# Abstract Algebra Problem

## Homework Statement

Let $A$ be an abelian group, written additively, and let $n$ be a positive integer such that $nx=0$ for all $x \in A$. Such an integer n is called an exponent for A. Assume that we can write $n=rs$, where r, s are positive relatively 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 $c \in A_{s}$. Hence $A=A_{r} \oplus A_{s}$.

## Homework Equations

Theorem #1
The abelian group A is a direct sum of subgroups B and C if and only if $A=B+C$ and $B \cap C = {0}$. This is the case if and only if the map $B*C \to A$ given by $(b,c) \mapsto b+c$ is an isomorphism.

## The Attempt at a Solution

So essentially, what needs to be shown is that $A=A_{r}+A_{s}$ and that $A_{r} \cap A_{s} = {0}$. I went ahead and used the latter form of Theorem #1, namely showing that $A_{r} * A_{s} \to A$ is an isomorphism. I already showed that $A_{r} * A_{s}$ is a homomorphism and that it is injective. So, the last step is to show that it is also surjective in order to establish that it is an isomorphism.

Conjecture #1: $A_{r}$ is a subgroup of $A$ with order r.
Proof: We know that $0 \in A_{r}$ since $r0=0$. Suppose $a_{1},a_{2} \in A_{r}$. Then $r(a_{1}+a_{2})=r a_{1} + r a_{2} = 0$ since r is some positive integer (not necessarily in A). It also follows that $-a_{1} \in A_{r}$ since $r(-a_{1})=0$. Therefore $A_{r}$ is a subgroup of $A$.

This next part of the proof of the conjecture is what I am concerned about. Suppose $m \in A_{r}$ such that $m \neq 0$. We know $m$ exists since for some $y \in A, rsy=ny=0$, so $m=sy$ is one possible choice, assuming $sy \neq 0$ for at least one $y$. This is clearly the case or we can continue down by descent, replacing $n$ with $s$ in the problem. Now out of the possible choices for $m$, choose the smallest one. I think $m$ generates $A_{r}$, but I am not sure how to prove this. Maybe it isn't even a generator, so the entire method is flawed.

However, if it is, then the rest of the proof follows, since we can then show that since we would know $m$ generates $A_{r}$ and $rm=0$, so $m$ has period $r$. It then follows that since $A_{r} \cap A_{s} = {0}$, and $A_{r} * A_{s}$ has order $rs=n$, $A_{r}*A_{s}$ must be surjective onto $A$ since $A$ also has order $n$, and $A_{r}, A_{s}$ were shown to be subgroups of $A$.

This is using the fact that the order of $A$ should be $n$ since $nx=0$ for all $x \in A$, which includes for its generators.

The joys of self studying algebra :D

Now that I think about I think I implicitly assumed that we are using the smallest possible exponent $n$, so things are awkward if we are not... for instance if $A$ were the group who elements you get by addition modulo 6. We could use $n=6, n=12,...$, and I assumed $n=6$ would be the choice. So that is another problem. Makes me think I am approaching the problem the wrong way.