Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ and $H$ be finite abelian groups. Show that if for any natural number $n$ the groups $G$ and $H$ have the same number of elements of order $n$, then $G$ and $H$ are isomorphic.

I know, that for an infinity group doesn't work : $ \Bbb Z_{27}$

It seems to me that I can use Finitely-generated abelian group

It is possible that this simple fact, but I would ask to write a proof .

share|improve this question
2  
Yes, use the theorem of finitely generated Abelian groups. –  Berci Dec 15 '13 at 2:07
    
I understand that you are not very good with English, but you should at least write everything you want to write. The second line is not even finished... –  tomasz Dec 15 '13 at 2:22

1 Answer 1

Since $G$ and $H$ are a direct product of cyclic groups of prime power order (fundamental theorem of finite Abelian groups), we just need to prove that if the number of elements of order $n$ are the same for $G$ and $H$, then both correspond to the same direct product.

Suppose $p^k$ divides both $|G|$ and $|H|$. The number of elements of order $p^k$ is $N\cdot \phi(p^k)$ where $N$ is the number of cyclic groups in the direct product of order $p^j$ for $j \ge k$. We can then easily determine $N$ for every $p^k$, and thus the whole direct product.

Since the information given is enough to completely determine $G$ and $H$, they must be isomorphic.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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