You need to know that both and are cyclic groups. Denote by 1 the element cl(1) in . Then 1 is a generator of Z30; any element is of the form for some integer m. So for any homomorphism f from to any group, f is completely determined by f(1) -- . So for a homomorphism f into , there are 10 choices for f(1). I leave it to you to show for any is well defined. However, for f to be onto, f(1) must be a generator of . So just ask yourself: how many generators of are there?