About conjugation map in second cohomology group I am reading the chapter Second cohomology groups of Continuation of the Notas de Matemàtica.
On page 13, the author defines conjugation map.
Let $G$ act on an abelian group $A$ and $H$ be a subgroup of $G$. Fix $g \in G$ and we choose $\alpha \in Z^2(H,A)$.
Define $c_g(\alpha) \in Z^2(gHg^{-1},A)$ by $c_g(\alpha)(x,y) = {^g \alpha(g^{-1}xg, g^{-1} y g)}$ where $x,y \in gHg^{-1}.$
I have problem in showing that $c_g(\alpha) \in Z^2(gHg^{-1},A). $
I need to verify :$$c_g(\alpha)(x,y)\,\,c_g(\alpha)(xy,z) = {^xc_g(\alpha)(y,z)}\,\, c_g(\alpha)(x,yz). $$
The lhs is $$c_g(\alpha)(x,y) c_g(\alpha)(xy,z) = {^g \alpha(g^{-1}xg, g^{-1}yg)}{^g \alpha(g^{-1}xy g, g^{-1} zg)} = {^g (\alpha(g^{-1}xg, g^{-1}yg) \alpha(g^{-1}xy g, g^{-1} zg) )} .$$
Thus, LHS = $${^g ( {^x \alpha(g^{-1}y g, g^{-1} zg)} \alpha(g^{-1} xg,  g^{-1} yz g))} = {^{gx} \alpha(g^{-1} yg, g^{-1}zg)} ^g\alpha(g^{-1}xg, g^{-1}yzg). $$
But the RHS is $${^xc_g(\alpha)(y,z)}\,\, c_g(\alpha)(x,yz) = {^{xg} \alpha(g^{-1} yg, g^{-1}zg)} ^g \alpha(g^{-1}xg, g^{-1}yzg).$$
Here, I am not getting how ${^{xg} \alpha(g^{-1} yg, g^{-1}zg)} ={^{gx} \alpha(g^{-1} yg, g^{-1}zg)} $?
 A: Let $A$ be a $G$-módule, let $a:H\times H\to A$ be a $2$-cocycle on $G$, and let $g$ be an element of $G$. We define $\bar a:gHg^{-1}\times gHg^{-1}\to A$ putting $$\bar a(x,y)=a(g^{-1}xg,g^{-1}yg)$$ for all choices of $x$ and $y$ in $gHg^{-1}$.
If $x$, $y$ and $z$ are in $gHg^{-1}$, then
\begin{align}
d\bar a(x,y,z)
&=x\cdot\bar a(y,z)-\bar a(xy,z)+\bar a(x,yz)-\bar a(x,y) \\
&=xg\cdot a(g^{-1}yg,g^{-1}zg)-a(g^{-1}xyg,g^{-1}zg)+a(g^{-1}xg,g^{-1}yxg)-a(g^{-1}xg,g^{-1}yg) \\
&=xg\cdot a(g^{-1}yg,g^{-1}zg)-g^{-1}xg\cdot a(g^{-1}yg,g^{-1}zg) +da(g^{-1}xg,g^{-1}yg,g^{-1}zg) \\
&=xg\cdot a(g^{-1}yg,g^{-1}zg)-g^{-1}xg\cdot a(g^{-1}yg,g^{-1}zg) \\
\end{align}
This tells us that $\bar a$ is a $2$-cocycle exactly when $$xg\cdot a(g^{-1}yg,g^{-1}zg)=g^{-1}xg\cdot a(g^{-1}yg,g^{-1}zg)$$
for all $x,y,z\in gHg^{-1}$ or, equivalently, when
$$gx\cdot a(y,z)=x\cdot a(y,z)$$
for all $x,y,z\in H$. In particular, if this holds we see that the cocycle $a$ takes values that are fixed by the action of $g$.
Certainly this is not the case in general, so what you want to do is simply impossible.

What one can do is to define a new $G$-module $B$: as an abelian group, it coincides with $A$, but on $B$ the elements $G$ act differently: if $b$ is an element of $B$ and $k$ one of $G$, we let $$k\triangleright b=gkg^{-1}\cdot b.$$ Here $\cdot$ is the action of $G$ on $A$, and $\triangleright$ is the action of $G$ on $B$. Now we define a map
$\bar a:gHg^{-1}\times gHg^{-1}\to B$ as before, but with values in $B$: we put $$\bar a(x,y)=a(g^{-1}xg,g^{-1}yg)$$ for all choices of $x$ and $y$ in $gHg^{-1}$. I claim that $\bar a$ is a $2$-cocycle on $gHg^{-1}$: if $x,y,z$ are in $gHg^{-1}$, then
\begin{align}
d\bar a(x,y,z)
&=x\triangleright\bar a(y,z)-\bar a(xy,z)+\bar a(x,yz)-\bar a(x,y) \\
&=gxg^{-1}\cdot a(g^{-1}yg,g^{-1}zg)-a(g^{-1}xyg,g^{-1}zg)+a(g^{-1}xg,g^{-1}yxg)-a(g^{-1}xg,g^{-1}yg) \\
&=da(g^{-1}xg,g^{-1}yg,g^{-1}zg) \\
&= 0.
\end{align}
Voilà.
This $G$-module $B$ is obtained from $A$ by «twisting» the action by $g$.
