# In how many different ways can one color the vertices of a regular pentagon into four colors?

I am trying to find the number of ways to color a pentagon with 4 colors up to symmetries. I know that I should be using Burnside's Theorem, and so far I know that the group $D_5$ should act on the set $X=$ {$1, 2, 3, 4, 5$}, the vertices of the pentagon.

I know that $D_5$ is generated by $(12345)$ and $(14)(23)$. From here I am unsure how to proceed. The only other thing I can think of is that there should be a function $f:X \mapsto Y =$ {$Q_1$,$Q_2$,$Q_3$,$Q_4$} that sends vertices of the pentagon to the four colors.

I suppose there must be a group $X'$ of all possible colorings, that is all possible functions $f$?

Edit: Up to four colors.

• I examined tried to count the number of $G'-$ equivalence classes and came up with a tentative solution. $X'_{1} = X'$ because the identity fixes every possible coloring. $|X'|= 4^{5}$. I think all of the rotations should be the same: $|X'_{(12345)}| = 4$ because $X'_{(12345)}$ should consist of all colorings that do not change under the permutation $(12345)$? – Lucky For Me Dec 1 '14 at 6:44
• Then $X'_{(25)(43)}$ implies that $f(2) = f(5)$ and that $f(4) = f(3)$. So $|X'_{(25)(43)}| = 4^{3}$? From this I get $\frac{1}{10}(4^{5}+4*4+5*4^{3}) = 136$. Can anyone give some feedback on this solution? – Lucky For Me Dec 1 '14 at 6:54
• I think that is correct. You said you were unsure how to proceed, but then you did proceed and I believe you got the right answer! – MJD Dec 1 '14 at 7:15
• Better use Polya's enumeration theorem. en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem – arindam mitra Dec 1 '14 at 10:29