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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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    @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?2011-05-18

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