How many ways can one paint the edges of a Petersen graph black or white?

I know that the symmetrygroup of the Petersen graph is $ [S5][1]$. Furthermore this this seems like a case where I should use Burnside's lemma. I'm sorry if the following is too verbose or uses non standard notation; I haven't been acquainted with graph theory.

S5 has 7 conjugacy classes, namely those with cycle types: (1,1,1,1,1),(1,1,1,2),(2,2,1),(2,3),(1,1,1,3),(4,1),(5). S5 has 15 edges so the identity (1,1,1,1,1) would leave $2^{15}$ different colorings fixed. The n-cycle (5) is a rotation of the whole graph and as such would leave $2^3$ colorings fixed. The "outside" could be white or black, the connecting edges and the "inside" edges could both be either white or black. Rotation around one "connecting" edge involves the (2,2,1) cycles. I won't tire you with the details but I found $2^9$ colorings.

From here I'm stuck however, I can't find any more symmetries than these. How do I find the colorings left fixed by the other conjugacy classes?


Addendum Mar 24 2016. A much improved solution to this problem is at the following MSE link which renders this thread obsolete.

  • $\begingroup$ Wow! Not exactly what I was looking for but very nicely done! $\endgroup$ – Lee Wang Jan 21 '13 at 20:55
  • $\begingroup$ Here is another interesting computation using cycle indices. $\endgroup$ – Marko Riedel Nov 10 '13 at 3:01

Using Burnside's lemma is the right idea.

Each element of $S_5$ determines a permutation of the 15 edges of the Petersen graph. If this permutation has exactly $r$ cycles on edges, then it fixes exactly $2^r$ 2-colorings of the edge set. If $a$ and $b$ are conjugate elements of $S_5$, then the permutations of the edges they determine have the same cycle structure. (To proof this, observe that the induced permutations are conjugate in $S_{15}$, and hence have the same cycle structure.)

So you just have to compute $r$ for one element in each conjugacy class, which is mildly tedious at worst, and then apply Burnside.

  • $\begingroup$ I'm not entirely sure if I follow your argument. How would one compute $r$ without a picture? My group theory book only deals with very simple visualizable cases. $\endgroup$ – Lee Wang Jan 20 '13 at 20:39
  • $\begingroup$ An automorphism of Petersen is a permutation of its 10 vertices, and hence is an element of $S_{10}$. The set of all automorphisms forms a permutation group isomorphic to $S_5$, but not equal to it. You are dealing with a case where you cannot readily associate each automorphism with a picture, it is necessary to view each automorphism as a permutation. $\endgroup$ – Chris Godsil Jan 20 '13 at 21:52
  • $\begingroup$ Ah that makes sense. But like I said my book only mentions cases where you can visualize the problem. I'm lost on how you would construct a nongeometric way to compte the $r$ for each conjugacy class. $\endgroup$ – Lee Wang Jan 21 '13 at 20:55
  • 1
    $\begingroup$ If $V(P)$ consists of pairs from $S=\{0,1,2,3,4\}$, then the edges correspond to partitions of $S$ with one cell of size 1 and two of size 2, e.g., $\{\{0\},\{1,2\},\{3,4\}\}$. Now the permutation $(01)(234)$ from $S_5$ maps this edge to $\{\{1\},\{0,3\},\{4,2\}\}$ and then to $\{\{0\},\{1,4\},\{2,3\}\}$ and so on. If we continue, we get a cycle of 6 edges. Choose an edge not in this cycle and repeat; eventually we arrange the 15 edges into cycles, and the number of these cycles is the value of $r$. Repeat for one permutation in each conjugacy class. $\endgroup$ – Chris Godsil Jan 21 '13 at 22:15
  • $\begingroup$ Thank you very very much! This was truly helpful. $\endgroup$ – Lee Wang Jan 22 '13 at 9:43

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