I am trying to prove that if $X(G,S) $ is a Cayley graph where $G$ is abelian and $|S|>2$, then $X(G, S)$ contains a $4$-cycle. I found an example proof at the following link:
http://exwiki.org/mw/index.php?title=Cayley_graphs_on_abelian_groups_have_a_4_cycle
I am not understanding the very first assumption, that is, we can always find distinct $x_1,x_2\in S$ s.t. $x_2x_1=e$. I understand that if there exists $x\in S$ s.t. $x\ne x^{-1}$ then $x^{-1}\in S$ and we have found two such elements, but what if $S$ contains only self inverses?
I think it should say $x_2x_1 \neq e$. In any case maybe it helps to think about the idea here. Cycles in Cayley graphs correspond to nontrivial relations in the generators. In general there may be very few relations: for example if $G$ is a free group then its generators satisfy no relations at all and so the Cayley graph is a tree. However $G$ being abelian calls a rather obvious relation to mind, namely that $xyx^{-1}y^{-1} = e$. If $x$ and $y$ are generators then this relation corresponds to a cycle $e,x,xy,y$, and we should just take care that these are four distinct elements, which just amounts to requiring $x,y,xy\neq e$ and $x\neq y$.