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$G$ 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).

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    $\begingroup$ It is finitely generated when G is. If S generates G then the s-1 with s in S generate the augmentation ideal. $\endgroup$ – Benjamin Steinberg Mar 29 at 23:01
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    $\begingroup$ Conversely if the augmentation ideal of a group ring is finitely generated so is the group. Some finite set of elements of the form g-1 must generate the augmentation ideal and by looking at the Cayley graph with respect to that finite set you can see it is connected and hence these elements generate. $\endgroup$ – Benjamin Steinberg Mar 29 at 23:06
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    $\begingroup$ @MarkSapir they generate as a left or right ideal. It is easier to see as a left ideal. Then the left ideal is the image of the boundary map from 1-chains to 0-chains for the right Cayley graph $\endgroup$ – Benjamin Steinberg Mar 29 at 23:20
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    $\begingroup$ Or algebraically ab-1=a(b-1)+ a-1 shows how to inductively build g-1 from an expression of g as a product of generators $\endgroup$ – Benjamin Steinberg Mar 29 at 23:24
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    $\begingroup$ Said differently the augmentation ideal is finitely generated iff the group is FP1 which is the same as finitely generated. $\endgroup$ – Benjamin Steinberg Mar 29 at 23:25
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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.

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  • $\begingroup$ Don't you also need the equality $(a-1)b=(a-1)(b-1)+(a-1)$ so that the right ideal is inside the left ideal? $\endgroup$ – user6976 Mar 30 at 1:28
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    $\begingroup$ @MarkSapir, the ideal is generated as an abelian group by the g-1 so my argument gets the entire ideal $\endgroup$ – Benjamin Steinberg Mar 30 at 1:35

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