Proposition 3

The homomorphism $u \to u/\overline{u} = u^2 / N(u)$ of $S^+$ into itself has $H$ as kernel, and defines, by passing to the quotient, an isomorphism of $S^+/H$ onto $O^+$.

In fact the relation $u/\overline{u} = 1$ is equivalent to $u = \overline{u}$, that is to say to $u \in A^* = H$; hence $H$ is the kernel of $u \to u/\overline{u}$. Since $N(u/\overline{u}) = 1$, $u/\overline{u}$ is a rotation (Remark 1). It remains to prove that every element $\varphi$ of $O^+$ is of the form $u/\overline{u}$ ($u \in S^+$). If $1 + \varphi$ is invertible, one may take $u = 1 + \varphi$, since the relation $N(\varphi) = \varphi \overline{\varphi} = 1$ implies $1 + \varphi = \varphi(1 + \overline{\varphi})$. Otherwise, one has $N(1 + \varphi) = (1 + \varphi)(1 + \overline{\varphi}) = 0$, that is to say, on putting $\varphi = a + b\omega$ ($a \in A,\ b \in A,\ \omega^2 = \delta \in A$), $2(1 + a) = 0$, whence $a = -1$; now the relations $a = -1$ and $N(\varphi) = a^2 - \delta b^2 = 1$ imply $b = 0$, whence $\varphi = -1$; since $\overline{\omega} = -\omega$, it is enough, in this case, to take $u = \omega$.

When $A(\Phi)$ is a field, Proposition 3 is a particular case of Hilbert's theorem (Chapter V, § 11, No. 5, Theorem 3).