Groups of finite cohomological dimension (over Z) are torsion-free

Question

As the title, I run into the statement and the material tells me it's obvious. But I can't justify it. (I know that finite cyclic groups have infinite cohomological dimension.)

Answer 1

We can do this by a similar argument to the one presented here. If $G$ is a group of finite cohomological dimension, then (either by definition or by a small argument) $\mathbb{Z}$ has a finite projective resolution $P_{\bullet}$ as a $\mathbb{Z}[G]$-module. If $H \le G$ is a subgroup of $G$ then the restriction of this resolution to $H$ remains a finite projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[H]$-module (because projective modules are direct summands of free modules and $\mathbb{Z}[G]$ is free as a $\mathbb{Z}[H]$-module), so it follows that $H$ also has finite cohomological dimension, and more generally that

$$\text{cd}(H) \le \text{cd}(G).$$

So if $G$ has torsion then it contains a subgroup of infinite cohomological dimension and hence must itself have infinite cohomological dimension.