Let $G$ be a finite group and $S$ be a subset of $G$. Let the Cayley graph of $G$ with respect to $S$ be defined as follows, provided that $1 {\not\in} S$ and $S$ is inverse closed.
"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 $X$ of a non-abelian group generated by elements $\{s,t\}$. Let $P=sts^2t^3$ (as an example) represent a path in the Cayley graph expressed in terms of the generating elements.
Suppose,
$v_1 \cdot sts^2t^3 = w_1$
$v_2 \cdot sts^2t^3 = w_2$
$v_3 \cdot sts^2t^3 = w_3$ etc.
for vertices $v_1,v_2,v_3, \cdots, w_1, w_2, w_3, \cdots \in V(X)$. If $v_1 \neq v_2 \neq v_3 \neq \cdots \neq v_n$, then will $w_1 \neq w_2 \neq w_3 \neq \cdots \neq w_n$, for any path $P$? ($n$ is the order of the Cayley graph).
The Cayley graphs are regular and we know that each vertex in a Cayley graph has similar environments. Thus, starting from any vertex we can move along edges corresponding to a similar set of generating elements. So it is clear that when any vertex is multiplied by $P$ it will reach another vertex. So my question expressed in another way is that, if $v_1,v_2,v_3, \cdots, v_n$ are distinct, then will $w_1,w_2,w_3, \cdots, w_n$ be also distinct for any arbitrary path $P$?
Thanks a lot in advance.
Yes, they will be distinct. This is a more basic and fundamental fact about groups that can be stated without mention of Cayley graphs.
Suppose $v_1 s = v_2 s$, where $v_1,v_2,s$ are any elements of a group. Then just multiply both sides by $s^{-1}$ to get that $v_1 = v_2$.
This shows that: "if $v_1 s = v_2 s$, then $v_1=v_2$".
What you are asking is the contrapositive of this statement, which is equivalent: "if $v_1 \neq v_2$, then $v_1 s \neq v_2 s$"