Editing Cayley graph (section)

=== Cayley integral simple group ===

A group <math>G</math> is Cayley integral simple (CIS) if the connected Cayley graph <math>\Gamma(G,S)</math> is integral exactly when the symmetric generating set <math>S</math> is the complement of a subgroup of <math>G</math>. A result of Ahmady, Bell, and Mohar shows that all CIS groups are isomorphic to <math>\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p^2\mathbb{Z}</math>, or <math>\mathbb{Z}_2 \times \mathbb{Z}_2</math> for primes <math>p</math>.<ref name="CIS">{{cite journal|last1=Ahmady|first1=Azhvan |last2=Bell|first2=Jason|last3=Mohar|first3=Bojan|authorlink3=Bojan Mohar|title=Integral Cayley graphs and groups|journal=[[SIAM Journal on Discrete Mathematics]]|volume=28|issue=2|pages=685–701|year=2014|arxiv=1307.6155 |doi=10.1137/130925487 |s2cid=207067134}}</ref> It is important that <math>S</math> actually generates the entire group <math>G</math> in order for the Cayley graph to be connected. (If <math>S</math> does not generate <math>G</math>, the Cayley graph may still be integral, but the complement of <math>S</math> is not necessarily a subgroup.)

In the example of <math>G=\mathbb{Z}/5\mathbb{Z}</math>, the symmetric generating sets (up to graph isomorphism) are
*<math>S = \{1,4\}</math>: <math>\Gamma(G,S)</math> is a <math>5</math>-cycle with eigenvalues <math>2, \tfrac{\sqrt{5}-1}{2},\tfrac{\sqrt{5}-1}{2},\tfrac{-\sqrt{5}-1}{2},\tfrac{-\sqrt{5}-1}{2}</math>
*<math>S = \{1,2,3,4\}</math>: <math>\Gamma(G,S)</math> is <math>K_5</math> with eigenvalues <math>4, -1,-1,-1,-1</math>
The only subgroups of <math>\mathbb{Z}/5\mathbb{Z}</math> are the whole group and the trivial group, and the only symmetric generating set <math>S</math> that produces an integral graph is the complement of the trivial group. Therefore <math>\mathbb{Z}/5\mathbb{Z}</math> must be a CIS group.

The proof of the complete CIS classification uses the fact that every subgroup and homomorphic image of a CIS group is also a CIS group.<ref name="CIS" />