Quaternions & the Mandelbrot set

The Mandelbrot set may be generalized by quaternion numbers
    Q_n+1 = Q_n² + Qc,   Q₀ = 0,  Qc = (c_r , c_i , c_j , c_k ).
As since (a + b i + c j + d k)² = a² - (b² + c² + d²) + 2ab i + 2ac j + 2ad k
due to terms like bc i j and cb j i cancel each other (bc i j = -cb j i). We get
    a = a² - (b² + c² + d²) + c_r
    b = 2ab + c_i
    c = 2ac + c_j
    d = 2ad + c_k
    a² + b² + c² + d² < BailOut²
The Quaternions Mandelbrot set is a 4 dimensional object. For Qc = (a, b, 0, 0 ) all points belongs to 2D (a, b ) plane and the set coincides with the classical M-set. For Qc =(a, b, c, 0 ) we get the 3D M-set.
We can rewrite the last equations as
    a = a² - |r|² + c_r
    r = 2ar + c
    a² + |r|² < BailOut²
where r = (b, c, d) and c = (c_i , c_j , c_k) are 3 dimensional vectors. By rotations of coordinate system any 3D vector c may be put into (|c|, 0, 0) or Qc = (a, |c|, 0, 0) - the classical Mandelbrot plane. Rotations conserve |r|. As since dynamic of the a - component and exiting condition depend on |r| only, therefore the QM-set is spherical symmetric in (b, c, d) subspace.

Due to the spherical symmetry we can get (a, b, c) 3D QM-set rotating the classical M-set relatively the real axis. The 3D QMandelbrot cactus consists of spheres and toruses so its cross-sections are circles and ellipses. Here you see the QM-set cross sections by (Re, Im, c, 0 ) planes for different c.
A few more questions:
Ouaternions generators (1, i, j, k) group has tight connection with the group of 3D space rotations. Is the QM-set symmetry consequence of this connection? Will we get similar symmetry for other generators groups? E.g. hyperbolic rotation symmetry for Dirac's (1, g⁰, g¹, g², g³) matrixes for which
    gⁱ g^j = - g^j gⁱ,    (g⁰)² = 1,    (g¹)² = (g²)² = (g³)² = -1

Sorry that quaternions are a bit "alien" here. Further we will return to the complex quadratic mappings.