This is from Neukirch's Class field theory - Bonn Lectures.
Let $A$ be a $G$ module. Given $G$ we have an exact sequence of $G$ modules
$\cdots\xrightarrow{d_3} X_2\xrightarrow{d_2} X_1\xrightarrow{d_1} X_0\xrightarrow{d_{0}} X_{-1}\xrightarrow{d_{-1}} X_{-2}\xrightarrow{d_{-2}} \cdots$
Where $X_q$ for $q\geq 1$ is a free $\mathbb{Z}G$ module generated by elements of $G\times G\times \cdots \times G~(q\rm{~times})$
Set $X_{-q}=X_{-q-1}$ for $q\leq -2$ and $X_0=X_{-1}=\mathbb{Z}G$
For a $G$ module $A$ with $A_q=\rm{Hom}(X_q,A)$ we have a complex
$\cdots\rightarrow A_{-2}\xrightarrow{\partial _{-1}} A_{-1}\xrightarrow{\partial _{0}} A_{0}\xrightarrow{\partial _{1}} A_{1}\xrightarrow{\partial _{2}} A_{2}\rightarrow \cdots$
We have $H^{-1}(G,A)=Ker(\partial_0)/Im(\partial _{-1})$
I do not want to write all the maps $d_i$ but i will write only maps which are needed in this case..
We consider $\rm{Hom}(X_{-2},A)\xrightarrow{\partial_{-1}}\rm{Hom}(X_{-1},A)\xrightarrow{\partial_{0}}\rm{Hom}(X_{0},A)$
See that $\rm{Ker} ~\partial_0=\{x:X_{-1}\rightarrow A: \partial_0(x)=0\}$
See that $\partial_0(x)=x\circ d_0$. This as a map is zero when $\partial (x)(1)=0$ i.e., when $x(d_0(1))=0$
$$x(d_0(1))=x\left(\sum_{g\in G}g\right)=\left(\sum_{g\in G}g\right) x(1)$$
See that $x(1)\in A$. So, with notation $N_G=\sum_{g\in G}g$ we have $$\rm{Ker} ~\partial_0=\left\{x:X_{-1}\rightarrow A: \left(\sum_{g\in G}g\right) x(1)=0\right\}=\{a\in A : N_Ga=0\}$$
Notation in this book is $\{a\in A : N_Ga=0\}=_{N_G}A$
I am stuck while determining the image of $\partial_{-1}$
See that $X_{-2}=X_{1}=\bigoplus_{\sigma\in G} \mathbb{Z}G (\sigma)$
Elements in $X_{-2}$ are $\{x: G\rightarrow A\}$ Image of $\partial_{-1}$ is $$\{\partial_{-1}(x): (x: G\rightarrow A)\}=\{x\circ d_{-1} : \mathbb{Z}G\rightarrow A\}$$
See that $$x(d_{-1})(1)=x(d_{-1}(1))=x\left(\sum_{\sigma\in G}\sigma^{-1}(\sigma)-(\sigma)\right)=\sum_{\sigma\in G}\sigma^{-1}x(\sigma)-x(\sigma)$$
I am not able to compute the Image and thus could not compute the cohomology group.
It has been stated that Image is $I_G A=\{\sum_{\sigma\in G}n_{\sigma}(\sigma a_{\sigma}-a_{\sigma}):a_{\sigma}\in A\}$
Help me to understand this better.
In the notes they identify $Hom_G(\mathbb{Z}[G], A)$ with $A$ via the identification $x \mapsto x(1)$,and the identify $Hom_G(\mathbb{Z}[G](\sigma_1), A)$ with $Hom_{Set}(G, A)$ via the identification $x \mapsto (\sigma \mapsto f((\sigma)))$. They determine the image of $\partial_{-1}$ under this identification.
Now you computed that
$(\partial_{-1}(x))(1) = (x \circ d_{-1})(1)=x(d_{-1}(1))=x\left(\sum_{\sigma\in G}\sigma^{-1}(\sigma)-(\sigma)\right)=\sum_{\sigma\in G}\sigma^{-1}x(\sigma)-x(\sigma)$.
Hence every element in $Im(\partial_{-1})$ is of the form $\sum_{\sigma\in G} \sigma^{-1}a_\sigma-a_\sigma$, for some $a_{\sigma} \in A$.
Now consider $d_{a,\sigma} \in Hom_{Set}(G,A)$ given by
$d_{a,\sigma}(\tau) = \begin{cases} 0 & \tau \neq \sigma^{-1} \\ a & \tau = \sigma^{-1} \end{cases}$
Then $d_{a,\sigma^{-1}} \mapsto \sigma a - a$, and hence every element of the form $\sum_{\sigma \in G} n_{\sigma} (\sigma a_{\sigma} - a_{\sigma})$ is in the image.