#19586: Add is_cayley_graph
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Reporter: jaanos | Owner:
Type: enhancement | Status: needs_info
Priority: major | Milestone: sage-7.1
Component: graph theory | Resolution:
Keywords: Cayley graphs | Merged in:
groups | Reviewers: Nathann Cohen
Authors: Janoš Vidali | Work issues:
Report Upstream: N/A | Commit:
Branch: | a58a7348bc022f39bf68383b70400e8b7f5b268b
u/jaanos/add_is_cayley_graph | Stopgaps:
Dependencies: |
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Comment (by dimpase):
Replying to [comment:78 jaanos]:
> Let's see... if the automorphism group has a regular subgroup `G`, then
we can pick any vertex `w` and construct a bijective mapping that maps a
group element `g` to the vertex `g^w`. Now, suppose we have an arc `uv`
such that group elements `g` and `h` map to `u` and `v`, respectively.
Then we may label the arc by `e = g^-1 h`, and any automorphism `a` from
`G` will map `uv = g^w h^w` to `(ag)^w (ah)^w`, i.e. an arc with the same
label. Since such a graph is necessarily vertex-transitive, it follows
that the (multi)set of labels on arcs starting in a vertex is independent
of the choice of said vertex. The Cayley graph of `G` with said (multi)set
as its generating set is then isomorphic to our graph.
> Do you find any flaw in the above argument?
it is certainly true that you can assume w=1. So you have the neighbours
of 1 in the graph (with its vertices labelled by the elements of G) say
g,,1,,,...,g,,k,,. Now you need an argument that g,,1,,,...,g,,k,,
generate G. I don't see it in what you wrote above.
> Also, can you provide the graphs which you claim to be counterexamples?
I'll try. (Or show me that they don't exist...)
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Ticket URL: <http://trac.sagemath.org/ticket/19586#comment:83>
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