yes, i know that the only subgroups of G are itself and the subgroup {e} which consists of the neutral element. This is because the only possibilities of the cardinalities of the subgroups are 1 or p.
It cannot be 1 because we assumed g was not equal to the identity. So the order of g must be p, and therefore G = {1 , g, g^{2}, ... , g^{p-1}} which is cyclic.