Let $K_n\times K_n$ be the Cartesian product of two complete graphs.

Is $K_n\times K_n$ is Cayley graph or not?

I know that I have to use this lemma:

A connected graph $G$ is Cayley if and only if there exists a subgroup $H\subset\operatorname{Aut}(G)$ which acts simply transitively(regularly) on $V (G)$.

but I don't know how?

Please advise me.


First, we observe that $K_n$ is a Cayley graph. Consider the cyclic group $C_n$. If we take as a generating set all of $C_n$, the resulting Cayley graph is $K_n$. Note that $C_n$ acts simply transitively on the graph: if $g$ is a generator for $C_n$, and the vertices are labeled $g^k$, the action is just multiplication (and since it is an abelian group, we don't have to specify on which side).

Now, consider the action of $C_n\times C_n$ on $K_n\times K_n$. It is straight forward that if $G$ acts on $X$ and $H$ acts on $Y$, then $G\times H$ acts on $X\times Y$. Moreover, if the actions are transitive or free, so is the resulting product action. Therefore, the product of two Cayley graphs is again a Cayley graph.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.