Theorem 1

Assume that W is irreducible. Then: (i) $m_1 = 1, m_l = h-1$. (ii) $\mathrm{Card}(\mathfrak{H}) = \frac{1}{2} lh$.

We retain the preceding notation. The restriction of $c = s's''$ to $P$ is the rotation with angle $2(\widehat{z'', z'})$ (§ 2, no. 5, Cor. of Prop. 6). Since $c$ is of order $h$, the $h$ elements $1, c, \ldots, c^{h-1}$ of $W$ are pairwise distinct; the elements $s', s'c, \ldots, s'c^{h-1}$ are thus pairwise distinct, and are distinct from the preceding elements since $c^i|P$ is a rotation and $s'c^j|P$ is a reflection. The set

is the subgroup $W'$ of $W$ generated by $s'$ and $s''$, and induces on $P$ the group $W''$ generated by the orthogonal reflections with respect to $Rz', Rz''$. The transform of $C$ by an element of $W'$ is either disjoint from $-C$ or equal to $-C$. Thus, the transform of $P \cap C$ by an element of $W''$ is either disjoint from $-(P \cap C)$ or equal to $-(P \cap C)$. Hence, for a suitable orientation of P, there exists an integer $m > 0$ such that $(\overline{z''}, z') = \frac{\pi}{m}$ (§ 2, no. 5, Cor. of Prop. 7). Moreover, the sets $g'(C)$, for $g' \in W'$, are pairwise disjoint; the sets $g''(P \cap C)$, for $g'' \in W''$, are thus pairwise disjoint; so $W''$ is of order $2h$. Hence $m = h$. By definition, $c|P$ is a rotation with angle $\frac{2\pi}{m}$, and so has eigenvalues $\exp \frac{2i\pi}{h}$, $\exp \frac{2i\pi(h-1)}{h}$. This proves that $m_1 = 1, m_l = h - 1$.

The transforms of $Rz'$ and $Rz''$ by $W'$ are $h$ lines $D_1, \ldots, D_h$ of P, and the points of $P - (D_1 \cup \cdots \cup D_h)$ are transforms by the elements of $W'$ of points of $P \cap C$. Thus, a hyperplane of $\mathfrak{H}$ necessarily cuts P along one of the lines $D_i$, and consequently is a transform by an operation of $W'$ of a hyperplane of $\mathfrak{H}$ containing $Rz'$ or $Rz''$.

Now, any $H \in \mathfrak{H}$ that contains $Rz'$ is one of the hyperplanes $H_1, \ldots, H_r$. Indeed, let $e_H$ be the unit vector orthogonal to H and on the same side of H as C. Then $e_H = \lambda_1 e_1 + \cdots + \lambda_l e_l$ with the $\lambda_i$ all $\geq 0$ (§ 3, no. 5, Lemma 6, (i)). Now $0 = (e_H|z') = \lambda_{r+1} a_{r+1} + \cdots + \lambda_l a_l$, so

Suppose that two of the $\lambda_i$ were non-zero, for example $\lambda_1$ and $\lambda_2$; since $e_1, \ldots, e_r$ are pairwise orthogonal, we would have and the coordinates of $s_1(e_H)$ would not be of the same sign, which is absurd (loc. cit.). Thus, $e_H$ is proportional to one of the vectors $e_1, \ldots, e_r$, which proves our assertion. Similarly, any $H \in \mathfrak{H}$ that contains $Rz''$ is one of the hyperplanes $H_{r+1}, \ldots, H_l$.

The number of elements of $\mathfrak{H}$ containing $Rz'$ or $Rz''$ is therefore $l$. If $h$ is even, $\mathrm{Card}(\mathfrak{H})$ is thus equal to $\frac{h}{2} l$. If $h$ is odd, $\mathrm{Card}(\mathfrak{H})$ is equal to $\frac{h-1}{2} l + r$, and also to $\frac{h-1}{2} l + (l - r)$; hence $r = l - r$, so $r = \frac{l}{2}$, and $\mathrm{Card}(\mathfrak{H}) = \frac{h-1}{2} l + \frac{l}{2} = \frac{h}{2} l$.