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**Homework Statement**

1. Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.

2. Suppose that G is a finite Abelian group. Prove that G has order p

^{n}, where p is prime, if and only if the order of every element of G is a power of p.

**The attempt at a solution**

1. By the Fundamental Theorem of Finite Abelian Groups, G must equal [itex]Z_{n_1} \oplus \cdots \oplus Z_{n_m}[/itex] where the n's are power's of primes. I also know that [itex]Z_{n_1} \oplus \cdots \oplus Z_{n_m}[/itex] is cyclic if the n's are relatively prime. I don't see how subgroups come into play here.

2. Since the order of every element in G must divide |G|, the order of every element must be some power of p where the power divides n. That takes care of one direction. In the other direction, if every element of G is a power of p, since these powers must divide |G|, then |G| = p

^{n}where n is a multiple of all the powers of the orders of the elements of G right? Is this enough to prove the statement?