Lemma 5

Every element $s$ of $G$ is of the form $zs'$, where $z$ is an invertible element of $Z$ and $s'$ belongs to $G \cap C^+(Q)$ or to $G \cap C^-(Q)$; the subgroup $G^+$ is of index 2 in $G$ when $E \neq \{0\}$.

The second assertion follows obviously from the first, since the nonsingular vectors belong to $G \cap C^-(Q)$. Suppose first that $\dim(E)$ is even, and let $s = t' + t''$, with $t' \in C^+(Q)$ and $t'' \in C^-(Q)$; by definition we have $sx = (\varphi(s).x)s$ for every $x \in E$; since $t'x$ and $(\varphi(s).x)t'$ (resp. $t''x$ and $(\varphi(s).x)t''$) are odd (resp. even) elements, we have $t'x = (\varphi(s).x)t'$, whence $s^{-1} t' x = xs^{-1} t'$ for every $x \in E$. It follows that $s^{-1} t' \in Z$ and, since $\dim(E)$ is even, $Z = A$ (No. 4, cor. of Theorem 2), hence $t' = as$, where $a \in A$. If $a \neq 0$, we therefore have $s = a^{-1} t'$ and $t' \in G \cap C^+(Q)$; if $a = 0$, $s = t'' \in G \cap C^-(Q)$ and the lemma is proved in this case. If $\dim(E)$ is odd, $A$ is of characteristic $\neq 2$, hence for every $s \in G$, $\varphi(s)$ is a product of symmetries with respect to nonsingular vectors $x_i (i = 1, 2, \ldots, h)$ (§ 6, No. 4, Proposition 5); if we put $s' = x_1 x_2 \ldots x_h$, we have $\varphi(s) = \varphi(s')$, hence $s = z s'$, where $z \in \mathbf{Z}$, and $s'$ belongs to $C^+(Q)$ or to $C^-(Q)$ according as $h$ is even or odd.

Suppose $\dim(E)$ is even. Since every element $u$ of $\mathbf{O}(Q)$ extends in one and only one way to an automorphism $\bar{u}$ of $C(Q)$ (Proposition 1), and since $C(Q)$ is central simple (Theorem 2), $\bar{u}$ is an inner automorphism (Chapter VIII, § 10, No. 1, Theorem 1). There therefore exists an element $s$ of $G$ such that $\varphi(s) = u$. On the other hand, the center of $C(Q)$ is contained in $C^+$, which entails that $\varphi(G)/\varphi(G^+)$ is isomorphic to $G/G^+$, hence that $\varphi(G^+)$ is of index 2 in $\varphi(G) = \mathbf{O}(Q)$ if $E \not= \{0\}$. This proves the first two assertions of c).

Suppose finally that $A$ is of characteristic $\not= 2$. Then every element $u$ of $\mathbf{O}(Q)$ is a product of symmetries with respect to hyperplanes orthogonal to nonsingular vectors $x_i (i = 1, \ldots, h)$ (§ 6, n° 4, prop. 5); consequently one has $u = (-1)^h \varphi(x_1 \ldots x_h)$ and $\det(u) = (-1)^h$. For $u$ to belong to $\mathbf{SO}(Q)$, it is necessary and sufficient that $h$ be pair, which shows that $\varphi(G^+) \supset \mathbf{SO}(Q)$. Since $\mathbf{SO}(Q)$ is of index 2 in $\mathbf{O}(Q)$ when $E \not= \{0\}$, one has $\varphi(G^+) = \mathbf{SO}(Q)$ if $E$ is of even dimension, which completes the proof of c). On the other hand, if the dimension of $E$ is odd, $\varphi(G)$ does not contain the orthogonal transformation $x \to -x$; indeed this extends to the principal automorphism $\alpha$ of $C(Q)$ (n° 1), and $\alpha$ is not an inner automorphism since the center $Z$ of $C(Q)$ contains a nonzero element of $C^-(Q)$ (th. 3). Thus $\varphi(G) \not= \mathbf{O}(Q)$, and, since $\varphi(G) \supset \varphi(G^+) \supset \mathbf{SO}(Q)$ and $\mathbf{SO}(Q)$ is of index 2 in $\mathbf{O}(Q)$, one has $\varphi(G) = \varphi(G^+) = \mathbf{SO}(Q)$. This proves d). Q.E.D.

The subgroup $\varphi(G^+)$ of $\mathbf{O}(Q)$, which is of index 2 if $E \not= \{0\}$, is called the group of rotations of $E$, and its elements are called rotations; it is denoted by $\mathbf{O}^+(Q)$. Notice that, if $A$ is not of characteristic 2, one has $\mathbf{O}^+(Q) = \mathbf{SO}(Q)$ (cf. exerc. 9).