# Natural action of $\mathbb{Z}G$ on $\mathbb{Z}$?

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).

• I think the main issue here is that you are looking for a non-trivial action. The map $\epsilon$ is probably the counit which indeed corresponds to the trivial action of $G$ on $\mathbb{Z}$ (not that no groups have non-trivial actions, it is just that not all do, and in this case it really is hinting at using the trivial action). – Tobias Kildetoft Apr 22 '16 at 20:45
• @TobiasKildetoft I see. It's just that the trivial action seems so.. trivial. If I say something is a $X$-module with trivial action, then why does it matter that it is a $X$-module? Anything could be $X$-module with trivial action. In my case, I thought I might just look $\epsilon$ as a map of groups, and not modules, since the "module" part would be simply a map from the action of $G$ on $\mathbb{Z}G$ to zero. Also, I never heard of the counit map. Is it another name for augmentation, or something completely different? – Henrique Augusto Souza Apr 22 '16 at 20:56
• Yes, it is another name for the augmentation (it is called the counit when the ring is considered as a Hopf algebra). Note that having a trivial action is not something all rings do (note that the trivial action is not the same as all elements acting trivially, just the the group elements do). – Tobias Kildetoft Apr 22 '16 at 21:06

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 $\mathbb{Q}$.

• The last comment is a bit misleading, as it also holds over the integers. The map just does not split the augmentation in that case. Good answer though. – Tobias Kildetoft May 20 '16 at 16:58
• You are correct. Maybe I should add here that even though the trivial representation is also a subrepresentation of regular representation over the integers, subrepresentations are not necessarily direct summands in this case. What is different for the rationals (or any field of characteristic zero) is that subrepresentations are direct summands. – Goa'uld May 20 '16 at 17:05
• @Goa'uld please help me to solve this problem math.stackexchange.com/questions/3025927/… thanks – neelkanth Dec 4 '18 at 18:32