Conjugation action group cohomology I have a question from Lang's Algebra (chapter twenty ex 6d). I think the vagueness confuses me as I am not even sure where to start
If $H$ is a normal subgroup of $G$, we have the cohomology groups $H^i (H,A)$. 
The question asks to describe how we can get an operation of $G$ on the cohomology function $H_G$ "by conjugation" and functorality.
Any advice or direction on what this operation would be or how to look for it would be greatly appreciated. 
 A: This is explained succinctly in Ken Brown's classic text Cohomology of Groups, pg48.
You have your projective resolution $F\to\mathbb{Z}$ for your group in question, and you just need to check that the chain map $\tau:F\to F$ given by $\tau(x)=g\cdot x$ is $G$-equivariant and gives you your action (fixed $g\in G$). Then it induces a well-defined map $\tau:F_G\to F_G$ which induces the desired map on homology.
A: Let $C^i(H,A)$ be the group of $i$-cochains, i.e., functions $H^i\rightarrow A$, which define the complex that computes the cohomology $H^i(H,A)$. For $g\in G$, you have the automorphism $\varphi_g:h\mapsto ghg^{-1}:H\rightarrow H$ because $H$ is normal in $G$, and you also have the abelian group automorphism $\psi_g:a\mapsto g^{-1}a:A\rightarrow A$, which is not usually $G$-equivariant, but is compatible with $\varphi_g$ in the sense that $\psi_g(\varphi_g(h)a)=h\psi_g(a)$. These maps give rise to maps $f\mapsto\psi_g\circ f\circ\varphi_g^i:C^i(H,A)\rightarrow C^i(H,A)$ which are compatible with the coboundary maps (here $\varphi_g^i:H^i\rightarrow H^i$ is the $i$-fold product of the map $\varphi_g$ with itself). So these maps descend to maps on cohomology $g^*:H^i(H,A)\rightarrow H^i(H,A)$ for all $i$. Then you get an action of $G$ on $H^i(H,A)$ by $g\cdot\kappa:=g^*(\kappa)$ (I may have the signs switched for a left action, i.e., you might need $\varphi_{g^{-1}}$ and $\psi_{g^{-1}}$ instead, but that works the same way). For example, on $H^0(H,A)=A^H$, this action is just given by $a\mapsto g^{-1}a$, so in particular, $H$ acts trivially on $H^0(H,A)$. Using an inductive argument with long exact sequences, it can be shown that $H$ acts trivially on $H^i(H,A)$ for all $i$, so you really have an action of $G/H$ on $H^i(H,A)$. This is important for defining the Hochschild-Serre spectral sequence $H^p(G/H,H^q(H,A))\Rightarrow H^{p+q}(G,A)$. 
