For any group $G$, cohomology can be viewed as a functor $$ H^\ast(G,-): G{\sf\text{-}mod}\to {\sf GrAbGrp}, $$ where $G{\sf\text{-}mod}$ denotes the category of (left) $\mathbb{Z}[G]$-modules and ${\sf GrAbGrp}$ denotes the category of (non-negatively) graded abelian groups.

It seems that there are non-isomorphic groups $G_1$ and $G_2$ whose integral group rings $\mathbb{Z}[G_1]$ and $\mathbb{Z}[G_2]$ are Morita equivalent (this is a result of Roggenkamp and Zimmermann). One could then ask the following question:

Do there exist non-isomorphic groups $G_1$ and $G_2$ for which there is an equivalence of categories $F: G_1{\sf\text{-}mod}\to G_2{\sf\text{-}mod}$ such that the functors $H^\ast(G_1,-)$ and $H^\ast(G_2,-)\circ F$ from $G_1{\sf\text{-}mod}$ to ${\sf GrAbGrp}$ are naturally isomorphic?

This is my (possibly naive) attempt to formalise the question in the title. It may be that the answer is trivially "no" by looking at $H^0$, ie the functor of coinvariants. In that case, I would like to know if there are other formulations for which the question becomes interesting.

`{\sf-mod}`

with`{\sf\text{-}mod}`

. Even when using the sans-serif font, the symbol`-`

in math mode is interpreted as a minus sign. Compare $G{\sf-mod}$ and $G{\sf\text{-}mod}$. $\endgroup$ – Zev Chonoles Nov 6 '13 at 14:32