Show: if an abelian group is a $G$-module, it is also a $\mathbb{Z}G$-module. Let $G$ be a finite group with identity $e$, $A$ an abelian group, and $\theta:G\to {\rm Aut}(A)$ a homomorphism.
Show that (via $\theta$) $A$ is a $\mathbb{Z}G$-module: $(\sum\limits_{g\in G}n_gg)\bullet a=\sum\limits_{g\in G}n_g\theta(g)(a)$.
I am not sure what this question is asking for.
My thoughts so far:
I am supposed to show $\theta:\mathbb{Z}G\to{\rm Aut}(A)$.
I am supposed to show that there is an extension of $\theta$ from $\mathbb{Z}G\to {\rm Aut}(A)$.
I know that to show $A$ is a $\mathbb{Z}G$-module, I need to show that there exists a homomorphism from $\mathbb{Z}G\to {\rm Aut}(A)$.  I am just not sure what the author means with the "$\bullet$" in this context, nor I am sure what I am expected to do with the summations -- is that equation supposed to be my homomorphism?
 A: There are two different ways to establish that something is a $\mathbb{Z}G$-module.  Either you provide a homomorphism $\Theta:\mathbb{Z}G\to \operatorname{Aut}(A)$, or you provide the action of $\mathbb{Z}G$ on $A$—that is, you define the expression $x\bullet a$ for every $x\in\mathbb{Z}G$ and $a\in A$ (satisfying certain axioms).
These two viewpoints are related as follows: if $\psi_x = \Theta(a)$ is the automorphism assigned to $x\in\mathbb{Z}G$, then $x\bullet a = \psi_x(a)$.  If you know what $\Theta$ is, you know what $\bullet$ is, and vice-versa.  So you only need to provide one or the other—I recommend whichever one you find most intuitive.
In any case, you have been provided with a homomorphism $\theta:G\to \operatorname{Aut}(A)$.  The problem provides you with the new $\mathbb{Z}G$-action in terms of $\theta$.
What is $\sum_{g\in G} n_g g$?  This is just an arbitrary element of $\mathbb{Z}G$, which is defined to be the collection of $\mathbb{Z}$-linear combinations of elements of $G$.  So the equation is taking an arbitrary $x\in\mathbb{Z}G$, and telling you where it should go when $a$ acts on it, i.e. $x\bullet a$, or $\Theta(x)$ applied to $a$.  If $x=\sum_{g\in G} n_g g$, this value should be $\sum_{g\in G} n_g \theta(g)(a)$.
If you think about this expression, it should make sense—$\Theta$ needs to agree with $\theta$ on $G$, and it needs to respect sums and integer coefficients, i.e $\Theta(2x)(a)=\Theta(x+x)(a)=2\Theta(x)(a)$.  All that remains is to make sure that this expression is well-defined, and a homomorphism.
