Timeline for A formula that proves that $G$ acts trivially on $H^*(G,M)$

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Oct 12, 2019 at 21:39	comment	added	Constantin-Nicolae Beli		This identity holds for arbitrary cochains, not only for cocycles. (As I wrote, $a\in C^n(G,M)$.) As a consequence, in the particular case when $a\in Z^n(G,M)$, we get $sa-a-db=c=0$ so $sa=a+db$, which implies $s[a]=[a]$, so we get an alternative proof of the fact that $G$ acts trivially on $H^n(G,M)$. However, this is not the reason why I need this result. I need it for arbitrary cochains, not for cocycles. (And only for $n=1$ or $2$ and $G$ commutative.) I know how to prove it, but if it is already somewhere, I would rather quote it than write it down. (It's a reference request.)
Oct 12, 2019 at 11:59	comment	added	Mark Wildon		Please could you clarify what your question is: it must be to prove an identity in cocycles (modulo coboundaries?). But on my reading, you seem to be asking for a proof that $G$ acts trivially on $H^n(G,M)$, which as you say, is in Brown.
Oct 11, 2019 at 20:16	history	asked	Constantin-Nicolae Beli	CC BY-SA 4.0