We return to the question of finding rotations in four dimensions. Conjugation doesn't do the job — as we saw in § 9.1.1, conjugation induces rotations in three dimensions, and single-sided multiplication isn't sufficient to induce rotations in four dimensions. Therein lies a clue; what about double-sided mlutiplication?

We compute \begin{equation} e^{i\alpha} \left(r_1 e^{i\theta} + r_2 e^{i\phi} j\right) e^{i\alpha} = r_1 e^{i\theta+2\alpha} + r_2 e^{i\phi} j \end{equation} corresponding to a rotation by $2\theta$ in the $1i$ plane. Again, we can repeat this construction with $i$ replaced by $j$ and $k$ to produce rotations in the $1j$ and $1k$ planes, respectively. Combining these rotations with those in the imaginary planes ($ij$, $jk$, and $ki$), we have constructed rotations in each of the 6 coordinate planes, from which arbitrary rotations can be constructed.

We have shown that every transformation in $\SO(4)$ is a combination of transformations that take $q$ either to $pq\bar{p}=pqp^{-1}$ or $pqp$. However, not every $\SO(4)$ transformation is necessarily of this form, so we write \begin{align} \SO(4) &= \langle\{ q \longmapsto pq\bar{p} : p,q\in\HH, |p|=1 \} \nonumber\\ &\qquad \hphantom{x}\cup \{ q \longmapsto pqp : p,q\in\HH, |p|=1 \}\rangle \end{align} where the angled brackets imply that $\SO(4)$ is generated by the given transformations.

Thanks to the associativity of the quaternions, we can rewrite these transformations in several ways. First of all, we can combine all of the factors on the left into a single factor, and the same on the right. Thus, \begin{equation} \SO(4) = \langle\{ q \longmapsto p_1qp_2 : p_1,p_2,q\in\HH, |p_1|=1=|p_2| \} \end{equation}

Furthermore, if we first conjugate $q$ by $p$, then multiply by $p$ on both sides, we could equivalently have multiplied by $p^2$ on the left; a similar construction yields multiplication on the right. Thus, instead of combining conjugation and two-sided multiplicaion, we could equivalently have combined left and right multiplication, that is \begin{equation} \SO(4) = \langle\{ q \longmapsto pq : p\in\HH, |p|=1 \} \cup \{ q \longmapsto qp : p\in\HH, |p|=1 \}\rangle \end{equation} But we have already seen that single-sided mlutiplication generates a version of $\SO(3)$. We have therefore established an equivalence between two copies of $\SO(3)$ and $\SO(4)$, which is customarily written as \begin{equation} \SO(4) \cong \SO(3) \times \SO(3) \end{equation}