Group theory question

Lately I was thinking about some matrices, and I wasn't quite sure how this fits in with what I've learned so far about group theory.

Normally, when you want to rotate 2 dimensional space, you might start with a basis of 2 orthogonal vectors:

       Y
       ^
       |
       |
        -----> X

You can do a 90 degree rotation, and turn (X,Y) into (Y,-X). You can do this operation using a matrix acting on the vector (X,Y):


  ( X )     =>   ( 0  1 ) ( X )
  ( Y )          ( -1 0 ) ( Y )

This can be turned into a continuous group that has the above discrete group as a subgroup:

   ( cos(a)      sin(a)  )
   ( -sin(a)     cos(a)  )

Now I was looking at a space in which UP is not quite the same as
-(DOWN), and LEFT not quite the same as -(RIGHT). It doesn't matter at
this stage why they are different, we will just write them a formally
distinct terms. In a specific coordinate system, our basis looks like this:

                 UP
                 ^
                 |
                 |
                 |
    LEFT <----------------> RIGHT
                 |
                 |
                 |
                 V
               DOWN

A 90 degree rotation of this basis could be described by a matrix (A)
acting on a 4-component vector:


     (RIGHT')     ( 0 0 0 1 )    (RIGHT)
     (UP'   )  =  ( 1 0 0 0 )    (UP   )
     (LEFT' )     ( 0 1 0 0 )    (LEFT )
     (DOWN' )     ( 0 0 1 0 )    (DOWN )


It seems pretty straight forward to rotate this space by multiples of
90 degrees. But what about rotating by other angles? It turns out that
we there is a continuous group that has the above matrix as a discrete subgroup. The matrix form is:

         (a b c d)
         (d a b c)
         (c d a b)
         (b c d a)

a, b, c and d can be expressed as weighted sums of the eigenvalues, and the eigenvalues are:

C1 =    cos(1a) + i* sin(1a)
C2 =    cos(2a) + i* sin(2a)
C3 =    cos(3a) + i* sin(3a)
C4 =    cos(4a) + i* sin(4a)

In this case, our group seems isomorphic to the ordinary 2D rotation group. But in 3D, you could also take a discrete subgroup of the rotation group, say the octahedron. You could look at this subgroup as a discrete subgroup of the rotation group in 3D, but you could also write it as a matrix acting on the vertices. For an octagon with 6 vertices, this would give a group of 6X6 matrices acting on

     (RIGHT')
     (UP'   )
     (LEFT' )
     (DOWN' )
     (FRONT )
     (BACK  )

Again, we can turn this into a continuous group of 6X6 matrices.

My question is, is this just a representation of SO(3)? Or am I making another group here?

Gerard

.