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A symmetry of a graph X is a permutation of the vertices that also happens to be a permutation of the induced edges. In particular, the distances between vertices are preserved by a symmetry. Show that the set of symmetries of X is a permutation group of V(X). Compute the cycle index of the group for the Petersen graph.

Please help. How do you show that something is a permutation group? I have never calculated the cycle index for a graph before and we don't have any examples like this in our notes. How do I do it for a graph?


marked as duplicate by Mike Pierce, Leucippus, user91500, Jack's wasted life, Eric Wofsey Mar 25 '16 at 6:10

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  • $\begingroup$ To show that you have a permutation group, you essentially only have to verify the usual group axioms (existence of identity, existence of inverse, associativity of composition and closure under composition). Then by Cayley's theorem the symmetry group is isomorphic to a permutation group (you just label the vertices appropriately, and express your symmetries as permutations switching up the nodes). As per the cycle index, once you have the permutation group in front of you, you compute the cycle index of it, as you would normally do. Does the task feel any clearer now? $\endgroup$ – A.Sh Mar 24 '16 at 23:40