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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?

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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 $i$.

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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
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

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