Showing torsion abelian groups have a unique $\widehat{\mathbb{Z}}$-module structure.

Question

Let $E$ be a torsion abelian group, we write it additively. Define the multiplication \begin{align} \widehat{\mathbb{Z}} &\times E \rightarrow E, \\ (a,g) &\mapsto a\cdot g:=a_ng, \end{align} where $n$ is the order of $g$ in $E$.

1. I wish to show this defines a scalar multiplication. I've shown all axioms hold barring that for all $a \in \widehat{\mathbb{Z}}$, and all $g,h \in E$, it holds $$a\cdot(g+h)=a\cdot g+a\cdot h. \quad(*)$$ If we denote the order of $g,h,g+h,$ respectively, by $n,m,k$, respectively, then I know $(*)$ follows from $a_k \equiv a_n \pmod{n}$ and $a_k \equiv a_m \pmod{m},$ which in turn holds if $n,m \mid k$ since $a \in \widehat{\mathbb{Z}}.$ However I haven't been able to show this.
2. I want to show that the above is the only $\widehat{\mathbb{Z}}$-module structure one can define on $E$. To this end, I know the following:
   • $\mathbb{Z}$ is dense in $\widehat{\mathbb{Z}}$ when the latter has the profinite topology,
   • $E$ has a unique $\mathbb{Z}$-module structure since $E$ is an abelian group,
   • If $E$ is given the discrete topology, then the above scalar multiplication is continuous.

So far I've been unable to combine these into a proof. This question is part of a homework assignment, so I don't want a full solution, just suggestions on what could be worthwhile to try.

Answer 1

Take $g\in E$ and $n=order(g)$.

For $a\in \hat{\Bbb{Z}}$ write $a= m+nb$ where $m\in 0\ldots n-1$ and $b\in \hat{\Bbb{Z}}$.

$$nb\cdot g = b\cdot \underbrace{(g+\ldots +g)}_n = 0$$ gives that $a\cdot g = m g$.

Of course $m = a_n$ in your notation.

Answer 2

For 1. it might be useful to show the following: If $m \in \Bbb N$ such that $mg=0$, then for any $a \in \widehat{\Bbb Z}$ $g^a=g^{a_m}$. Now to show that $a(g+h)=ag+ah$, apply this with $m=\mathrm{ord}(g)\cdot\mathrm{ord}(h)$ and then continue.