# Special groups, special resolutions and group cohomology

$$\newcommand{\Z}{\mathbf{Z}}$$ 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.

• 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 $0\rightarrow R\xrightarrow{\ \times (t-1)\ }R\rightarrow \mathbb{Z}\rightarrow 0$.
– abx
Sep 29, 2022 at 16:02
• @abx oups, I will edit my question, to avoid commutativity. Thanks for your remark.
– GSM
Sep 29, 2022 at 16:04
• 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).
– Z. M
Sep 29, 2022 at 16:21
• @Z.M let say that free module in this question is given by a finite direct sum of Z[G]. The multiplication on the right and on the left are the obvious ones.
– GSM
Sep 29, 2022 at 16:24
• @GSM I just realised that my answer doesn’t work. If you unaccept it then I’ll delete it. Oct 2, 2022 at 21:24