From this question, we have that the Cayley graph of direct product of two groups is a cartesian product of some cayley graphs on the factor groups. But, I do not see this translation easily.
Specifically, let us say I take the group $\mathbb{Z}_2\times\mathbb{Z}_8$ and the cycle graph on it. The cycle graph can be written as a cayley graph by taking any non-involution element and its inverse as the generating set in the group $\mathbb{Z}_2\times\mathbb{Z}_8$. But, how is it a cartesian product of some graphs on the indivdual groups $\mathbb{Z}_2$ and $\mathbb{Z}_8$? Is the ususal definition of cayley graphs not used while making the product graph? That is, can the cycle on the product group elements be written as a cartesian product of two $1$-factors on the individual factor group elements? Any hints? Thanks beforehand.
Best Answer
$\mathbb{Z}/2 \times \mathbb{Z}/8$ is not cyclic, so it needs at least $2$ generators to get the whole group. So if you take the "cycle graph" as you've defined it, you will not get a cayley graph at all.
If instead you take the cayley graph with two generators (one of $\mathbb{Z}/8$ and one of $\mathbb{Z}/2$) you'll find you do get the product of the respective cayley graphs.
I hope this helps ^_^