Every element with finite conjugates in the ring of real quaternions is a real number

Question

Let $H$ be the ring of real quaternions and let $x$ be a member of $H$ having finite conjugates. Prove that $x$ is a real number.

I worked a lot on this question, but no progress! :|

Answer 1

Real quaternions are in the center of quaternion ring, and they are not affected by conjugations. Therefore it's sufficiently to concern oneself with purely imaginary quaternions. Indeed, for conjugation of a quaternion with real part $a$ and imaginary part $b$ by $q\in H$ you have $q(a+b)q^{-1}=a+qbq^{-1}$, so the number of conjugates does not depend on $a$.

Intuitively, purely imaginary quaternions represent 3-D rotations, and they are all conjugate to each other if and only if they have the same magnitude. Also conjugation is a transitive property, so it's enough to show that every purely imaginary quaternion of absolute value 1 is conjugate to $i$.

So take a quaternion $ai+bj+ck$ with $a^2+b^2+c^2=1$ and try to find a purely imaginary quaternion $xi+yj+zk$ that would transform $i$ into it:

$(xi+yj+zk)i=(ai+bj+ck)(xi+yj+zk)$

The rest is trivial: simplify and find x, y, and z.