Let $G$ be a finite group of order $n= ord(G)$ and consider the exact sequence provides group extension $E$ by $0 \to M \to E \to G \to 0$ where $M$ has the $G$-module structure.
This gives rise for building (in canonical way) group cohomologies $H^r(G,M), r \ge 1$.
It's obviously clear that because $n= ord(G)$ the canonical multiplication by $n$ map $H^r(G,M) \xrightarrow{*n} H^r(G,M)$ is a zero map.
My question is why is this equivalently to say that $H^r(G,M)\otimes \mathbb{Z}/n\mathbb{Z} =0$?
Here I can't use the trick $1 = n/n$ like by showing that it vannishing by tensoring with $\mathbb{Q}$?