TITLE: Short proof for the non-Hamiltonicity of the Petersen Graph
QUESTION [5 upvotes]: It is well known that the Petersen Graph is not Hamiltonian. I can show it by  case distinction, which is not too long - but it is not very elegant either.
Is there a simple (short) argument that the Petersen Graph does not contain a Hamiltonian cycle?

REPLY [4 votes]: Another easy argument is by using the edge chromatic number which is 4 for the Petersen graph.If it Hamiltonian, then removing the Hamiltonian cycle leaves a perfect matching. In 
this case 3 colors would be sufficient for edge coloring: 2 for the cycle and 1 for the 
matching.