Show that the categories $G$-mod and $\mathbb{Z}G$-mod are equivalent.

Question

I have another basic question inspired from reading the sixth chapter of Weibel's "An Introduction to Homological Algebra".

First version of the question: a bit ambiguous At the first paragraph, there is written that the category of $G$-mod can be identified with the category of $\mathbb{Z}G$-mod, but the isomorphism is not so evident at my eyes. Please, could you help me in finding it?

A "corollary" of my main question, is the following another question. Let consider the ring $\mathbb{Z}$ as a trivial $G$-module (yes, there is a connection with a previous question I asked here - A clarification about the meaning of "Let $\mathbb{Z}$ be *the* trivial $G$-module".). Now, let's see $Z$ as a $\mathbb{Z}G$-module using the previous identification.

Does $\mathbb{Z}$ become here a trivial $\mathbb{Z}G$-module?

Second version of the question: let's be more precise! It seems that my question could be a bit ambiguous. Indeed we can interpret the writing "$\mathbb{Z}G$-module" in two ways, because $\mathbb{Z}G$ is a group and a ring, and the consequently question "Does $\mathbb{Z}$ become here a trivial $\mathbb{Z}G$-module?" assumes two quite different meaning. Let me give the opportunity to clarify what I am searching for.

1. First interpretation. We can think at $\mathbb{Z}G$-mod as the category $G$-mod (https://en.wikipedia.org/wiki/G-module) choosing as group $G= \mathbb{Z}G$. In this context, the two categories $G$-mod and $\mathbb{Z}G$-mod are equivalent as essentially shown by user D. Burde (thanks! :) ). Furthermore, the property "being a trivial $\mathbb{Z}G$-module" here means "to be an abelian group on which $\mathbb{Z}G$ acts trivially". Under this interpretation, my answer mean: if we consider $Z$ as $G$-module and then wee see it as a $\mathbb{Z}G$-module (and we can because the two categories are equivalent), does $Z$ continue to be a trivial $\mathbb{Z}G$-module? As explained still by D. Burde, the answer is yes and the problem is solved.

2. Because $\mathbb{Z}G$ is also a ring, we can consider the category $\mathbb{Z}G$-mod in the sense "an ordinary $R$-mod choosing as ring R=$\mathbb{Z}G$ (in this ordinary sense - https://en.wikipedia.org/wiki/Module_%28mathematics%29 - just to avoid any kind of misunderstanding). Now, the books of Weibel claims that the two categories $G$-mod and $\mathbb{Z}G$-mod are equivalent. I am sure Weibel refers to this kind of interpretation, because we want to use this fact to guarantee for the category $G$-mod the existence of enough projectives and injectives (a property true in every $R$-mod category). Consequently my true question is:

   How can I find an equivalence between these two categories?

As in the previous "version" of the answer, a "corollay-question" is: let's consider $\mathbb{Z}$ as a trivial $G$-mod, and then see $\mathbb{Z}$ as a $\mathbb{Z}$G-module (of course, second interpretation).

How is defined the multiplication of $\mathbb{Z}G$ on $\mathbb{Z}$?

On a first time, I was expecting to find again a sort of "trivial module", but another post of mine (What exactly is a trivial module?) seems to suggest something different.

Finally, please notice how essentially the entirely chapter 6 of Weibel's book (the treatment about Group (Co)homology) is "based" on the fact that $G$-mod has enough projectives/injectives, and to find the equivalence I am asking for should be the main way to prove this fact (as suggested by the author himself). Consequently, let me underline again how to understand this point is absolutely important from my point of view to feel safe every time I speak about projective resolutions in $G$-mod (just to say an example) :-)

I hope you appreciate my effort to express my motivation, and of course thank you very much for any kind of helps! Cheers

Answer 1

The intended interpretation is absolutely that a $G$-module $M$ becomes a module over the ring $\mathbb{Z} G$. As a group, $\mathbb{Z} G$ is simply a free abelian group of the same rank as the order of $G$, so that structure remembers nothing about how $G$ acted on $M$.

As to the trivial $G$ module $\mathbb{Z}$, the action of $\mathbb{Z} G$ is $(a_1 g_1+...+a_ng_n) m=a_1(g_1m)+...+a_n(g_n m)=a_1m+...+a_nm=(a_1+...+a_n)m$. While this actuion is more or less trivial, one does not call $M$ a "trivial" $\mathbb{Z}G$ module. Wolfram Mathworld-an inconsistent resource at best-to the contrary, there is no action of an arbitrary ring on an arbitrary abelian group, so there is nothing deserving the general title of trivial module.

EDIT Looks like you still need a construction of the isomorphism between $G$ modules and $\mathbb{Z}G$ modules. Given $M\in G-\underline{Mod}$, define $M'\in \mathbb{Z}G-\underline{Mod}$ by the action $(\sum a_ig_i)m=\sum a_i(g_i m)$, where $\sum a_i g_i\in \mathbb{Z} G,m\in M, g_i$ acts on $m$ via the $G$-module structure, and $a_i$ acts on $g_im$ by repeated addition. In the other direction, given $N\in\mathbb{Z}G-\underline{Mod}$ define $N'\in G-\underline{Mod}$ by $gn=gn$, where on the right hand $g$ is an element of $\mathbb{Z}G$ and on the left hand an element of $G$. I leave it to you to check $M'$ and $N'$ are modules with the given structures. Given a map $f:M_1\to N_1$ of $G$-modules, we get a map $f':M_1'\to N_1'$ of $\mathbb{Z}G$-modules by $f'(m)=f(m)$, and similarly in the other direction. It's obvious from the definition that these operations are mutually inverse, since I'm literally using the same functions on both sides, so this gives the desired isomorphism of categories.

Answer 2

The identification is as follows. Let $M$ be a $G$-module. Then $M$ becomes a $\mathbb{Z}[G]$-module by $$ \left( \sum_gn_g g\right)m=\sum_g n_g(gm). $$ Conversely if $M$ is a $\mathbb{Z}[G]$-module, then $M$ becomes a $G$-module by $$ gm=(1g)m. $$ Since $R=\mathbb{Z}[G]$, viewed as a left $R$-module, is a projective object , $\mathbb{Z}[G]$-mod, and hence $G$-mod, is an abelian category with enough projectives.