Lemma 1

(i) Let $w \in W$ and $t \in T$. The number $(-1)^{n(s, t)}$ has the same value $\eta(w, t)$ for all sequences $s = (s_1, \ldots, s_q)$ of elements of S such that $w = s_1 \ldots s_q$.

(ii) For $w \in W$, let $U_w$ be the map from R to itself defined by

The map $w \mapsto U_w$ is a homomorphism from W to the group of permutations of R.

(2) Recall that two elements (resp. two subsets) of a group W are said to be conjugate if there exists an inner automorphism of W that transforms one into the other.

For $s \in S$, define a map $U_s$ from $R$ to itself by

where $\delta_{s,t}$ is the Kronecker symbol. It is immediate that $U_s^2 = \mathrm{Id}_R$, which shows that $U_s$ is a permutation of $R$.

Let $s = (s_1, \ldots, s_q)$ be a sequence of elements of $S$. Put $w = s_q \ldots s_1$ and $U_s = U_{s_q} \ldots U_{s_1}$. We shall show, by induction on $q$, that

This is clear for $q = 0, 1$. If $q > 1$, put $s' = (s_1, \ldots, s_{q-1})$ and Using the induction hypothesis, we obtain But $\Phi(s) = (\Phi(s'), {w'}^{-1} s_q w')$ and $n(s, t) = n(s', t) + \delta_{w', -1} s_q w', t$, proving formula (14).

Now let $s, s' \in S$ be such that $p = ss'$ has finite order $m$. Let $s = (s_1, \ldots, s_{2m})$ be the sequence of elements of $S$ defined by $s_j = s$ for $j$ odd and $s_j = s'$ for $j$ even. Then $s_{2m} \ldots s_1 = p^{-m} = 1$ and formula (11) implies that

Since $p$ is of order $m$, the elements $t_1, \ldots, t_m$ are distinct and $t_{j+m} = t_j$ for $1 \leq j \leq m$. For all $t \in T$, the integer $n(s, t)$ is thus equal to 0 or 2 and (14) shows that $U_s = \mathrm{Id}_R$. In other words, $(U_s U_{s'})^m = \mathrm{Id}_R$. Thus, by the definition of Coxeter systems, there exists a homomorphism $w \mapsto U_w$ from $W$ to the group of permutations of $R$ such that $U_s$ is given by the right-hand side of (13). Then $U_w = U_s$ for every sequence $s = (s_1, \ldots, s_q)$ such that $w = s_q \ldots s_1$ and Lemma 1 follows immediately from (14).