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We say that a $G$-module $I$ is induced if $$I\cong L\otimes\mathbb{Z}G$$ where $L$ is an abelian group and the action on $L\otimes\mathbb{Z}G$ is given by the action of $G$ only on the second component, so that $$g(l\otimes h)=l\otimes gh$$ Here comes my question: is it true that if $$H^k(G,\mathbb{Z}G)=0$$ then $$H^k(G,I)=0$$ for any induced $G$-module $I$?

Thanks a lot, bye

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What is the motivation for this question? You're evidently looking at infinite groups, since for finite groups induction is the same as coinduction and the Eckmann-Shapiro lemma tells us that $0 = H^{k}(\{e\},L) = H^{k}(G,I)$ for $k \geq 1$ independently of $H^{k}(G,\mathbb{Z}G)$. Generally, I don't see why one would expect induction to interact well with cohomology. –  t.b. May 17 '11 at 14:52
@TheoBuehler Yes, I am looking for infinite groups: I am trying to understand why in Harer's paper "The virtual cohomological dimension of the mapping class group of an orientable surface", Inv.Math. 84, 157-176 (1986), at page 173 he says that in order to verify that a group is a duality group is enough to check that it is of type FP-n and that $H^k(G,\mathbb{Z}G)\neq 0$ for a unique value of k. If you take the original paper by Bieri and Eckmann, you'll see that they require G to have this property for every induced module and not only for $\mathbb{Z}G$. I don't get it... –  fatoddsun May 17 '11 at 22:28
Oh, but this sounds like the Johnson-Wall condition (On groups satisfying Poincare duality, Ann. of Math. 96 (1972), 592–598). Mike Davis has a nice survey on PD-groups (available here), where he outlines a proof why the definitions by JW and by BE are equivalent. But I must admit that I've never gone through that stuff in detail myself. –  t.b. May 17 '11 at 22:41
The book Topological methods in group theory, Springer GTM 243 by Ross Geoghegan contains a short section on PD groups and points to Bieri's notes Homological dimensions of discrete groups, Queen Mary College Mathematical Notes, 1982. Maybe these references contain some insights you're looking for. –  t.b. May 18 '11 at 10:01
@TheoBuelher thanks a lot for your reply. Following the material you suggested I found a nice proof of this fact, but I have still a question... Take a finite projective resolution $0\to P_n\to\ldots\to P_0\to \mathbb{Z}\to 0$ over $\mathbb{Z}G$. According to the text, the hypotheses $H^k(G,\mathbb{Z}G)=0$ for all $k\neq n$ should imply that we can form a new projective resolution $\ldots\to P^*_{n-1}\to P^*_n\to H^n(G,\mathbb{Z}G)\to 0$. Why is it true? –  fatoddsun May 18 '11 at 15:00

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