Please Register or Log in to view the hidden image! Is there another way to show the Petersen graph doesn't contain a Hamilton cycle other than showing every 3-regular graph of order 10 which is Hamiltonian, has smaller cycles in it than the Petersen?
If a graph has a Hamilton path, that means it must have a completely connected subgraph where every vertex is order 2. Every vertex is order 3 to start, so to get such a subgraph, you would need to pick 5 pairs of connected vertices and sever the edges between them. To maintain complete connectedness, given a partition of the graph into two groups of five, all pairs cannot be made by grabbing one point from each half of the partition. But given how the Petersen graph is defined by linking disjoint subsets, any selection of 5 pairs to disconnect will break the graph into two unconnected subsets. I'm sure a math major could make this more formal. Please Register or Log in to view the hidden image!
