Exercise 7, § 2

Let $G$ be a group operating on the right on a set $X$ and let $A$ be a subset of $G$. The Schreier quiver $\mathscr{G}_G(X,A)$ is the quiver $(S,F, o, t)$ defined by:

– The set of its vertices is S = X ;

– The set of its arrows is $F = X\times A$ ;

– The mappings $o$ and $t$ are given by $o(x, a) =x$ and $t(x, a) =x\cdot a$ for all $(x, a)\in X\times A$.

When X = G on which G operates by right multiplication, this quiver is denoted $\mathscr{G}(G,A)$ and is called the Cayley quiver of G with respect to A.

The Schreier graph (resp. Cayley graph) is called the graph associated with this quiver.

a) Prove that the quiver $\mathscr{G}_G(X,A)$ is connected if and only if the subgroup H of G generated by A has at most one orbit on X.

b) Suppose that A generates G. Now prove that the Cayley graph $\mathscr{G}(G,A)$ is connected and that it is bipartite if and only if there exists a group homomorphism $\varphi : G\rightarrow \mathbf{Z}/2\mathbf{Z}$ such that $\varphi (a) = 1$ for all $a\in A$.

c) Now prove that there exists a unique action of G on the Cayley graph $\mathscr{G}(G,A)$ such that the action of G induced on the vertices of this graph is the action of G on itself by left multiplication.