Using Burnside's lemma for a triangle Find the number of distinguishable ways the edges of an equilateral triangle can be
painted if four different colors of paint are available, assuming only one color is used on each edge, and the same color may be used on different edges. I'm not sure how to apply Burnside's lemma to this question.
Any help is greatly appreciated.
 A: Let $G=D_3$ (the dihedral group), and let $X$ be the set of all possible ways to paint the triangle. We define an action of $G$ on $X$ like this: the painting $g.x$ is gained by painting the triangle with painting $x$ and then using the element $g$ on the triangle. It is easy to check that it is indeed an action of a group on a set. Also note that $|X|=4^3$ because in general we have $4$ options to choose the color for each edge. But we are interested only in the number of paintings up to symmetry, which in the terms of group theory is the number of orbits in the action we defined. And here we can use Burnside's lemma. 
For each $g\in G$ we define $X^g=\{x\in X: g.x=x\}$, and Burnside's lemma tells us that the number of orbits is exactly $\frac{1}{|G|}\sum_{g\in G} |X^g|$. Obviously $|G|=|D_3|=6$, so now we only need to find $|X^g|$ for each $g\in G$. So let's do that.
For the identity $e$ it is obvious that $e.x=x$ for all $x\in X$. This must be the case even by the definition of an action. Hence $|X^e|=|X|=4^3$. 
Now let $g$ be the rotation by angle $\frac{2\pi}{3}$ counterclockwise. We have to understand what is $X^g$ in that case. Let's choose an order for the edges and assume that for a painting $x\in X$ the edges are painted in colors $a,b,c$. After the rotation the edges move $a\to b\to c\to a$. So now we get a painting where the edges are painted in colors $c,a,b$. In which case will that painting be the same as painting $x$? Only when $a=b=c$, i.e when $x$ is a painting where all edges have the same color. There are only $4$ such paintings, so $|X^g|=4$. In the same way you can show that $|X^{g^2}|=4$.
Now let's move to reflections. Let $h$ be any reflection, doesn't matter which. Any reflection on the triangle simply switches the place of two edges. So let $x$ be a painting. We will have $x=h.x$ only when the two edges that switched places were painted in the same color. How many such paintings $x$ there are? We have $4$ options to choose the color of the edge which stays in its place and $4$ options to choose the common color for the two edges that change places. So there are $16$ such paintings. Hence $|X^h|=4^2$. And this is true for each of the three reflections.
Now by Burnside's lemma the number of orbits is $\frac{4^3+4+4+4^2+4^2+4^2}{6}=20$. 
