Finite-group cohomology is annihilated by the group order

If $G$ is finite and $n>0$, multiplication by $|G|$ annihilates $H^n(G,M)$. Indeed, restriction to the trivial subgroup is zero in positive degree, while the restriction-corestriction composite is multiplication by $|G|$.

Let $K$ be a finite-index subgroup of a group $G$ and $M$ a $G$-module. The restriction map in group cohomology followed by the corestriction map in group cohomology satisfieson $H^n(G,M)$. The corestriction map in group cohomology sums the coset contributions, and the inner-conjugation maps on the original group cohomology are the identity.

The opposite composite has a different domain. When $K$ is normal, ${\mathrm{Res}}_K^G{\mathrm{Cor}}_K^G$ on $H^n(K,M)$ is the sum of the conjugation actions of the cosets in $G/K$. Those actions need not be trivial. For finite $G$ and $K=1$, positive-degree group cohomology of $K$ vanishes, so the first composite proves that $|G|$ annihilates $H^n(G,M)$ for $n>0$.