Sufficient Condition for Quaternion Multiplication to Commute

Theorem
In general, quaternion multiplication does not commute.

But, for $\mathbf x,\mathbf y \in \H$, $\mathbf x \times \mathbf y = \mathbf y \times \mathbf x$ if any one of the following conditions hold:

Proof of $\paren 1$
It follows directly from Complex Numbers form Subfield of Quaternions and Complex Multiplication is Commutative.

Proof of $\paren 2$
Let $\mathbf x \in \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k: a\in \R}$.

Let $\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k: e, f, g, h \in \R$.

Then:

The above is the proof of $\paren {2a}$, and the proof of $\paren {2b}$ is similar.