Unique Decomposition of Elements in an Abelian Group

[UNChaneul]

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

 

[UNChaneul]

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.

 

[UNChaneul]

Nevermind, solved my own problem. Guess writing things up in a different form can be useful haha.