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.

I have an exercise. I cannot solve. Please help me to solve it:

Prove that any periodic abelian group is the direct sum of its maximal $p$-subgroups.

share|improve this question
2  
Hint: Take an element, and (assuming you write the Abelian group additively), write it a sum of elements of prime power order. –  Geoff Robinson Aug 25 '11 at 9:39

1 Answer 1

Let $G$ be a periodic abelian group. For each prime number $p$, Let $G_p$ = {$x \in G$; $p^n x = 0$ for some integer $n \geq 0$}.

It suffices to prove that $G$ is the direct sum of $G_p$, where $p$ runs through all the prime numbers.

Let $a \in G$. Let $n$ be its order. Let $n = p_1^{m_1}\cdots p_r^{m_r}$ be the prime decomposition. Let $n_i = p_1^{m_1}\cdots p_r^{m_r}/p_i^{m_i}$. Since gcd($n_1, \dots n_r$) = 1, $n_1k_1 + \cdots n_rk_r = 1$ for some integers $k_1, \dots k_r$ Let $b_i = n_ik_ia$ for $i = 1, \dots r$. Then $a = b_1 + \cdots + b_r$. Since $n_ip_i^{m_i} = n$, $p_i^{m_i}b_i = 0$. Hence $b_i \in G_{p_i}$.

Next we prove that this decomposition is unique. It suffices to prove that if $c_1 + \cdots + c_s$ = 0, then each $c_i = 0$, where $c_i \in G_{p_i}$ and $p_i \neq p_j$ for $i \neq j$. Since $c_1 = -(c_2 + \cdots + c_s)$, $mc_1 = 0$ for some integer $m$ relatively prime to $p_1$. Since $c_1 \in G_{p_1}$, $c_1$ must be 0. Similarly $c_i = 0$ for all $i$ and we are done.

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.