I tried to solve the following exercise, please could somebody tell me if I did it right?:
Prove that non-real elements $x,y \in \mathbb H$ commute if and only if their imaginary parts are parallel, that is, $\Im(x) = \lambda \Im y$ for some $\lambda \in \mathbb R$.
For this I would like to use the following (without proof): Two vectors $x,y \in \mathbb R^3$ are parallel if and only if $x \times y = 0$. Also, it is easy to see that $\Im (xy) = x \times y$ and $x \cdot y = -\Re(xy)$.
Now first let $x,y \in \mathbb H$ be two purely imaginary quaternions. Therefore they equal their imaginary parts: $x = \Im x, y = \Im y$.
Assume the imaginary parts of $x$ and $y$ are parallel. Then
$$ xy = \Im (xy) + \Re(xy) = \Re(xy) = - x \cdot y = - y \cdot x = \Re(yx) = yx$$
Next assume that $xy = yx$. Then $$x \times y = \Im (xy) = \Im (yx) = y \times x = - x \times y$$
hence $x \times y = 0$ and therefore $x,y$ are parallel.
Since the claim holds for purely imaginary quaternions and since the real component of a quaterion commutes with all the other parts the claim follows.
Yes, everything is correct.
Maybe you should also spell out for the last part that $a+u$ commutes with $b+v$ (with $a,b\in\Bbb R$ and $u,v$ purely imaginary), iff $u$ and $v$ commutes.