Take the tour ×  Does there exist a quaternion $q$ on the unit
sphere such that, given the vanilla complex plane $\mathbb{C}$,
$q\mathbb{C}q^{-1} = \bar{\mathbb{C}}$?  Motivation: ordinarily,
the plane is rotated by multiplying it by a complex number in
the unit circle. Taking the conjugate amounts to flipping the
plane along the real axis, which can be viewed as a rotation.
share|improve this question This looks similar to the definition
of a normal subgroup. –  PEV Mar 18 '11 at 20:16 add comment  2
Answers  up vote 2 down vote accepted  As a variation on
Joseph's answer:  Since $j \cdot 1 \cdot j^{-1} = 1$ while $j
\cdot i \cdot j^{-1} = -i,$ we see that $j z j^{-1} =
\overline{z}$ for any complex number $z$.  [This is a special
case of a more general theorem, by the way: if $A$ is a central
simple algebra over a field $K$, and $L$ is a field extension of
$K$ contained in $A$, then any field automorphism of $L$ over
$K$ can be realized via conjugation by an invertible element of
$A$. This particular case