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Let $\Gamma_1=\Gamma_1(G_1,S_1)$ (respectively $\Gamma_2=\Gamma_2(G_2,S_2)$) be a Cayley graph for the group $G_1$ and the finite generating set $S_1$ (respectively $G_2$ and the finite generating set $S_2$).

  • If we assume $\Gamma_1,\Gamma_2$ are isospectral, what restrictions does this place on the underlying groups $G_1$ and $G_2$?

  • Conversely, given two groups $G_1$ and $G_2$ is there a choice of $S_1$ and $S_2$ such that $\Gamma_1$ and $\Gamma_2$ are isospectral? If the answer is in general no, are there known conditions on $G_1$ and $G_2$ (other than $G_1=G_2$) that guarantee an affirmative answer?

(As noted in the comments below, the converse has an obvious answer if the groups are finite, so for these questions we can assume that $G_1$ and $G_2$ are infinite, while keeping the restriction that $S_1$ and $S_2$ are finite).

Finally,

  • What can be said about the expansion properties of $\Gamma_1$ and $\Gamma_2$ if they are isospectral?
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  • $\begingroup$ I apologise in advance for this trivial comment: the complete graph is a Cayley graph for every group of the appropriate cardinality. Hence, even knowning that two groups have isomorphic Cayley graphs doesn't tell you much about the groups. You'll probably need to make your question more precise. $\endgroup$
    – verret
    Sep 25, 2014 at 3:59
  • $\begingroup$ @verret Indeed, the Cayley graph of a finite group is completely uninteresting - the OP can correct me if I'm wrong, but I think it is implicit that the groups in question are infinite. Indeed, there are quite a few results about the spectral theory of Cayley graphs of infinite discrete groups. $\endgroup$ Sep 25, 2014 at 4:46
  • $\begingroup$ @PaulSiegel I have added some details about the question. Can you kindly explain why the Cayley graph of a finite group is uninteresting? And what is known about spectral theory of infinite discrete groups? $\endgroup$
    – user6818
    Sep 25, 2014 at 4:54
  • $\begingroup$ @verret I guess you are referring to the fact that non-isomorphic groups can have isomorphic Cayley graphs - right? Any examples? $\endgroup$
    – user6818
    Sep 25, 2014 at 5:12
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    $\begingroup$ @verret Let's be clear: the "Cayley graph of a group" is not well defined: you must consider the Cayley graph of a group together with a (finite) generating set. As your comment pointed out, you can always choose the generating set of a finite group to be the whole group in which case you get the complete graph, and hence there are no nontrivial invariants of "the Cayley graph" of a finite group. You can correct me if I'm missing something, but I don't see how this is inconsistent with what you said (let alone inflammatory!) $\endgroup$ Sep 25, 2014 at 12:04

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