Consider the Cayley graph of $\mathbb{Z}_3 \times \mathbb{Z}_5$ generated by the generating set $S=\{g_1=(1,0), g_2=(0,1)\}$. Then $|g_1|=3, |g_2|=5$.
Consider a Hamiltonian cycle, $ABCDEFGHIKJLMNO$. Then the Hamiltonian cycle can be expressed in terms of the generating elements as follows.
$g_1^2 g_2 g_1^{-1} g_2 g_1 g_2 g_1^{-1} g_2 g_1^2 g_2^{-4} = e$
Since the group is abelian, we can simplify it to be as follows, right?
$g_1^3 g_2^0 = e$
Then can we adjust the $e$ on the right hand side in terms of the generating elements as, $g_1^3 g_2^0 = g_1^0 g_2^0$ or as $g_1^3 g_2^0 = g_1^3 g_2^5$ (because the orders of the elements are 3 and 5 respectively)?
Thanks a lot in advance.