A proof of a conjecture of Sabidussi on graphs idempotent under the lexicographic product

Abstract

Given a graph G, with each vertex x of G is associated a graph Gx. The vertex sets V (Gx) are supposed to be pairwise disjoint. The lexicographic sum G[Gx;x∈V(G)] of the graphs Gx over the graph G is defined on the union ∪x∈V(G)V(Gx) as follows. For every u∈∪x∈V(G)V(Gx), the unique vertex x of G, such that u ∈ V (Gx), is denoted by x(u). For any u,v∈∪x∈V(G)V(Gx), uv is an edge of G[Gx; x ∈ V (G)] if either x(u)≠x(v) and x(u)x(v)∈E(G) or x(u)=x(v) and uv∈E(Gx(u)). If all the graphs Gx are isomorphic to a same graph H, then the lexicographic sum G[Gx; x ∈ V (G)] is called the lexicographic product of H by G and is denoted by G[H]. A graph G is then said to be idempotent under the lexicographic product if G[G] is isomorphic to G. Since a graph G with an unique vertex satisfies this property, it is added that an idempotent graph under the lexicographic product possesses at least two vertices, and hence is infinite. In another vein, given sets V and V′, consider permutation groups Γ on V and Γ′ on V′. The wreath product Γ≀Γ′ of Γ′ by Γ is the permutation group on V×V′ defined in the following way. Given a permutation ϕ of V×V′, ϕ∈Γ≀Γ′ if there exist an element g of Γ and a function ε from V into Γ′ such that for every (x,y)∈V×V′, (x,y)ϕ=(xg,y(xε)). It is easy to verify that Aut(G) ≀ Aut(H) is a subgroup of Aut(G[H]) for any graphs G and H. In 1960, Sadidussi conjectured the following. If G is a graph idempotent under the lexicographic product, then Aut(G) ≀ Aut(G) is a proper sugroup of Aut(G[G]). The answer is positive. For such a graph G, there exist a vertex x of G, a subset X of V(G) and an automorphism ϕ of G[G] such that ({x}×V(G))ϕ=X×V(G) and the subgraph of G induced by X is isomorphic to G. The argument is short and formal; it is essentially based on the associativity of the lexicographic product.