G finite abelian group
1 subgroup of order d for each divisor d of |G|.
Prove G cyclic.
What if we proved the contrapositive. Suppose that $\displaystyle G$ is not cyclic. Try constructing subgroups of $\displaystyle G$ which are distinct but have the same order (hint: you're going to want to construct these subgroups to be cyclic groups [math}C_n,C_m[/tex] say with some prime $\displaystyle p\mid m,n$)