Proposition 2

The subgroup of $\mathbf{GL}(E)$ generated by $\sigma_s$ and $\sigma_{s'}$ is a dihedral group of order $2m(s, s')$.

Since $\sigma_s$ and $\sigma_{s'}$ are of order 2, and are distinct, it is enough to show that their product $\sigma_s \sigma_{s'}$ is of order $m(s, s')$. When $m(s, s')$ is infinite, this follows from $a$ above. When $m(s, s')$ is finite, it follows from Prop. 1 that E is the direct sum of $E_{s,s'}$ and its orthogonal complement $V_{s,s'}$; since $\sigma_s$ and $\sigma_{s'}$ are the identity on $V_{s,s'}$, and since the restriction of $\sigma_s \sigma_{s'}$ to $E_{s,s'}$ is of order $m(s, s')$ by $b$, the order of $\sigma_s \sigma_{s'}$ is indeed equal to $m(s, s')$.