Non-real quaternion generates subalgebra isomorphic to $\mathbb{C}$

I am trying to understand the quaternion multiplication geometrically.

Following this answer, on What is a good geometric interpretation of quaternion multiplication?, I am trying to see how:

1. Every non-real quaternion generates a subalgebra isomorphic to $\mathbb{C}$
2. How $\mathbb{H}$ splits as this subalgebra, and it's orthogonal compliment

So far, I have been unable to construct an isomorphism needed by (1); I have been hoping that this isomorphism was a simple multiplication by an element in $\mathbb{H}$, that depends on the generating-element.

• @LordSharktheUnknown Perhaps I misunderstood the original answer. By non-real (or unreal), I thought this meant it looked like $ai + bj +ck + r$ where not all $(a, b, c) \neq (0, 0, 0)$ (i.e: it was not synonymous with "pure" quaternion). In this case of unit quaternions, this would mean it is not 1. – Abel May 26 '18 at 18:59

The quaternions form an algebra over $\mathbb{R}$. If $q$ is a nonreal quaternion, then the subalgebra it generates is $\mathbb{R}[q]$, which is commutative. Now notice that $q+q^*$ and $qq^*=q^*q$ are both real and that $q$ is a root of the polynomial $x^2-(q+q^*)x+qq^*$.
Therefore $\mathbb{R}[q]$ is a $2$-dimensional algebra over $\mathbb{R}$. Since it has no zero divisors, it is a field and thus it it isomorphic to $\mathbb{C}$.
Alternatively, writing $q$ in polar form as $r\exp(\theta\mathbf{u})=r\cos(\theta)+r\sin(\theta)\mathbf{u}$, in the $\mathbb{R}$-algebra generated by $q$, we can obtain $\mathbf{u}$ from $q$ and vice-versa, so $\mathbb{R}[q]=\mathbb{R}[\mathbf{u}]$. Since $\mathbf{u}$ is a square root of negative one, it functions just as $i$ in $\mathbb{C}$ does, so there is an isomorphism $\mathbb{R}[\mathbf{u}]\cong\mathbb{C}$ given by sending $a+b\mathbf{u}\mapsto a+bi$.