Theorem 1

Let $C$ be a chamber and let $S$ be the set of reflections with respect to the walls of $C$.

(i) The pair $(W, S)$ is a Coxeter system.

(ii) Let $w \in W$ and let $H$ be a wall of $C$. The relation $l(s_H w) > l(w)$ implies that the chambers $C$ and $w(C)$ are on the same side of $H$.

(iii) For any chamber $C'$, there exists a unique element $w \in W$ such that $w(C) = C'$.

(iv) The set of hyperplanes $H$ such that $s_H \in W$ is equal to $\mathfrak{H}$.

Every element of $S$ is of order 2 and Lemma 2 shows that $S$ generates $W$. For any wall $H$ of $C$, denote by $P_H$ the set of elements $w \in W$ such that the chambers $C$ and $w(C)$ (which do not meet $H$) are on the same side of $H$. We shall verify conditions (A'), (B') and (C) of Chap. IV, § 1, no. 7.

(A') $1 \in P_H$: Trivial.

(B') $P_H$ and $s_H.P_H$ are disjoint: Indeed, $w(C)$ and $s_H w(C)$ are on opposite sides of $H$, so if $w(C)$ is on the same side of $H$ as $C$, it is not on the same side as $s_H w(C)$.

(C) Let $w \in P_H$ and let $H'$ be a wall of $C$ such that $w s_{H'} \notin P_H$; then $w s_{H'} = s_H w$: By assumption, $w(C)$ is on the same side of $H$ as $C$ and $w s_{H'}(C)$ is on the other side. Thus, $w s_{H'}(C)$ and $w(C)$ are on opposite sides of H; hence, the chambers $s_{H'}(C)$ and C are on opposite sides of the hyperplane $w^{-1}(H)$. Let $a$ be a point of the face of C with support $H'$. The point $a = s_{H'}(a)$ is in the closure of the two chambers C and $s_{H'}(C)$ that are contained respectively in the two open half-spaces bounded by $w^{-1}(H)$; thus, $a \in w^{-1}(H)$, so $H' = w^{-1}(H)$. From this we deduce that $s_{H'} = w^{-1}s_Hw$, hence $ws_{H'} = s_Hw$.

Assertions (i) and (ii) follow from this and Prop. 6 of Chap. IV, § 1, no. 7. Moreover, we have (loc. cit., condition (A))

Lemma 2 shows that W acts transitively on the set of chambers. Moreover, if $w \in W$ is such that $w(C) = C$, then $w \in P_H$ for every wall H of C, hence $w = 1$ by (3). This proves (iii).

Finally, let H be a hyperplane such that $s_H \in W$. If H did not belong to $\mathfrak{H}$, there would be at least one chamber $C'$ meeting H (§ 1, no. 3, Prop. 7). Every point of $H \cap C'$ is invariant under $s_H$, and thus belongs to the chambers $C'$ and $s_H(C')$; thus, $C' = s_H(C')$, which contradicts (iii) since $s_H \neq 1$.