Theorem 4

Put, for $s \in G$ and $x \in E$, $\varphi(s).x = sx s^{-1}$.

a) The mapping $\varphi$ is a homomorphism of $G$ into the orthogonal group $O(Q)$ of $Q$ and its kernel is the set of invertible elements of the center $Z$ of $C(Q)$.

b) The set $E \cap G$ is the set of nonsingular vectors of $E$; for $x \in E \cap G$, $-\varphi(x)$ is the symmetry with respect to the hyperplane orthogonal to $x$.

c) If $\dim(E)$ is even, one has $\varphi(G) = O(Q)$, $\varphi(G^+)$ is of index 2 in $O(Q)$ if $E \neq \{0\}$, and is equal to $SO(Q)$ if $A$ is of characteristic $\neq 2$.

d) If $\dim(E)$ is odd (which implies that $A$ is of characteristic $\neq 2$), one has $\varphi(G) = \varphi(G^+) = SO(Q)$.

Indeed, one has $Q(sxs^{-1}) = (sxs^{-1})^2 = sx^2 s^{-1} = Q(x)$ for $s \in G$ and $x \in E$, which shows that $\varphi(s) \in O(Q)$. For $\varphi(s) = 1$ to hold, it is necessary and sufficient that $s$ commute with the elements of $E$, that is to say, belong to the center $Z$ of $C(Q)$. This proves a).

For an element $x$ of $E$ to belong to $G$, it is necessary that it be invertible, that is to say, that it be a nonsingular vector (since $x^2 = Q(x)$). If this is so, one has $x^{-1} = Q(x)^{-1} x$, whence, for every $y \in E$,

this proves b) (§ 6, no 4).