I am currently reading Cohomology of Groups by Brown and I am stuck on page 58. They do an example of the computation of the cohomology group of a finite cyclic group $G=\langle t\rangle$ of order $n$.
My first question is a about the free resolution (I.6.3):
I see the maps $t-1$ and $N=1+t+\ldots+t^{n-1}$ as endomorphisms of $M$. How do these precisely induce maps $\mathbf{Z}[G]\longrightarrow\mathbf{Z}[G]$? The natural way would be to define something like $\sum_{g\in G} a_g g\longmapsto \sum_{g\in G}a_g g(t-1)$. However, this does not seem correct, (or maybe it is an enormous abuse of notation?), because $t-1$ is not an element of $G$.. I thought of the identification $\mathbf{Z}[G]\cong \mathbf{Z}[X]/(X^n-1)$, but this boggles my mind, like the fact that this sequence is exact. We have to prove that $\operatorname{Ker}(t-1)=\operatorname{Im}N$? But how is this defined? And why aren't all the cohomology groups trivial, as the $t-1$ and $N$ stay the same in the second sequence?
The map $\mathbf{Z}[G]\longrightarrow \mathbf{Z}$ he writes down is the augmentation map $\sum_{g\in G} a_g g\longmapsto \sum_{g\in G}a_g$. The kernel $I_G$ of $\pi$ is the subgroup of $\mathbf{Z}[G]$ generated by all elements of the form $i_g=g-1$ where $g$ runs over $G$.
Why is the image of $t-1$ equal to $I_G$? The inclusion $\subset$ is clear, but I don't see why the other inclusion is true.
Any help is appreciated.
The maps $\mathbb{Z}[G] \to \mathbb{Z}[G]$ are given by multiplication, so $$ \sum a_i g_i \mapsto (t-1) \sum a_i g_i. $$ and similarly for the norm. You are unhappy that $t - 1$ isn't in $G$, but it's unquestionably in the ring $\mathbb{Z}[G]$, so there's no problem.
The fact that the image of $(t-1)$ is $I_G$ is an old trick. As you say, the one containment is clear.
For the other, suppose $\sum a_i t^i$ is in $I_G$. Then $\sum a_i = 0$. Thus $\sum a_i t^i = \sum a_i (t^i - 1)$, and you can factor a $t-1$ out of the right hand side.