Show that the normal covering spaces of \(S^1 \vee S^1\) are precisely the graphs that are Cayley graphs of groups with two generators. More generally, the normal covering spaces of the wedge sum of \(n\) circles are the Cayley graphs of groups with \(n\) generators.
Solution
Let \[ R_n=\bigvee_{i=1}^n S^1, \qquad F_n=\pi_1(R_n)=\langle x_1,\dots,x_n\rangle. \]
1. Let\[ p:\widetilde R\to R_n \]be a connected normal covering corresponding to a normal subgroup\[ H\triangleleft F_n. \]Then the vertex set of \(\widetilde R\) is naturally the quotient group\[ Q=F_n/H. \]
The fiber over the unique vertex of \(R_n\) is the set of right cosets \(H\backslash F_n\). Since \(H\) is normal, these cosets form the group \(F_n/H\). Connectedness makes every fiber point a vertex of the covering graph.
2. For each \(q\in Q\) and each generator \(x_i\), the lift of the \(i\)th circle gives an oriented edge\[ q\longrightarrow q\bar x_i, \]where \(\bar x_i\) is the image of \(x_i\) in \(Q\).
This is the standard coset action of \(F_n\) on the fiber by lifting loops. The endpoint of the lift of \(x_i\) beginning at the vertex \(q\) is the vertex represented by \(q\bar x_i\).
3. Therefore \(\widetilde R\) is exactly the Cayley graph\[ \operatorname{Cay}(Q;\bar x_1,\dots,\bar x_n). \]
The vertices and labelled oriented edges in <1>1–<1>2 are precisely the defining data of this Cayley graph.
4. Conversely, let \(Q\) be any group generated by elements\[ q_1,\dots,q_n. \]Then its Cayley graph with these generators is a connected normal covering of \(R_n\).
The epimorphism \[ F_n\twoheadrightarrow Q, \qquad x_i\mapsto q_i, \] has normal kernel \(H\). The covering corresponding to \(H\) is normal, and by 3 its graph is exactly \[ \operatorname{Cay}(Q;q_1,\dots,q_n). \]
5. Hence the normal connected covers of a wedge of \(n\) circles are precisely Cayley graphs of groups with \(n\) specified generators.
4.For \(n=2\) this is the first assertion of the exercise.