Factor Group Lemma of Cayley Graph Factor Group Lemma: Suppose that 1.$N$ is a cyclic, normal subgroup of group $G$. 2.$(s_1,s_2,\ldots,s_m)$ is a hamiltonian cycle in $Cay(G/N;S). 3.The product $s_1s_2\cdots s_m$ generates $N$. Then $Cay(G;S)$ has a Hamiltonian cycle. I can show the above theorem, but I cannot understand how we can conclude the following corollary. Suppose that 1.$N$ is a cyclic, normal subgroup of $G$, such that |N| is a prime power. 2. $\langle s^{-1}t\rangle=N$ for some $s,t\in S\cup S^{-1}$. 3. There is a hamiltonian cycle in $Cay(G/N;S)$ that uses at least one edge labelled $s$. Then there is a hamiltonian cycle in $Cay(G;S)$.

Let $(s_1,s_2,\ldots,s_{m-1},s)$ be a Hamiltonian cycle in $\Gamma(G/N,S)$. If $N = \langle s_1s_2 \cdots s_{m-1}s \rangle$, we are done by the lemma, and otherwise $N = \langle (s_1s_2 \cdots s_{m-1}s)(s^{-1}t) \rangle = \langle (s_1s_2 \cdots s_{m-1}t \rangle$, so we can apply the lemma to the Hamiltonian cycle $(s_1,s_2,\ldots,s_{m-1},t)$ of $\Gamma(G/N.S)$.