Let me do it for and you think about . The polynomial can be factored as where . Therefore, the splitting field is . Now is equal to the degree of the minimal polynomial of over . Notice is irreducible over (I am assuming you know this result) so . Let and thus . This tells us that we are looking at a Galois group that has elements. If is an automorphism of then it is completely determined by its value of . Remember that automorphisms of permute the zeros of polynomials, this means that must be one of the zeros of , thus, . We see that there are at most automorphism of , however we know that which forces us to conclude that there is an automorphism such that for each . Therefore, . It is easy to see that behaves like the group i.e. . Thus, has a generator for example i.e. .