counting problem using fixed points of a group action The question states i have square which is made up of 8 congruent triangular tiles.
[imagine a square with the midpoints of each side joined up]
each tile can be coloured white or black.
I have to use the counting theorem to determine how many different such squares can be made,if 2 squares are regarded as the same when a rotation or reflection takes one to the other.
To do this i need to find the number of fixed points for each symmetry of the square.
for the identity this is just $$2^8$$
But for the rotations and the reflections in the horizontal and vertical i cant seem to convince myself the fixed points have the same number. I  am saying that for any of these symmetries i have freedom to colour 2 of the 4 quadrants [if we just look at the square with horizontal and vertical lines]
This gives me a possible 4 triangles to colour so in total $$2^4$$ fixed points for each of the 5 symmetries.
for the diagonal reflections i think have can colour 3 quadrants so 6 triangles which gives me $$2^6$$ for both the diagonal reflections. 
So my final answer is, by counting theorem is
$$ (2^8+5.2^4+2.2^6)/8 $$
$$= 58$$
Is this close?
Thanks
 A: There is quite a number of these at
MSE Meta
written by  different users using  several equivalent variants  of the
standard notation.

Basically you need to compute the  cycle index $Z(G)$ of the action on
the eight  tiles.  You can continue  with Polya or  Burnside once this
has been computed. We now compute $Z(G).$
There is the identity, which contributes $$a_1^8.$$
There  are two  rotations  by  $90$ degrees  and  $270$ degrees  which
contribute $$2a_4^2.$$
The rotation by $180$ degrees contributes $$a_2^4.$$
The horizontal and the vertical reflections about the two axes passing
through the center contribute $$2a_2^4.$$
The two reflections about the  two diagonals passing though the center
fix  the four tiles  on the  diagonal for  a contribution  of $$2a_1^4
a_2^2.$$
This yields for the cycle index
$$Z(G) = \frac{1}{8}
\left(a_1^8 + 2a_4^2 + 3a_2^4 + 2a_1^4a_2^2\right).$$
Using PET the answer is given by
$$\left.Z(G)(B+W)\right|_{B=1, W=1}
= \frac{1}{8}
(2^8 + 2\times 2^2 + 3\times 2^4 + 2\times 2^4 2^2)
= 55.$$
Remark.  This  is  the  subsituted  cycle  index  (which  was  not
essential to finding an answer to this problem):
$$Z(G)(B+W) =
1/8\, \left( B+W \right) ^{8}+1/4\, \left( B+W \right) ^{4}
 \left( {B}^{2}+{W}^{2} \right) ^{2}\\+3/8\, \left( {B}^{2}+{W}^{2
} \right) ^{4}+1/4\, \left( {B}^{4}+{W}^{4} \right) ^{2}.$$
It  has the  advantage  that  it produces  the  classification of  the
colorings according to the numbers of colors present which is
$${B}^{8}+2\,{B}^{7}W+7\,{B}^{6}{W}^{2}+10\,{B}^{5}{W}^{3}+15\,{B}
^{4}{W}^{4}\\+10\,{B}^{3}{W}^{5}+7\,{B}^{2}{W}^{6}+2\,B{W}^{7}+{W}
^{8}.$$
E.g. the  coefficient on $B^7  W$ is two  because with just  one white
tile there  are only two possibilities,  either it is  incident on the
center or it isn't.

If Burnside rather than Polya is  asked for then we can simply observe
that we  have two  choices of picking  a color  that is constant  on a
cycle and continue as before.
