In the excellent Algebraic Graph Theory book by Godsil and Royle, they show that you can construct an infinite family of directed Cayley Graph (so vertex transitive) that are not Hamiltonian. The backbone of the prove is from a clever use of a theorem in the following paper by Swan, "A Simple Proof of Rankin's Campanological Theorem" (http://www.jstor.org/stable/2589053), which says that:
Theorem 1:
Let E={a,b} be a subset of a finite group G. Suppose that G has a partition into r disjoint E-cycles. If $ c=ab^{-1} $ has odd order then $ r \equiv |G:<a>|\equiv |G:<b>| \mod 2$
(Another good reference on the subject is a paper by Ruskey, Jiang, Weston "The Hamiltonicity of directed $\sigma - \tau$ Cayley graph")
Based on Theorem 1, they show first an example that the vertex set V(X) of the directed Cayley Graph X=X(Sym(4), {a,b}), where a=(1,2), b=(1,2,3,4) ({a,b} generates Sym(4)) can be partitioned in an even number of disjoint directed cycles, and this in particular means that it does not have a directed Hamilton Cycle. (Then they generalize to an infinite family of Directed Cayley Graph).
Now I have 2 questions:
How do you infer that "since V(X) can be partitioned in an even number of directed (disjoint) cycles" then "it does not have a directed Hamilton Cycle"?
Would it be true that given any vertex transitive graph X, if V(X) can be partitioned in an even number of directed (disjoint?) cycles then it does not have a directed Hamiltonian Cycle? Or it needs to be a directed Cayley graph?
Could you please point me to the right direction?