Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.
Use the fundamental theorem for finite Abelian groups twice. Once to show that it is sufficient to prove the result for -groups, and the second time to investigate said -groups.
Note also that if divides the order of a group then that group contains an element of order .
You would have been better posting this in the abstract algebra forum, btw...
this time suppose is a finitely generated abelian group such that for any positive integer there exists at most one subgroup of with what can we say about ?
Assume the result holds for all groups generated by less than or equal to elements. Let be a subgroup generated by elements. Then, is of finite index as contains only torsion elements. Thus, is generated by less than or equal to elements (this can be seen by viewing as an -dimensional vector space). Is it of the form we wish? Let . Then as we are in an Abelian group these subgroup as normal and so we can apply the correspondence theorem to get that they are normal subgroups of : . Then, , and so they are the same subgroups. Thus, is of the correct form. As it consists only torsion elements it is cyclic.
Viewing our group as a vector space we can glue these two bits together - their intersection is trivial and and so we have a direct sum, and so a direct product. Thus, .
I think we cannot do anything with . Clearly it is just , but conforms to the conditions stipulated.