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## Homework Statement

Let [itex]A[/itex] be an abelian group, written additively, and let [itex]n[/itex] be a positive integer such that [itex]nx=0[/itex] for all [itex]x \in A[/itex]. Such an integer

*n*is called an exponent for A. Assume that we can write [itex]n=rs[/itex], where

*r*,

*s*are positive relatively prime integers. Let [itex]A_{r}[/itex] consist of all [itex]x \in A[/itex] such that [itex]rx=0[/itex], and similarly [itex]A_{s}[/itex] consist of all [itex]x \in A[/itex] such that [itex]sx=0[/itex]. Show that every element [itex]a \in A[/itex] can be written uniquely in the form [itex]a=b+c[/itex], with [itex]b \in A_{r}[/itex], and [itex]c \in A_{s}[/itex]. Hence [itex]A=A_{r} \oplus A_{s}[/itex].

## Homework Equations

Theorem #1

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

## The Attempt at a Solution

So essentially, what needs to be shown is that [itex]A=A_{r}+A_{s}[/itex] and that [itex]A_{r} \cap A_{s} = {0}[/itex]. I went ahead and used the latter form of Theorem #1, namely showing that [itex]A_{r} * A_{s} \to A[/itex] is an isomorphism. I already showed that [itex]A_{r} * A_{s}[/itex] 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:

**[itex]A_{r}[/itex] is a subgroup of [itex]A[/itex] with order r.**

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

**This next part of the proof of the conjecture is what I am concerned about.**Suppose [itex]m \in A_{r}[/itex] such that [itex]m \neq 0[/itex]. We know [itex]m[/itex] exists since for some [itex]y \in A, rsy=ny=0[/itex], so [itex]m=sy[/itex] is one possible choice, assuming [itex]sy \neq 0[/itex] for at least one [itex]y[/itex]. This is clearly the case or we can continue down by descent, replacing [itex]n[/itex] with [itex]s[/itex] in the problem. Now out of the possible choices for [itex]m[/itex], choose the smallest one.

**I think [itex]m[/itex] generates [itex]A_{r}[/itex], 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 [itex]m[/itex] generates [itex]A_{r}[/itex] and [itex]rm=0[/itex], so [itex]m[/itex] has period [itex]r[/itex]. It then follows that since [itex]A_{r} \cap A_{s} = {0}[/itex], and [itex]A_{r} * A_{s}[/itex] has order [itex]rs=n[/itex], [itex]A_{r}*A_{s}[/itex] must be surjective onto [itex]A[/itex] since [itex]A[/itex] also has order [itex]n[/itex], and [itex]A_{r}, A_{s}[/itex] were shown to be subgroups of [itex]A[/itex].

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

The joys of self studying algebra :D