Since this was the first hit when I Googled my homework, I'll resurrect this thread with my thoughts.
This holds trivially for prime numbers.
Claim: For n "square-free", [tex]\mathbb{Z}_n[/tex] is (up to isomorphism) the unique abelian group of order n.
We extend the fact that [tex]\mathbb{Z}_{ab} \cong \mathbb{Z}_a \times \mathbb{Z}_b \Leftrightarrow gcd(a, b) = 1[/tex] by induction to an arbitrary amount of factors. Then the factors (which are also the subscripts of Z, and the orders of the groups) are pairwise coprime, and we have our result.
Thoughts?