I'm studying projective modules and I'm having problem coming up with (or understanding) examples of non-free projective modules. I got that when a ring is a direct sum $R = A \oplus B$, both $A$ and $B$ are non-free projective $R$-modules via the action:
$$\psi: R\times A \to A, ((a, b), a') \mapsto aa'$$
Then I tried to attack the following question:
For what groups $G$ is $\mathbb{Z}$ a projective $\mathbb{Z}G$-module? (Hint: When does $\epsilon: \mathbb{Z}G \to \mathbb{Z}$ split as a map of $G$-modules?)
But I fail to see how would $\mathbb{Z}$ be a $\mathbb{Z}G$-module with a non-trivial action. On the hint, I also fail to see how $\mathbb{Z}$ would be a $G$-module other than with the trivial action. Is it for me to consider them as the modules with trivial action? I can't see a natural way for either $\mathbb{Z}G$ or $G$ to act on $\mathbb{Z}$.
Also, is the map $\epsilon$ referring to the augmentation of the group ring $\mathbb{Z}G$?
I believe that my deeper problem is with looking $\mathbb{Z}$ (the 'small' structure) as a $\mathbb{Z}G$-module (the 'big' structure), and not the other way. Is very simple to see $\mathbb{Z}G$ as a $\mathbb{Z}$ module, once $\mathbb{Z}$ is a substructure of $\mathbb{Z}G$. But I can't see how and why I would look the big structure acting on the smaller, substructure one! In the first example, when a ring is a direct sum, I was able to understand the action, but I don't know why it would be interesting to look at things this way and not the other way around ($R$ as a $A$-module or a $B$-module).
It only holds for the trivial group.
As it was mentioned in the comments, we consider $\mathbb{Z}$ as a trivial $\mathbb{Z}G$-module and in this way the augmentation map $\epsilon \colon \mathbb{Z}G \to \mathbb{Z}$ is a morphism of $\mathbb{Z}G$-modules.
Let $ f \colon \mathbb{Z} \to \mathbb{Z}G$ be a morphism of $\mathbb{Z}G$-modules and say $ f(1) = \sum_{h \in G} n_h h$ where all the $n_h=0$ except for a finite number of them. Then we have for any $g \in G$,
$$ g \cdot f(1) = f(g \cdot 1) = f(1)$$
because $f$ is a morphism of $\mathbb{Z}G$-modules and $\mathbb{Z}$ is a trivial $\mathbb{Z}G$-module. Therefore
$$ \sum_{h \in G} n_h gh = \sum_{h \in G} n_h h $$
If $\epsilon \circ f$ is the identity, then $f$ can not be identically zero, so there must be an $h$ in the group such that $n_h \neq 0$.
Pick an $h$ such that $n_h \neq 0$ and choose $g=h^{-1}$. In this case, the equality tells us that the coefficient of the identity element $e$ must be $n_h$, that is, $n_h = n_e$. Similarly, given $k$ in $G$ we can choose $g=k$ and we would conclude that $n_k = n_e$. And therefore setting $n = n_e$
$$ f(1) = \sum_{g \in G} n g $$
with $n \neq 0$. For this to make sense as an element of $\mathbb{Z}G$, the group $G$ has to be finite. But then $\epsilon f(1) = n|G|$, so the only way this can be the identity is if $n=1$ and $|G|=1$, that is, $G$ is the trivial group. Of course, if $G$ is the trivial group, then $\mathbb{Z}G = \mathbb{Z}$ and certainly $\mathbb{Z}$ is projective over $\mathbb{Z}$, it is free.
So that answers your question, but just for fun, this would have been different if we had asked the same question about $\mathbb{Q}G$. If $G$ is a finite group, $\mathbb{Q}$ is a proyective $\mathbb{Q}G$-module, for in the argument above we could have set
$$ f(1) = \sum_{g \in G} \frac{1}{|G|}g $$
and $\epsilon f(1) = 1$. If you are familiar with representation theory of finite groups, this is a disguised form of saying that the regular representation of $G$ over $\mathbb{Q}$ contains a copy of the trivial representation of $G$ over $\mathbb{Q}$.