How come the non-abelian property is lost in this isomorphism of quaternions? Identify $q \in \mathbb{H}$ by $q \iff \begin{pmatrix} a & b \\ -\overline{b} & \overline{a} \end{pmatrix}$ where $a = t + ix$ and $b = (y + zi)j$.
I just noticed that if $q = j$ for example, then $q \iff \begin{pmatrix} 0 & 1 \\ -1 & 0  \end{pmatrix}.$ But notice if I conjugate $e^{i\phi_1} j e^{-i\phi_2} = -e^{i(\phi_1 + \phi_2)}j$. Yet if I do the matrix multiplication, I get  \begin{pmatrix} 0 & e^{i(\phi_1 - \phi_2)} \\ -e^{i(\phi_2 - \phi_1)} & 0 .\end{pmatrix}
How come the non-abelian property is lost in this isomorphism of quaternions?
 A: This should be a comment, but it’s too long.
First, you’re not doing a “conjugation”. Second, after the word “conjugate”, I get
$$
e^{i\phi_1}je^{-i\phi_2}=e^{i(\phi_1+\phi_2)}j\,.
$$
Third, I don’t believe you did your matrix multiplication correctly. To make make things easy for myself, I’m going to write $\cos\phi_1=c_1$, $\sin\phi_1=s_1$, similarly for $\phi_2$, as well as $\cos(\phi_1+\phi_2)=c$, $\sin(\phi_1+\phi_2)=s$. Then my matrix computation goes:
\begin{align}
&\pmatrix{c_1+is_1&0\\0&c_1-is_1}
\pmatrix{0&1\\-1&0}
\pmatrix{c_2-is_2&0\\0&c_2+is_2}\\
&=\pmatrix{c_1+is_1&0\\0&c_1-is_1}
\pmatrix{0&c_2+is_2\\-c_2+is_2&0}\\
&=\pmatrix{0&c_1c_2-s_1s_2+i(c_1s_2+c_2s_1)\\-c_1c_2+s_1s_2+i(c_1s_2+c_2s_1)&0}\\
&=\pmatrix{0&c+is\\-c+is&0}=\pmatrix{c+is&0\\0&c-is}j\,,
\end{align}
which is exactly what the other computation gave.
Fourth, I don’t see how your computation, if done correctly, would have said anything in particular about commutativity. If you had had $\phi_1=\phi_2$, then the computation would have been a conjugation, and you would have found that $e^{i\phi}$ conjugates $j$ into $e^{2i\phi}j$, not particularly surprising.
