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Let $G$ be a finite group, and $V$ a (say complex) finite dimensional representation of $G$.

Let me view $V$ as a $G$-module in the obvious way. Is it true that $$H^n(G;V)=0$$ for $n\geq 1$? I suspect that the answer is no, but I haven't been able to find a counterexample. If the answer (surprisingly) turns out to be yes, I would love to see a proof!

JeCl
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    Yes, it is true. One can prove it via "transfer" from the trivial subgroup. You can find a proof for instance in K.Brown "Cohomology of groups" or in other books on homological algebra. – Moishe Kohan Feb 13 '24 at 17:03
  • Oh very cool! For the precise reference, look at corollary 10.2 of chapter 3 of K. Brown's book. – JeCl Feb 13 '24 at 18:30

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