Color Vertices of a Regular Pentagon with Two Colors

Question

I'm trying to figure out how many distinct ways there are to color the five vertices of a pentagon two different colors. I know this requires the use of Burnside's theorem, but am struggling a bit with the implementation. Here's what I have so far.

I believe that the logic of the theorem is that, because we can rotate the figure to produce a configuration of colorings that are effectively identical, we want to sort out how many configurations of colorings are fixed for all of the rotations and reflections of the figure and then take the average, i.e., normalizing by the number of such rotations and reflections.

In light of this, and what I believe to be the symmetries of the pentagon, I believe we have:

(1) Rotation that changes nothing (i.e, the identity): this fixes everything, i.e., all possible configurations. This makes sense when trying to select which vertices to color and which to leave alone, but with two colors, I'm a bit lost here as to what the total would be.

(2), (3), (4), and (5) are simply rotations through $72$ degrees each. Then, what I'll label (6), (7), (8), (9), and (10) are reflections through the five corners of the pentagon.

In other words, I think I understand the general process of how to go about the problem, but framing this in terms of $k$ colorings is throwing me off a bit.

Any helpful comments or insights would be greatly appreciated.

Answer 1

This probably isn't how you need to format your own solution, but I think it provides a useful insight and is too long for a comment.

If I am correct in my interpretation of "distinct" then the following treatment should give the correct answer ...

Let each vertex be coloured either red or yellow

Let $N(r)$be the number of distinct colourings having $r$ red vertices and $5-r$ yellow vertices.

Clearly $N(r)=N(5-r)$ $$N(0)=1\\N(1)=1\\N(2)=2$$ could have both reds together or separated with one yellow on one side and two yellows on the other.

$$ N_{tot}=\sum_{r=0}^5N(r)=2\sum_{r=0}^2N(r)=8 $$

Answer 2

This is similar to: http://mathworld.wolfram.com/Necklace.html The above link gives the general formula as $$N(n, a)=\frac 1n\sum_{i=1}^{v(n)}{\phi(d_i)a^{n/d_i}}$$ Where $n$ is the number of beads (in this case vertices) and $a$ is the number of colors. $v(n)$ is the number of divisors, $d_i$, of $n$ and $\phi$ is the totient function. In the formula rotations are considered indistinct but reflections are not.