Consider the fact that and .

Applying these relations we have the following:

It is clear that since 4 does not divide 3 or 5 there will be no elements of order 4 in or respectively (Lagrange). Furthermore, it is obvious that 1 and 3 have order 4 in . Thus any combination of these elements in the 3 copies of will work so long as you choose 0 in the rest of the places. Note obviously not ALL of the componants can be 0 or else it is the identity and has order 1. I claim that there will be elements, you oughta be able to prove why there are no more and no less.