Proposition 4

a) The map $s \mapsto g_s$ from $S$ to $W$ is injective.

b) Each of the $g_s$ is of order 2.

c) If $s, s' \in S$, $g_s g_{s'}$ is of order $m(s, s')$.

Assertion a) follows from the fact that the composite map

from $S$ to $\mathbf{GL}(E)$ is injective.

For b) (resp. c)), we remark that the order of $g_s$ (resp. the order of $g_s g_{s'}$) is at most 2 (resp. at most $m(s, s')$). Since we have seen in no. 2 that the order of $\sigma_s$ (resp. of $\sigma_s \sigma_{s'}$) is 2 (resp. $m(s, s')$), we must have equality.

In view of a), $S$ can be identified with a subset of $W$ by means of the map $s \mapsto g_s$.

¹ This means that, if $L_S$ denotes the free group on $S$, $W$ is the quotient of $L_S$ by the smallest normal subgroup of $L_S$ containing the elements $(ss')^{m(s, s')}$, for $m(s, s') \neq +\infty$.