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?
$\endgroup$ 01 Answer
$\begingroup$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 $i$.
