How to form Cayley graph from group Define a Cayley graph as follows:
$G$ is finite group. $C \subseteq G$ such that $C$ does not contain identity element of $G$ and $g^{-1} \in C$ for all $g \in C$. Cayley graph $X(G,C)$ is formed with vertices $V(X)=G$, edges $E(X)=\{(a,b): a,b \in G, ab^{-1} \in C\}$.
a. Prove that the graph $X(C,G)$ is well-defined.
b. A graph is vertex transitive if $\text{Aut}(X)$ acts transitively on vertices. Show that $X(G,C)$ is vertex transitive.
 A: It is not clear to me what "proving the graph is well-defined" entails, since the definition does not involve any sort of actions or maps on equivalence classes. Perhaps it wants you to show that if the segment $(a,b)$ is an edge of $X(G,C)$ then so is $(b,a)$, or something (this follows from the hypothesis on $C$ that it is closed under taking inverses). I'll let a graph theorist comment.
For part (b), show that right-multiplication by any element $g\in G$ induces a graph automorphism on $X(G,C)$. First of all this is a bijection on the vertices, and second of all you want to show $(a,b)$ is an edge of $X(G,C)$ if and only if $(ag,bg)$ is an edge of $X(G,C)$, for all $g\in G$. The utility of this is that the group $G$ acts transitively on itself by (both left and right) multiplication, so to create an automorphism of $X(G,C)$ that sends vertex $a$ to the vertex $b$ (here with $a,b\in G$ arbitrary), just pick a $g\in G$ such that $ag=b$, i.e. consider right-multiplication by $a^{-1}b$.
