Let $G$ be a finite group and $S$ be a subset of $G$. We 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 $g \in G, \, s \in S$.
I have written down several cycles in the above Cayley graph as follows.
((1), (2), (3) cycles sketched in red, green and pink respectively).
All (1), (2), (3) shows relationships between the elements $g_1$ and $g_2$.
(3) is the relationship given by a longest cycle. However, the same relationship is given by equation (2) which is not related to a longest cycle. When considering (3), when thinking about the powers of $g_1$ and $g_2$, it seems as if the powers satisfy,
$3 \equiv 0 (mod 3) \rightarrow (i)$
$0 \equiv 0 (mod 5) \rightarrow (ii)$
and when considering (4), it seems to satisfy,
$0 \equiv 0 (mod 3) \rightarrow (iii)$
$5 \equiv 0 (mod 5) \rightarrow (iv)$
So, in a situation where $|g_1|=p$ and $|g_2|=q$ where $p, q$ are distinct primes and $g_1^m g_2^n =e$, will it satisfy, $m \equiv 0 (mod p)$ and $n \equiv 0 (mod q)$?
The L.H.S. of both $(i), (ii)$ have got a value congruent to the equivalence class $0$ and in $(iii), (iv)$ also, it is the same. One has got the value equal to $p$ or $q$ in the modulus and the other always $0$. Is it a characteristic occupied by the longest cycle? I mean can I explain that it is due to the longest cycle?
If so, whenever I have a relation $m \equiv 0 (mod p)$ and $n \equiv 0 (mod q)$, can I conclude that the relation corresponding to a longest cycle is given by the solutions either,
$(m=0,n=q)$ or $(m=p, n=0)$?
Is there a way to mention a specific characteristic occupied by the longest cycle to these powers of the generating elements?