# Complement of a Cayley graph

Let $X(G,S)$ be the (undirected) Cayley graph, with $G$ group and $S \subseteq G$ such that $1_G \notin S$ and $S=S^{-1}$. Is the complement of $X$ a Cayley graph?

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It seems to me that the ideal answer to this question has been given (but not accepted). I have voted to close the question as no longer relevant -- I don't think we need to see it popping back up again ad infinitum. –  Pete L. Clark May 9 '10 at 8:45

The complete graph on $|G|$ vertices is a Cayley graph for $S=G\setminus\{1_G\}$. Its complement, the graph without edges, is not a Cayley graph.
Isn't it the Cayley graph (under tbg's defintion) for $S=\emptyset$? –  Robin Chapman Apr 25 '10 at 7:00
@Robin Chapman: usually, $S$ is assumed to be a generating set. –  Benoît Kloeckner Apr 25 '10 at 8:13
Amongst people who actually work with the things, Cayley graphs are not required to be connected. For example, there is Sabidussi's theorem which asserts that a graph is a Cayley graph for the group $G$ if and only if $G$ acts regularly on its vertices. –  Chris Godsil Apr 25 '10 at 18:54