Orbit counting lemma hexagon

Question

Given its a hexagon with 60 degrees rotation i know rotational symmetry group C6 thus = {e, a, a^2, a^3, a^4, a^5} and there would be three reflections about the central axis {e, b, b^2}. What i don't know is how to apply the diagonal lines into the mix? I'm guessing |Fix(a)|= 3 given only choice of three colours and all the sides being the same colour along with |Fix(a^-1)|=3, |Fix(e)|=19683 given 9 sides and 3 colour thus 9^3=729. Could someone help me finish answering the question properly?

Answer 1

Label the lines requiring coloring: $s_0, s_1, s_2, s_3, s_4, s_5$ are the sides, in order. $d_i$ is the diagonal from $s_i$ to $s_{i+3}$ for $i = 0, 1, 2$.

Now examine the orbits of each of the elements of $C$:

• $e$ carries everything to itself, so each line is its own orbit.
• $a$ has orbits $(s_0s_1s_2s_3s_4s_5)$ and $(d_0d_1d_2)$.
• $a^2$ has orbits $(s_0s_2s_4), (s_1s_3s_5), (d_0d_2d_1)$.
• $a^3$ has orbits $(s_0s_3), (s_1s_4), (s_2s_5), (d_0), (d_1), (d_2)$.
• $a^4$ has orbits $(s_0s_4s_2), (s_1s_5s_3),(d_0d_1d_2)$.
• $a_5$ has orbits $(s_0s_5s_4s_3s_2s_1), (d_0d_2d_1)$.

Elements in the same orbit have to be colored the same. But colors for different orbits can be picked freely. Do you see how to do the rest?