# A finite abelian group has order $p^n$, where $p$ is prime, if and only if the order of every element of $G$ is a power of $p$

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.

I tried the following route, but got stuck. Using the fundamental theorem of finite Abelian groups, the problem reduces to proving Cauchy's theorem for a cyclic abelian group. If G is a cyclic group, and p divides G, then G has an element of order p whether p is prime or not. If we regard G as the integers mod p, then we can notice that if $|G| = kp$ then the integer k has order p in G.

• Do you know the structure theorem for finite Abelian groups? – Bey Apr 5 '15 at 21:42
• What have you tried so far? Can you prove at least one direction of implication? – Omnomnomnom Apr 5 '15 at 21:46
• The claim is true also without the "abelian" bit. – Timbuc Apr 5 '15 at 21:47
• I tried the following route, but got stuck. Using the fundamental theorem of finite Abelian groups. The problem reduces to proving Cauchy's theorem for a cyclic abelian group. If G is a cyclic group, and p divides G, then G has an element of order p whether p is prime or not. If we regard G as the integers mod p, then we can notice that if |G| = kp then the integer k has order p in G – frierfly Apr 5 '15 at 22:59

Assume $p$ divides the order, and $q$ is some other prime that also divides the order. By Cauchy's theorem there is an element of order $q$.

Therefore, if the order is not a power of a prime then all the elements can't be of order a power of the same prime.

Assume that the order of the group is $p^n$. Then the order of an element $a$ must divide the order.

To be exact, see theorem 1.1, it relies on Cauchy's Theorem. As already pointed out, the "abelian" part is not necessary.