I have been reading on Undergraduate Algebra by Serge Lang, and not so far deep in I got myself stuck already.
The problem says:
If $A$ is an abelian group, written additively, and $n$ is a positive integer such that $\forall x\in A$ it holds $nx=0$. Assume that we can write $n=rs$ where $r$ and $s$ are positive relatively prime integers. Let $A_r$ consist of all $x\in A$ such that $rx=0$, and similarly $A_s$ of all $x \in A$ such that $sx=0$. Show that $A$ is direct sum of $A_s$ and $A_r$.
I have tried the following :
If $nx=0$ and $x \neq 0$ then $nx=(rs)x=0$. Then it holds that either $rx=0$ or $sx=0$ or both.
Now I claim that for nonzero $x$, both can not be true at the same time, thus making $A_r$ and $A_s$ mutually disjoint. If both held at the same time then $rx=sx$ then it holds that $(r-s)x=0$ then since $x$ is nonzero it must hold that $r=s$. Thus since they are relatively prime they must be equal to 1, thus group would be trivial thus x would have to be zero, which is contradiction. Thus the 2 defined sets must be disjoint if the group is nontrivial.
Now since $r$ and $s$ are relatively prime there are $p$ and $m$ such that $pr+ms=1$.
Now I claim that every element in group $A$ can be written as sum of elements of $A_r$ and $A_s$. Since: $$a=1a=(pr+ms)a=pra+msa.$$
But from above arguments it is obvious that each elements of $A$ must belong to either $A_r$ or $A_s$.
What is unclear to me is whether my proof is correct or not. Especially the part where I prove that every element is a sum of elements of subsets seems really sketchy to me.
This problem is to be found in the book Undergraduate Algebra by Serge Lang, third edition, on page 40, problem 15.
Your proof is mostly correct; however, there are some (important!) details that should be corrected.
We need to establish two facts: $A_r \cap A_s = \{0\}$ and $A = A_r + A_s$.
The second part, where you prove that every element of $A$ is a sum of elements from $A_r$ and $A_s$ is clear. You show that $$a = pra + msa,$$ and it is obvious that $pra \in A_s$, while $msa \in A_r$. This shows that $A = A_r + A_s$.
The bit that needs correcting is the first part of the proof, where you show that $A_r \cap A_s = \{0\}$. (Notice, they are not disjoint, as they share a common element - $0$. It is more precise to say their intersection is trivial).
In a general Abelian group setting, the fact that $(r-s)x = 0$ does not imply that $(r-s) = 0$.
A way around this problem is to use $pr + ms = 1$ again: if $x \in A_r \cap A_s$, we have: $$x = 1x = prx + ms x = 0 + 0 = 0.$$
This shows that $A_r \cap A_s = \{0\}$, as desired.
Apart from that, the proof is ok. There is just one (common) mistake that you should be careful about. You claim the following:
"If $nx=0$ and $x\neq0$ $then nx=(rs)x=0$. Then it holds that either $rx=0$ or $sx=0$ or both."
Actually, this isn't true; we can have $rx \neq 0$ and $sx \neq 0$ but still $rsx = 0$ (try to think of an example). Fortunately, we don't even need this conclusion to complete our proof.