linguistic:translation, structural:reformatting
Normal and Eulerian generating sets

Given a general group G {\displaystyle G}, a subset S ⊆ G {\displaystyle S\subseteq G} is normal if S {\displaystyle S} is closed under conjugation by elements of G {\displaystyle G} (generalizing the notion of a normal subgroup), and S {\displaystyle S} is Eulerian if for every s ∈ S {\displaystyle s\in S}, the set of elements generating the cyclic group ⟨ s ⟩ {\displaystyle \langle s\rangle } is also contained in S {\displaystyle S}.A 2019 result by Guo, Lytkina, Mazurov, and Revin proves that the Cayley graph Γ (G, S) {\displaystyle \Gamma (G,S)} is integral for any Eulerian normal subset S ⊆ G {\displaystyle S\subseteq G}, using purely representation theoretic techniques.

The proof of this result is relatively short: given S {\displaystyle S} an Eulerian normal subset, select x 1, …, x t ∈ G {\displaystyle x_{1},\dots,x_{t}\in G} pairwise nonconjugate so that S {\displaystyle S} is the union of the conjugacy classes Cl (x i) {\displaystyle \operatorname {Cl} (x_{i})}. Then using the characterization of the spectrum of a Cayley graph, one can show the eigenvalues of Γ (G, S) {\displaystyle \Gamma (G,S)} are given by { λ χ = ∑ i = 1 t χ (x i) | Cl (x i) | χ (1) } {\textstyle \left\{\lambda _{\chi }=\sum _{i=1}^{t}{\frac {\chi (x_{i})\left|\operatorname {Cl} (x_{i})\right|}{\chi (1)}}\right\}} taken over irreducible characters χ {\displaystyle \chi } of G {\displaystyle G}. Each eigenvalue λ χ {\displaystyle \lambda _{\chi }} in this set must be an element of Q (ζ) {\displaystyle \mathbb {Q} (\zeta)} for ζ {\displaystyle \zeta } a primitive m t h {\displaystyle m^{th}} root of unity (where m {\displaystyle m} must be divisible by the orders of each x i {\displaystyle x_{i}}). Because the eigenvalues are algebraic integers, to show they are integral it suffices to show that they are rational, and it suffices to show λ χ {\displaystyle \lambda _{\chi }} is fixed under any automorphism σ {\displaystyle \sigma } of Q (ζ) {\displaystyle \mathbb {Q} (\zeta)}. There must be some k {\displaystyle k} relatively prime to m {\displaystyle m} such that σ (χ (x i)) = χ (x i k) {\displaystyle \sigma (\chi (x_{i}))=\chi (x_{i}^{k})} for all i {\displaystyle i}, and because S {\displaystyle S} is both Eulerian and normal, σ (χ (x i)) = χ (x j) {\displaystyle \sigma (\chi (x_{i}))=\chi (x_{j})} for some j {\displaystyle j}. Sending x ↦ x k {\displaystyle x\mapsto x^{k}} bijects conjugacy classes, so Cl (x i) {\displaystyle \operatorname {Cl} (x_{i})} and Cl (x j) {\displaystyle \operatorname {Cl} (x_{j})} have the same size and σ {\displaystyle \sigma } merely permutes terms in the sum for λ χ {\displaystyle \lambda _{\chi }}. Therefore λ χ {\displaystyle \lambda _{\chi }} is fixed for all automorphisms of Q (ζ) {\displaystyle \mathbb {Q} (\zeta)}, so λ χ {\displaystyle \lambda _{\chi }} is rational and thus integral.

Consequently, if G = A n {\displaystyle G=A_{n}} is the alternating group and S {\displaystyle S} is a set of permutations given by { (12 i) ± 1 } {\displaystyle \{(12i)^{\pm 1}\}}, then the Cayley graph Γ (A n, S) {\displaystyle \Gamma (A_{n},S)} is integral. (This solved a previously open problem from the Kourovka Notebook.) In addition when G = S n {\displaystyle G=S_{n}} is the symmetric group and S {\displaystyle S} is either the set of all transpositions or the set of transpositions involving a particular element, the Cayley graph Γ (G, S) {\displaystyle \Gamma (G,S)} is also integral.