Proposition 5

Let M be a monoid (with unit element 1) and f a map from S to M. For s, s' \in S, let m(s, s') be the order of ss' and put

If $a(s, s') = a(s', s)$ whenever $s \neq s'$ are in S, there exists a map g from W to M such that

for all $w \in W$ and any reduced decomposition $(s_1, \ldots, s_q)$ of w.

For any $w \in W$, let $D_w$ be the set of reduced decompositions of w and $F_w$ the map from $D_w$ to M defined by

We are going to prove by induction on the length of w that $F_w$ is constant, which will establish Prop. 5. The cases $l(w) = 0, 1$ being trivial, we assume that $q \geq 2$ and that our assertion is proved for the elements w with $l(w) < q$. Let w be of length q and $s, s' \in D_w$; by Lemma 4 it suffices to prove that $F_w(s) = F_w(s')$ in cases a) and b) of that lemma.

a) The formula

for $w' = s_1 \ldots s_{q-1}$ and $w'' = s_2 \ldots s_q$ and the induction hypothesis show that $F_w(s) = F_w(s')$ if $s_1 = s'_1$ or $s_q = s'_q$.

b) Suppose that there exist two elements s and s' of S such that $s_j = s'_k = s$ and $s_k = s'_j = s'$ for j odd and k even. It suffices to treat the case $s \neq s'$. The sequences s and s' are then two distinct reduced decompositions of w in the dihedral group generated by s and s'. By the Remark in no. 2, the order m of ss' is necessarily finite and, in the notation of that remark, $s = s_m$ and $s' = s'_m$. Consequently, $F_w(s) = a(s, s')$ and $F_w(s') = a(s', s)$ and hence