In all that follows, let $k$ be a field and $G$ be a finite group.

It is well-known that the order of $G$ is invertible in $k$ iff the group ring $k[G]$ is semisimple, which is equivalent, *inter alia*, to the fact that $\operatorname{Ext}^1_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules (= $k$-linear representations of $G$), or indeed, $\operatorname{Ext}^i_{k[G]}(V,W)$ for all $i\geq 1$, viꝫ. $k[G]$ has (left) global dimension zero.

Today I learned that this is also equivalent to the (*a priori* weaker) condition that $k[G]$ be (left-)hereditary, which is equivalent, *inter alia*, to the fact that $\operatorname{Ext}^2_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules. (See, e.g., Dicks, “Hereditary Group Rings”, *J. London Math. Soc.* **20** (1979) 27–38, theorem 1.)

This suggests the following question: what can be said about $G$ and $k$ if $\operatorname{Ext}^{d+1}_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules for a given $d\geq 2$? In other words, if we assume $k[G]$ has (left) global dimension $\leq d$? Does $k[G]$ having finite global dimension imply that the order of $G$ is invertible in $k$?