}^{k\left(w\right)}{m}_1^{\mu \left(w\right)}\text{.}$ This completes the verification of our conjecture in case $\ell =2\text{.}$

$\square$