Proposition 9

Let $a \in E$. The following conditions are equivalent: (i) We have $W = W_a.L$; (ii) The restriction of the homomorphism $U$ to $W_a$ is an isomorphism from $W_a$ to $U(W)$; (iii) For every hyperplane $H \in \mathfrak{H}$, there exists a hyperplane $H' \in \mathfrak{H}$ parallel to $H$ and such that $a \in H'$.

It is clear that (i) $\Leftrightarrow$ (ii), since $L$ is the kernel of $U$ and $L \cap W_a = \{1\}$.

Assume (i) and let $H \in \mathfrak{H}$; then $s_H \in W_a.L$ so there exists a vector $t \in \Lambda$ such that $a = s_H(a) + t$; the vector $t$ is orthogonal to $H$, and if $H' = H + \frac{1}{2} t$ then $s_{H'}(x) = s_H(x) + t$ for all $x \in E$ (cf. § 2, no. 4, Prop. 5). Since $t \in \Lambda$ and $s_H \in W$, we have $s_{H'} \in W$, and so $H' \in \mathfrak{H}$; we also have $a = s_{H'}(a)$, and so $a \in H'$. Thus, (i) implies (iii).

Assume (iii). Let $H \in \mathfrak{H}$; take $H'$ as in (iii). Then $s_{H'}(a) = a$, so $s_{H'} \in W_a$; since $H$ is parallel to $H'$, the element $w = s_{H'} s_H$ of $W$ is a translation (§ 2, no. 4, Prop. 5), so $w \in L$; then $s_H = s_{H'} w \in W_a . L$. Since $W$ is generated by the family $(s_H)_{H \in \mathfrak{H}}$, it follows that $W = W_a . L$ and hence (iii) implies (i).