This is not a nice problem. Let be the Sylow -subgroups of respectively. Note that they are both normal, and so their product is normal in and of of order . Now, note that trivially if denotes the Sylow subgroup of then , where both follow from order arguments. Thus, . Thus, we are left to classify homomorphisms which induce different semidirect products. But, note that , since any group of order is cyclic, and so isomorphic to , splits accross the product of cyclic groups, and for every prime . And, since we can count the number of possible homomorphisms by counting the number of possible homomorphisms . But, this clearly corresponds to the number of elements of order two in which is four. Thus, there are at most four homomorphisms and so at most four non-isomorphic groups of order . That said, the groups (the only possible cyclic group), , , and are four non-isomorphic groups of . Thus, these are all the possible groups of order .