Let $G$ be a group and $H$ be a normal subgroup of $G$. Let $M$ be $G$-module. Let $H^1(H,M)$ be first group cohomology.
$G$ acts on $H^1(H,M)$ by $(\sigma*f)(g):=gf({\sigma}^{-1}g{\sigma})$・・・①.
My question is, how $G/H$ acts on $H^1(H,M)$ ?
My thought : Let $\sigma \mathrm{mod}H\in G/H$. Action ① is trivial for $H$ modulo coboundary.In other words, if $\sigma_1≡\sigma_2 \text{mod} H, (\sigma_1 \text{mod}H)*X) (f)(g)-(\sigma_2 \text{mod}H)*X) (f)(g) \in B^1(H,M)$.
(See Ehud Meir's comment in https://mathoverflow.net/questions/212636/the-term-h1n-ag-n-in-the-inflation-restriction-exact-sequence), so we can define $(\sigma \text{mod}H *f)(g):=\sigma f({\sigma}^{-1}g{\sigma})$. But I often see $\sigma \in G/H$ acts on $H^1(H,M)$ by just $(\sigma*f)(g):=gf({\sigma}^{-1}g{\sigma})$(For example, see https://mathoverflow.net/questions/461065/tate-shafarevich-group-and-sigma-phic-phi-sigmac-for-all-c-in-oper).Is it correct to understand that the explanation here is omitted, and it is simply written as if $\sigma$ is acting directly?
