Is it possible to know the corresponding graph labeling after using "relabel()"?
I am trying to generate the following cayley graph
G=AlternatingGroup(5)
S=[(1,2,3),(1,2,4),(1,2,5)]
C=G.cayley_graph(generators=S, simple=True)
U=C.to_undirected()
U.vertices()
[(), (3,4,5), (3,5,4), (2,3)(4,5), (2,3,4), (2,3,5), (2,4,3), (2,4,5), (2,4)(3,5), (2,5,3), (2,5,4), (2,5)(3,4), (1,2)(4,5), (1,2)(3,4), (1,2)(3,5), (1,2,3), (1,2,3,4,5), (1,2,3,5,4), (1,2,4,5,3), (1,2,4), (1,2,4,3,5), (1,2,5,4,3), (1,2,5), (1,2,5,3,4), (1,3,2), (1,3,4,5,2), (1,3,5,4,2), (1,3)(4,5), (1,3,4), (1,3,5), (1,3)(2,4), (1,3,2,4,5), (1,3,5,2,4), (1,3)(2,5), (1,3,2,5,4), (1,3,4,2,5), (1,4,5,3,2), (1,4,2), (1,4,3,5,2), (1,4,3), (1,4,5), (1,4)(3,5), (1,4,5,2,3), (1,4)(2,3), (1,4,2,3,5), (1,4,2,5,3), (1,4,3,2,5), (1,4)(2,5), (1,5,4,3,2), (1,5,2), (1,5,3,4,2), (1,5,3), (1,5,4), (1,5)(3,4), (1,5,4,2,3), (1,5)(2,3), (1,5,2,3,4), (1,5,2,4,3), (1,5,3,2,4), (1,5)(2,4)]
Here, I used the "relabel()" function and I got the following vertices
U.relabel()
V= U.vertices()
V
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59]
Is it possible to know the corresponding labels? Like for example what is the corresponding permutation for 0?