inner automorphism induces identity map on group cohomology I'm taking a course in group cohomology,  and the professor taught the functorial property of the construction of cohomology.  he mentioned that an inner automorphism induces exactly the identity map over cohomology groups; he then said we can easily use this fact to obtain similar facts for cohomology of profinite groups (discrete modules). 
I guess this is a trivial fact but I'm a beginner and have no idea with this.  can anyone tell me is this a theorem?  or as an exercise can anyone give me some hint?  thanks! 
 A: If you are studying group cohomology from the point of view of topology, you are probably used to writing $H^n(G,A)$ where $A$ has a trivial $G$-action (often even $A=\mathbb{Z}$). If you are studying it from an arithmetical point of view (say in the context of class field theory) then usually $A$ will be a non-constant $G$-module. The first setting is of course a special case of the second one, so I will assume that you are using non-trivial modules, and you can adapt the statements for the simpler case.
To induce a map $H^n(G,A)\to H^n(H,B)$, you need a group morphism $\varphi: H\to G$ and a group morphism $f:A\to B$ which are compatible, meaning that $f(\varphi(h)a) = hf(a)$. If $G=H$ and $A=B$, then for any $g\in G$ you get an automorphism $\psi_{g,A}^n: H^n(G,A)\to H^n(G,A)$ by taking $\varphi:G\to G$ to be conjugation by $g$, and $f:A\to A$ is given by $a\mapsto ga$. Of course if $A$ is a constant module then $f$ is just the identity.
Note that this is functorial in the sense that if $A\to B$ is any $G$-morphism, you get a commutative square
$$\require{AMScd}
\begin{CD}
H^n(G,A) @>{\psi_{g,A}^n}>> H^n(G,A)\\
@VVV @VVV \\
H^n(G,B) @>{\psi_{g,B}^n}>> H^n(G,B).
\end{CD}$$
Now to prove that $\psi_{g,A}^n$ is trivial for all $A$, $g$ and $n$, one possible way is to proceed by induction on $n$. It is of course essentially trivial for $n=0$ (I leave you to check that). Now you can always embed $A$ into some module $I$ such that $H^r(G,I)=0$ for all $r>0$ (for instance an injective module). The short exact sequence
$$0\to A\to I\to B\to 0$$
of $G$-modules (where $B=I/A$) gives the commutative diagram with exact lines:
$$\require{AMScd}
\begin{CD}
0 @>>> H^n(G,B) @>>> H^{n+1}(G,A) @>>> 0\\
@VVV @V{\psi_{g,B}^n}VV @V{\psi_{g,A}^{n+1}}VV @VVV\\
0 @>>> H^n(G,B) @>>> H^{n+1}(G,A) @>>> 0.
\end{CD}$$
Now by induction $\psi_{g,B}^n$ is the identity, and the middle horizontal morphisms, which are the same on both rows, are isomorphisms by exactness, so $\psi_{g,A}^{n+1}$ has to be the identity.
Note that for $n=0$ here this does not exactly work since you only get the zero on the right, but actually surjectivity of the maps $H^n(G,B)\to H^{n+1}(G,A)$ is enough to make the proof work.
