Finite Subgroups of the Mobius Group

Every finite group of Mobius transformations is isomorphic (i.e., 
conjugate) to a group of rotations of the extended complex plane.  
For any positive integer g let n(g) denote the number of conjugacy 
classes of Mobius transformations each member of which generates a 
cyclic group of order g. For what values of g is n(g)=1 (mod 6)?

It turns out that n(g)=1 (mod 6) if and only if g = p^(2d) or 2p^(2d), 
where d is any positive integer and p is any prime of the form 12k+11.  
This is closely related to the roots of polynomials with coefficients 
taken from a diagonal of Pascal's triangle.  For example, the diagonal

                 1
                 1  1
                 1  2  1
                 1  3  3  1
                 1  4  6  4  1
                 1  5 10 10  5  1/
                 1  6 15 20 15/
                 1  7 21 35/
                 1  8 28/
                 1  9/
                 1/

gives the quintic polynomial

          x^5 + 9x^4 + 28x^3 + 35x^2 + 15x + 1  =  0

which has the roots
                          / k pi \
          x  =  -4 cos^2 |  ----  |      k=1,2,3,4,5
                          \  11  /

It can be shown that a Mobius transformation (az+b)/(cz+d) generates 
a cyclic group of order 11 iff (a+d)^2/(ad-bc) equals one of these 5 
roots.  This also defines the 5 conjugacy classes of transformation 
groups of order 11.  In general, the number of conjugacy classes of 
order g is phi(g)/2, where phi is the Euler totient function.  See 
the note Congruences Involving the Totient Function for a discussion 
of why this implies g = p^(2d) or 2p^(2d) if n(g)=1 (mod 6).

Also, for more on this subject see Linear Fractional Transformations.