If $G$ is a group and $M$ a $G$-module then for $n\geq 0$ we have an action of $G$ on the cochains from $C^n(G,M)$. If $s\in G$, $a\in C^n(G,M)$ then $(sa)_{s_1,\ldots,s_n}=sa_{s^{-1}s_1s,\ldots,s^{-1}s_ns}$, $\forall s_1,\ldots,s_n\in G$.
This action induces an action of $G$ on $H^n(G,M)$, which is known to be trivial. (See, e.g., Brown, Chapter III, Proposition 8.3.)
I'm interested in the following identity.
Let $a\in C^n(G,M)$ and $s\in G$ and let $b\in C^{n-1}(G,M)$, where $$b_{s_1,\ldots,s_{n-1}}=\sum_{k=0}^{n-1}(-1)^ka_{s_1,\ldots,s_k,s,s^{-1}s_{k+1}s,\ldots,s^{-1}s_{n-1}s}~\forall s_1,\ldots,s_{n-1}\in G.$$ Then we have $sa-a-db=c$, where $$c_{s_1,\ldots,s_n}=\sum_{k=0}^n(-1)^kda_{s_1,\ldots,s_k,s,s^{-1}s_{k+1}s,\ldots,s^{-1}s_ns}~\forall s_1,\ldots,s_n\in G.$$
As a consequence, if $a\in Z^n(G,M)$ then $da=0$ so $c=0$ so in $H^n(G,M)$ we have $[sa-a]=[sa-a-db]=[c]=[0]$ so $s[a]=[a]$ so the action of $G$ on $H^n(G,M)$ is trivial.
The proof is straightforward, but it's quite annoying to write it down. I checked it for $n\leq 3$ and it is quite easy to see the pattern. My question is whether somebody saw this somewhere. I looked for it in a couple of books, but didn't find it.
In the paper I'm writing I only need the cases $n=1$ and $2$, with $G$ commutative, so those annoying $s^{-1}s_is$ are replaced by $s_i$. This is easy to write down, only a few lines, but I would rather quote the general result, provided it is written somewhere.
