In the theory of cohomology of groups, I came across some problem. It might be silly question, but I get puzzled around it long time.
Let $G$ be a group and $A$ abd $B$ be abelian group with action of $G$ on each of them.
Then $A$ and $B$ are $G$-modules (or $\mathbb{Z}[G]$ modules).
Consider the set $\mbox{Hom}(A,B)$ of homomorphisms of abelian group $A$ into $B$. Given action of $G$ on this set by (given $x\in G$ and $a\in A$) $$f^x(a):=(f(a^{x^{-1}}))^x \hskip1cm (*).$$ My question is simple, I don't know where is my fault of understanding.
Can't we consider the action by $$f^x(a):=(f(a))^x\hskip1cm (**)$$ or $$f^x(a):=f(a^x) \hskip1cm (***)?$$ In these cases, the action of $G$ is essentially considered only on one module. But What is the reason to consider the action of $G$ on $\mbox{Hom}(A,B)$ as in ( * ) in the Cohomology of groups? Also, why inverse comes in (*)?
In the classroom, I was writing in notebook the action of $G$ as in $(***)$, but as soon as the speaker wrote $(*)$, I puzzled with above question.
A nice thing about this action is that the fixed points $\mbox{Hom}(A,B)^G$ are precisely the $G$-linear homomorphisms $\operatorname{Hom}_G(A,B)$.
Also note that $f^x(a):=f(a^x)$ is a right action not a left action, because under this definition $(f^x)^y(a)=f(a^{yx})=f^{yx}(a)$, that's why you include the inverse.