Let $G$ be a finite group and $S$ be a subset of $G$. Let the Cayley graph of $G$ with respect to $S$ be $Cay(G,S)$, provided that $1 {\not\in} S$ and $S$ is inverse closed.
Consider the Cayley graph of a finite group, as an example say, $Cay(\mathbb{Z}_3 \times \mathbb{Z}_5, \{g_1,g_2\})$, where $g_1=(0,1)$ and $g_2=(1,0)$, $|g_1|=5$, $|g_2|=3$. A rough sketch of the Cayley graph is as follows.
Some Hamiltonian cycles starting with $(0,0)$ vertex in the Cayley graph are,
$g_2^{-1} g_1 g_2 g_1 g_2^{-1} g_1 g_2^{2} g_1^{-4} g_2^2 g_1$, which can be simplified to $g_1^4 g_1^{-4} g_2^{5} g_2^{-2}$
$g_1^4 g_2 g_1^{-3} g_2 g_1^4 g_2^{-2}$, which can be simplified to $g_1^8 g_1^{-3} g_2^2 g_2^{-2}$
$g_1 g_2 g_1^{-1} g_2 g_1^2 g_2^{-2} g_1 g_2^2 g_1 g_2^{-2} g_1$, which can be simplified to $g_1^6 g_1^{-1} g_2^4 g_2^{-4}$
So, the Hamiltonian cycles are in the form $g_1^{x_1} g_1^{-x_2} g_2^{x_3} g_2^{-x_4}$ for some values $x_1, x_2, x_3, x_4$.
I observed that,
a) The product of the generating elements in each of the above Hamiltonian cycles became equal to $(0,0)$.
b) $x_1+x_2+x_3+x_4=15$.
c) $x_1 - x_2 \equiv 0 (mod 5)$ and $x_3 - x_4 \equiv 0 (mod 3)$.
Now, from $a, b, c$ conditions, thinking back about the Hamiltonian cycles, I can get the values $(4,4,5,2), (8,3,2,2), (6,1,4,4)$ as possible solutions for $(x_1, x_2, x_3, x_4)$ respectively (satisfying the $a, b, c$ conditions). But I also get $(3,3,6,3)$ as a possible solution, but I am unable to find a Hamiltonian cycle for such values of $x_1, x_2, x_3, x_4$ (i.e. generating element $g_1$ is appearing 3 times in the cycle etc. which seems not possible).
Q1: So, there is no cycle having the values $(3,3,6,3)$, am I right? Or have I missed that cycle?
Q2: If there is no such cycle, then it seems like I can not obtain values for $x_1, x_2, x_3, x_4$ using only $a, b, c$ conditions. Can any suggestion be given to include more conditions so that I can get only the values corresponding to Hamiltonian cycles?