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.

Please help me with this.

Let $G$ be abelian group, and let $A_k$ be a family of subgroups of $G$. Prove that $G=\sum A_k$ (internal) if and only if every non-zero element $g\in G$ has a unique expression of the form $g=a_{k_1}+...+a_{k_n}$, where $a_{k_i} \in A_{k_i}$, the $k_i$ are distinct and each $a_{k_i}\neq 0$.

I have a hunch that the term "$g$ has a unique expression" here means that $A_k \bigcap A_j =\{0\}$. If so, how do we reason it and proved it to be so?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

No, it does not mean that. It means that $\displaystyle A_i\cap(\sum_{j\neq i}A_j)=0$ for all $i$.

Later: Well, actually, it is equivalent to that. That «$g$ has a unique expression of the form $a_1+\cdots+a_n$ with $a_i\in A_i$» means exactly that

  • first, there exist $a_1,\dots,a_n$ with $a_i\in A_i$ for each $i$ such that $g=a_1+\cdots+a_n$, and

  • second, that whenever you have elements $a_1,\dots,a_n, b_1,\dots,n_n$ with $a_i,b_i\in A_i$ for each $i$ such that $=a_1+\cdots+a_n=b_1+\cdots+b_n$, then $a_i=b_i$ for all $i$.

share|improve this answer
    
Oh ok. but what's the difference? And also how do we show that uniqueness implies $A_i\cap(\sum_{j\neq i}A_j)=0$. –  Seoral Jan 14 '11 at 6:19
2  
Consider the following subgroups of $\mathbb Z^2$: $A_1=\langle(1,0)\rangle$, $A_2=\langle(1,1)\rangle$, $A_3=\langle(0,1)\rangle$. Now show that for all $i\neq j$ you have $A_i\cap A_j=0$, yet it is not true that $A_1\cap(A_2+A_3)=0$, for example. –  Mariano Suárez-Alvarez Jan 14 '11 at 6:22
    
As for how to prove it: you should try to do it yourself for a while first... –  Mariano Suárez-Alvarez Jan 14 '11 at 6:24
    
Ok, i think i know how to prove it already.thanks. –  Seoral Jan 14 '11 at 7:23
add comment

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.