# Do there exist non-isomorphic groups with the same cohomology?

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.

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A minor formatting suggestion: I would recommend replacing {\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 $G{\sf\text{-}mod}$. – Zev Chonoles Nov 6 '13 at 14:32
Thanks Zev, it does look better now. – Mark Grant Nov 6 '13 at 15:03

The answer would seem to be yes if I understood your question properly. The paper http://www.m-hikari.com/ija/ija-password-2009/ija-password9-12-2009/ladraIJA9-12-2009.pdf shows that if $\mathbb ZG$ is isomorphic to $\mathbb ZH$, then one can choose an isomorphism which is augmentation preserving. They then show that if $f\colon \mathbb ZG\to \mathbb ZH$ is the augmentation-preserving isomorphism, then viewing a $\mathbb ZH$-module as a $\mathbb ZG$-module gives isomorphisms in homology and cohomology. I didn't check if their proof gives naturality of the isomorphisms, but I would be surprised if it didn't.