Can a Hamiltonian cycle of an undirected Cayley graph contain inverses of the generating elements? Let $G$ be a finite group and $S$ be a subset of $G$. Let us define the Cayley graph of $G$ with respect to $S$ as follows, provided that $1 {\not\in} S$ and $S$ is inverse closed.
Definition:
The Cayley graph of $G$ with respect to $S$, $Cay(G,S)$ is the graph whose vertices are the elements of $G$ and $g$ is adjacent to $gs$ for all $g \in G, \, s \in S$.
Consider a Cayley graph of a group $G$, $X=Cay(G,S)$ where $S=\{s,t\}$. When we consider a Hamiltonian cycle in any such Cayley graph, it can be expressed using the generating elements as (as an example),
$s t s s t^{-1} ...$ 
Is there a speciality as, when we consider the Hamiltonian cycle of an undirected graph, "it should definitely contain the inverses of $s$ and $t$(i.e. $s^{-1}$ and $t^{-1}$) as well as $s$ and $t$" ?. 
I mean can there be Hamiltonian cycles with only $s$ and $t$ terms with out $s^{-1}$ and $t^{-1}$ ?
Thanks a lot in advance.
 A: Not necessarily.
For instance, if $G$ is a cyclic group of order $n\ge 3$ with a generating elements $s$ and $S=\{s,s^{-1}\}$  then a sequence $e,s,s^2,\dots,s^{n-1}$ of vertices is a Hamiltonian cycle in $Cay(G,S)$ which is obtained going each time along an $s$-edge.
A bit more complex example is a Cartesian product $G=G_1\times G_2$ of two such groups of a common order $n$, $G_1$ with a generator $s_1$ and $G_2$ with a generator $s_1$ of order $n_2$. Put $\bar s_1=(s_1,e)$, $\bar s_2=(e,s_2)$, and $S=\{\bar s_1, \bar s_2, \bar s_1^{-1},\bar s_2^{-1}\}$. Then a sequence
$$\begin{matrix}
e, & \bar s_1, & \bar s_1^2, & \dots &\bar s_1^{n-1},\\
\bar s_2\bar s_1^{n-1}, & \bar s_2\bar s_1^{n}, &\dots, & \dots, &\bar s_2\bar s_1^{n-2},\\
\dots & \dots & \dots & \dots & \dots \\
\bar s_2^{n-1}\bar s_1, & \bar s_2^{n-1}\bar s_1^2, &\dots, &\dots, &\bar s_2^{n-1}\bar s_1^0,
\end{matrix}$$
is a Hamiltonian cycle in $Cay(G,S)$ which is obtained going each time along an $\bar s_1$- or
$\bar s_2$-edge.
Moreover, for a (finite) group $G$ there exists a Hamiltonian cycle in a graph $Cay(G,S)$ for some generating inverse closed subset $S$ of $G$ with the required property iff the elements of $G$ can be ordered in a cyclic sequence $g_0,g_1,\dots, g_n=g_0$ such that there are no indices $i,j$ with $g_i^{-1}g_{i+1}g_{j}^{-1}g_{j+1}=e$. The latter condition gives a lot of freedom for an order, so it seems to be possible for a lot of groups.
