Find Orbits and Stabilizers

Question

The question is:

Let $S$ denote the set of possible black-or-white colorings of the edges of an equilateral triangle. The triangle's symmetry group $D_6$ acts naturally on $S$.

(a) How many orbits are there?

(b) By listing an element from each orbit show there are 10 orbits if three colors were used.

So I have sort of done the first part but my answers are contradicting when I list out the orbits vs when I used the formula

$$|D_6| = |\operatorname{Stab}(S)| \cdot|\operatorname{Orb}(S)|$$

And I am also stuck in the second part. Any help is appreciated 🙂

Answer 1

The formula you are using is incorrect. It is not used for the whole set, but for one element only $$\forall\ s\in S,\lvert D_6\rvert=\lvert\text{Stab}(s)\rvert\cdot\lvert\text{Orb}(s)\rvert$$

That's the problem you are having and also $\lvert\text{Orb}(s)\rvert$ is not the number of orbits, but the size of the orbit of $s$.

Knowing that the stabilizers must be subgroups and knowing that the non trivial subgroups of $D_6\cong S_3$ are $C_2,C_3$ we can count the orbits.

• $\text{Stab}(s)\cong D_6$, then all sides must have the same colour.
• $\text{Stab}(s)\cong C_3$, it is impossible, all sides must have the same colour, but that means that $\text{Stab}(s)\cong D_6$.
• $\text{Stab}(s)\cong C_2$, then two sides must have the same colour.
• $\text{Stab}(s)\cong\{e\}$, then all sides must have different colour.

Then counting all elements

Stabilizer	Size of orbit	2 colours	3 colours
$D_6$	$1$	$2$	$3$
$C_2$	$3$	$2\cdot1\cdot{3\choose2}=6$	$3\cdot2\cdot{3\choose2}=18$
$\{e\}$	$6$	$0$	$3\cdot2\cdot1=6$
Total		$2^3=8$	$3^3=27$

And counting all orbits, by dividing the number of elements by the size of the orbit

Stabilizer	2 colours	3 colours
$D_6$	$2$	$3$
$C_2$	$2$	$6$
$\{e\}$	$0$	$1$
Total	$4$	$10$

So for 3 colours you can list (using red, blue and green): RRR, BBB, GGG, RRB, RRG, BBR, BBG, GGR, GGB, RGB.