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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
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1 Answer

up vote 10 down vote accepted

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.

The paper http://www.jstor.org/stable/3062112 shows there are nonisomorphic finite groups with isomorphic integral group rings.

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In fact, it appears that the example in your second link (Martin Hertweck's article "A counterexample to the isomorphism problem for integral group rings") already gives an isomorphism which commutes with the augmentation. I think the relevant statement there is the definition of a group basis in the second phrase of the article. –  Ricardo Andrade Nov 6 '13 at 15:33
@RicardoAndrade, thanks. I hadn't read Hertweck's article. I had heard a talk on integral group rings over a year ago in Brazil stating what the example is and was convinced that this should solve the OP's question. So I did a quick search to see if isomorphism of group rings implies that the augmentation must be preserved and found the first link. –  Benjamin Steinberg Nov 6 '13 at 15:36
Thanks, Benjamin. This seems to answer my question. It now seems obvious that the integral group ring completely determines cohomology. I wasn't aware of this counter-example to the isomorphism problem for integral group rings finite groups (but should have been). By the way, do you know the status of this problem for torsion-free groups? –  Mark Grant Nov 6 '13 at 15:46
@MarkGrant, I'm pretty sure the status of this question was addressed at the talk I attended in Brazil, but I cannot remember what the answer was. Group rings is not really my area but the counterexample and the history leading up to it caught my attention. –  Benjamin Steinberg Nov 6 '13 at 15:58
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