$G$ is a finitely presented group (but not a finite group), and $\mathbb{Z}G$ is the corresponding group ring. 
$I$ is the kernel of the augmentation morphism $\mathbb{Z}G\rightarrow \mathbb{Z}$.


Is $I$ (always) a finitely generated $\mathbb{Z}G$-module (let say right module). 



augmentation ideal is always finitely generated?
The augmentation ideal is finitely generated as a left (or right) ideal if and only if the group is finitely generated. It is obvious the augmentation ideal is generated as an abelian group by all elements of the form $g-1$ with $g\in G$. Then from the computation $ab-1=a(b-1)+a-1$ one easily deduces by induction on word length that if $S$ generates $G$, then the elements of the form $s-1$ with $s\in S$ generates the augmentation ideal as a left module. 


For the converse, note that if the augmentation ideal is finitely generated as a left ideal, then there must be some finite subset $S$ of $G$ such that the elements $s-1$ with $s\in S$ generate the augmentation ideal (just have $S$ be the elements you need to write your finite generating set in terms of the $g-1$ generating set). Now if one takes the right Cayley graph of $G$ with respect to the set $S$, then the augmentation ideal is precisely the image of the boundary map from $1$-chains to $0$-chains and so the Cayley graph has vanishing reduced homology in dimension $0$ and hence is connected. Thus $S$ generates the group.


