Proposition 4

Let $F$ be an octonion algebra of type $(-1, 0, -1, -1)$ over $\mathbf{A}$. There exists a vector space $V$ of dimension 3 over the field with two elements $\mathbf{Z}/2\mathbf{Z}$ and a bijection $\lambda \mapsto e'_\lambda$ of $V$ onto the basis $(e_i)_{0 \leq i \leq 7}$ such that

for all $\lambda, \mu$ in $V$. In order that $e'_\lambda (e'_\mu e'_\nu) = (e'_\lambda e_\mu) e'_\nu$, it suffices that, in $V$, $\lambda, \mu, \nu$ be linearly dependent over $\mathbf{Z}/2\mathbf{Z}$; this condition is necessary if $2 \neq 0$ in $\mathbf{A}$.

We preserve the notation at the beginning of this no. It follows from formulae (33) of § 2, no. 5 that the set $S$ consisting of the elements $\pm e_0, \pm e_1, \pm e_2, \pm e_3$ is stable under multiplication. Moreover, for $x, y, y'$ in $E$, by formula (22) of § 2, no. 5,

so that the set $T$ consisting of the elements $\pm e_i$ ($0 \leq i \leq 7$) is stable under multiplication; moreover, its multiplication table is independent of the ring $\mathbf{A}$.

In particular, let $\mathbf{A}''$ be the field $\mathbf{Z}/2\mathbf{Z}$ with two elements and let $E''$ be the quaternion algebra of type $(1, 0, 1)$ over $\mathbf{A}''$ and $F''$ the algebra of octonions of type $(1, 0, 1, 1)$ over $\mathbf{A}''$; let $(e''_i)_{0 \leq i \leq 7}$ be the basis of $F''$ formed as described above. Since $-e''_i = e''_i$, the set $T''$ of $e''_i$ has 8 elements and is stable under multiplication; moreover, it follows immediately from the above that the mapping $\theta : T \to T''$ such that $\theta(e_i) = \theta(-e_i) = e''_i$ for $0 \leq i \leq 7$ is a homomorphism for multiplication. Moreover the quaternion algebra $E''$ is in this case commutative and hence $F''$ is associative (§ 2, no. 5, Proposition 5); further, conjugation in $F''$ is in this case the identity. Hence $T''$ is a group and formulae (8) show that it is commutative; these formulae and formulae (33) of § 2, no. 5 show that the square of every element of $T''$ is the unit. If $V$ is used to denote the group $T''$ written additively, $V$ can be given a unique vector space structure over $\mathbf{Z}/2\mathbf{Z}$, necessarily of dimension 3 since

For all $\lambda \in V$, let $e'_\lambda$ then denote the element of $(e_i)_{0 \leq i \leq 7}$ such that $\theta(e'_\lambda) = \lambda$; then $e'_0 = e_0$; moreover, as $\theta$ is a homomorphism and the relation $\theta(x) = \theta(y)$ is equivalent to $x = \pm y$, $e'_\lambda e'_\mu = \pm e'_{\lambda + \mu}$. If $\lambda, \mu, \nu$ are linearly independent over $\mathbf{Z}/2\mathbf{Z}$, they form a basis of $V$ and hence all the elements $e_i$ ($0 \leq i \leq 7$) would belong to the subalgebra generated by $e'_\lambda, e'_\mu$ and $e'_\nu$; when $2 \neq 0$ in $\mathbf{A}$, $e'_\lambda (e'_\mu e'_\nu) = (e'_\lambda e'_\mu) e'_\nu$ is therefore impossible, for $F$ would be associative and hence $E$ would be commutative (§ 2, no. 5, Proposition 5), which contradicts relations (33) of § 2, no. 5. On the other hand, if $\lambda, \mu, \nu$ are linearly dependent in V, the three elements $e'_\lambda, e'_\mu, e'_v$ belong to a subalgebra with 2 generators of F, which is therefore associative (no. 1, Proposition 1); whence the conclusion.