How to get generators for a finite point group? How do I find generators of Octahedral group when viewing it as a subgroup of $S_8$?
Update: Alex Bartel's suggestion gives the following permutations and Cayley graph



Update 2: second suggestion by Alex gives following Cayley graph, which seems to be isomorphic to the Cayley graph of symmetric group given elsewhere, check!


 A: Observation number one: the group we are talking about is $S_4$.
Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.
So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.
Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.
Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.
