# Direct sum of subsets of an commutative group

Let A be an abelian group, written additively, and let $n$ be a positive integer such that $nx=0$ for all $x\in A$. Assume that we can write $n=rs$, where $r,s$ are positive relative prime integers. Let $A_r$ consist of all $x\in A$ such that $rx=0$, and similarly $A_s$ consist of all $x\in A$ such that $sx=0$. Show that every element $a\in A$ can be written uniquely in the form $a=b+c$ with $b\in A_r$, and $c\in A_s$.

I defined $f:A_r \times A_s \rightarrow A$ by $(b,c) \mapsto b+c$, I showed that $A_r, A_s$ are subgroups of $A$ and that this map defines an isomorphism.

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Even though I agree with Arturo's answer which is nice, your approach is very good too (and perhaps more algebraic-intuitive than number-theoretic-intuitive). –  Patrick Da Silva Jan 29 '12 at 1:08
yeah, I would have never thought of that approach. I am glad he posted it :-). –  user2467 Jan 29 '12 at 1:21

Given $a\in A$, let $m$ be the least positive integer such that $ma=0$. Since $na=0$, then $m|n$. Since $\gcd(r,s)=1$, we have: $$\gcd(m,r)\gcd(m,s) = \gcd(m,rs) = \gcd(m,n) = m.$$ Let $d = \gcd(m,r)$, and let $e=\gcd(m,s)$. We also have $d|r$ and $e|s$, hence $\gcd(d,e)=1$. Hence there exist $\alpha$ and $\beta$ such that $\alpha d + \beta e = 1$.
Now verify that $s(\alpha da) = r(\beta ea) = 0$, and that $a=\alpha da +\beta ea$.
This gives existence. To get uniqueness, suppose that $a=b+c =b'+c'$. Then $b-b'$ has order dividing $r$, $c'-c$ has order dividing $s$; since $b-b'=c-c'$, then $b-b'$ has order dividing $\gcd(r,s)$.