1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you know the structure theorem for finite Abelian groups? $\endgroup$ – Bey Apr 5 '15 at 21:42
  • 1
    $\begingroup$ What have you tried so far? Can you prove at least one direction of implication? $\endgroup$ – Omnomnomnom Apr 5 '15 at 21:46
  • 1
    $\begingroup$ The claim is true also without the "abelian" bit. $\endgroup$ – Timbuc Apr 5 '15 at 21:47
  • $\begingroup$ 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 $\endgroup$ – frierfly Apr 5 '15 at 22:59
5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

See: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cauchyapp.pdf

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.