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$\newcommand{\Z}{\mathbf{Z}}$ Let $G$ be a non-abelian group. And let $\Z$ be the ring of integers. Under which condition on the group $G$ can we find a free resolution $F_{\bullet}\rightarrow \Z$ of $\Z$ as left $\Z[G]$-module satisfying the conditions ?

  1. $F_{\bullet}\rightarrow \Z$ is a finite resolution I.e. each $F_{i}$ is finitely generated free left $\Z[G]$-module. And $F_{i}=0$ for $i$ enough big.
  2. $F_{\bullet}\rightarrow \Z$ is also a resolution of $\Z[G]$-bimodules. I.e. for each $i$, $F_{i+1}\rightarrow F_{i}$ and $F_{0}\rightarrow \Z$ are maps of $\Z[G]$-bimodule.

In this question each free $\Z[G]$-module is seen as $\Z[G]$-bimodule in the obvious way.

As far as I know, there is a big class of groups satisfying condition 1. I will be happy if one can provide a non-abelian example satisfying 1 and 2. Or a proof that such non-abelian group does not exist.

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    $\begingroup$ No. Take $G=\mathbb{Z}$. Then $\mathbb{Z}[G]$ is the ring $R=\mathbb{Z}[t,t^{-1}]$ of Laurent polynomials, and $\mathbb{Z}$ admits the resolution $0\rightarrow R\xrightarrow{\ \times (t-1)\ }R\rightarrow \mathbb{Z}\rightarrow 0$. $\endgroup$
    – abx
    Commented Sep 29, 2022 at 16:02
  • $\begingroup$ @abx oups, I will edit my question, to avoid commutativity. Thanks for your remark. $\endgroup$
    – GSM
    Commented Sep 29, 2022 at 16:04
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    $\begingroup$ How do you view a free left $R:=\mathbb Z[G]$-module as a $R$-bimodule? Note that the "obvious way" is not well-defined since it depends on how you see a module to be free. For example, let $M$ be the free left $R$-module $R$ of rank 1. For a unit $u\in R$, the right multiplication $u\colon R\to M$ is another way to see $M$ as a free $R$-module, while the "obvious" right multiplications do not coincide (if $u$ is not in the center). $\endgroup$
    – Z. M
    Commented Sep 29, 2022 at 16:21
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    $\begingroup$ @GSM I just realised that my answer doesn’t work. If you unaccept it then I’ll delete it. $\endgroup$
    – Jeremy Rickard
    Commented Oct 2, 2022 at 21:24
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    $\begingroup$ @JeremyRickard Let $\Delta$ denote the set of elements of $G$ that have finitely many conjugates: this is a characteristic subgroup and is (locally finite)-by-abelian. Since our group $G$ in the question has finite cohomological dimension over $\mathbb Z$, it follows $\Delta$ is torsion-free abelian of finite rank. So some subgroup $H$ of finite index in $G$ centralises $\Delta$. Now your original line of reasoning works for $H$ so I think you successfully show $G$ must be virtually abelian. That is an interesting reduction. $\endgroup$
    – Peter Kropholler
    Commented Dec 15, 2022 at 18:56

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