Corollary 1

*For any subset $X$ of $S$, the subgroup $W_X$ of $W$ consists of the elements $w$ of $W$ such that $S_w \subset X$.

If $w = s_1 \ldots s_q$ with $s_1, \ldots, s_q$ in $S$, then $w^{-1} = s_q \ldots s_1$; hence

Prop. 4 of no. 5 shows that $S_{sw'} \subset \{s\} \cup S_{w'}$ for $s \in S$ and $w' \in W$, which implies the formula

by induction on the length of $w$. By (23) and (24), the set $U$ of $w \in W$ such that $S_w \subset X$ is a subgroup of $W$; we have $X \subset U \subset W_X$, hence $U = W_X$.