Sufficient Condition for Quaternion Multiplication to Commute

Theorem
In general, quaternion multiplication does not commute.

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

Proof of $\left({1}\right)$
It follows directly from Complex Numbers form Subfield of Quaternions and Complex Multiplication is Commutative.

Proof of $\left({2}\right)$
Let $\mathbf x \in \left\{ {a\mathbf 1+0\mathbf i+0\mathbf j+0\mathbf k:a\in \R}\right\}$.

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 prove of $\left({2a}\right)$, and the prove of $\left({2b}\right)$ is similar.