Consider a finite group $G=\mathbb{Z}_3 \times \mathbb{Z}_5$. Let the undirected Cayley graph of the group be gererated by $\{s,s^{-1}, t, t^{-1}\}$, where $|s|=3, |t|=5$. Then different cycles in the Cayley graph can be written as,
$s^2 t s^{-2} t^{-1}$, (this is a cycle of length 6)
$s^2 t^2 s^{-2} t^{-2}$, (this is a cycle of length 8)
$s^2 t s^{-1} t s t s^{-1} t s^2 t^{-4}$ (this is a cycle of length 15) etc. The latter one, i.e. $s^2 t s^{-1} t s t s^{-1} t s^2 t^{-4}$ is a Hamiltonian cycle. Since the group is abelian I can simplify it and write as,
$s^3 t^0$ (and the other two cycles can be simplified like that too).
And because the above are cycles we can write them equating to $e$, where $e$ is the identity of $G$ as an example if I write the Hamiltonian cycle, it would be like $s^3 t^0=e \rightarrow (1)$.
It is in the form $s^m t^n=e \rightarrow (2)$
It represents a relationship between the generating elements of the Cayley graph. In other words, the cycles represent relationship between generating elements of the Cayley graph.
Thinking about the elements $s$ and $t$, since $s^0=e$ and $t^0=e$, I can get $s^0 t^0=e$ and rewrite (2) as
$s^m t^n=s^0 t^0 \rightarrow (3)$
Thinking about $(3)$ and the powers of the elements, can I write as $m \equiv 0 (mod 3)$ and $n \equiv 0 (mod 5)$ because m is present for an element of order 3 and n for an element of order 5 and on the right hand side we have $e$, where we can think that $e$ is composed of $s^0 t^0$ so that it will result in the above two congruence relations?
Suppose in some situation I know only the congruence relations $m \equiv 0 (mod 3)$ and $n \equiv 0 (mod 5)$, and does not know $m, n$ values nor the Hamiltonian cycle. Then if I solve the congruence relations by taking maximum (maximum because I'm finding the one corresponding to a longest cycle, which is the Hamiltonian cycle) as,
{Max,
$m \equiv 0 (mod 3)$
$n \equiv 0 (mod 5)$}
then can it give the above same values of $m, n$? If there are multiple solutions, then at least as one of the solutions? Can we solve such a system by taking maximum like that?