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If $G$ is a group and $I$ the augmentation ideal $I=Ker(\mathbb{Z}G\rightarrow \mathbb{Z})$ suppose that:

  1. $I$ is a flat (right) $\mathbb{Z}G$-module.
  2. $I$ is a finitely generated (right) $\mathbb{Z}G$-module.

If $J$ is a finitely generated flat (right) $\mathbb{Z}G$-submodule of $\mathbb{Z}G$, Does it follow that there exists an other $L$ finitely generated flat (right) $\mathbb{Z}G$-submodule of $\mathbb{Z}G$ such that $J\oplus L\cong I$ as $\mathbb{Z}G$-module?

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    $\begingroup$ You are asking about finitely generated groups of homological dimension 1. There is a lot of literature on this. $\endgroup$
    – Benjamin Steinberg
    Mar 31, 2020 at 23:01
  • $\begingroup$ Do you mean equality or isomorphism? For instance of $G$ is an infinite cyclic group, then the augmentation ideal is free on one generator. If we view the ring as Laurent polynomials in the variable $x$ then the augmentation ideal is generated by $x-1$. If I consider the ideal generated by $(x-1)^2 $ it is free on one generator but is not literally a direct summand in the augmentation ideal but it is isomorphic as a module. $\endgroup$
    – Benjamin Steinberg
    Mar 31, 2020 at 23:21
  • $\begingroup$ @BenjaminSteinberg i do mean isomorphism! $\endgroup$
    – ABC
    Apr 1, 2020 at 10:52
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    $\begingroup$ Do you know the answer for free groups? $\endgroup$
    – Benjamin Steinberg
    Apr 1, 2020 at 13:56
  • $\begingroup$ Probably the answer is no. Take G to be a free group of rank 2. The augmentation ideal is then a free module of rank 2. If what you want is true then all flat finitely generated ideals are 2-generated. I suspect there are free submodules of any rank. I would guess the kernel of the projection to the group ring of Z/nxZ/n is free and the rank should be finite and increasing in $n$. $\endgroup$
    – Benjamin Steinberg
    Apr 1, 2020 at 20:16

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