It is well known that given a short exact sequence $1\to H \to G \to G/H \to 1$ the transgression map
$$
H^{p-1}(G/H, H^1(H,A)) \to H^{p+1}(G/H,A^H)
$$
in the inflation-restriction sequence is in fact the cup product with the opposite of the class of the extension $1 \to H^{ab} \to G/H' \to G/H \to 1$ in $H^2(G/H, H^{ab})$, given that the action of $H$ on the module $A$ is trivial.


What is known about the transgression map in case $A$ is a non-trivial module? Is it possible then to describe transgression in similar terms?



transgression in terms of cup product in case of non-trivial action of the group on the coeffecients module
I can offer the following generalization for non-trivial coefficients $A$: Write $Q = G/H$. The cap product $$H\_1(H;\mathbb{Z}) \otimes H^1(H;A) \to \mathbb{Z}\otimes\_H A = A\_H$$ is $G$-linear with trivial $H$-action and induces a cup product 
$$\cup: H^p(Q;H\_1(H;\mathbb{Z})) \otimes H^q(Q;H^1(H;A))\xrightarrow{} H^{p+q}(Q;A\_H).$$
Let $u \in H^2(Q;H\_1(H;\mathbb{Z}))$ be the class that corresponds to the extension 
$$1 \to H^{ab} \to G/H' \to Q \to 1$$ 
and let $\kappa: A^H \to A\_H$ be the canonical homomorphism. 



> 
> **Theorem:** The compostion 
>  $$H^p(Q;H^1(H;A)) \xrightarrow{d\_2}H^{p+2}(Q;A^H)\xrightarrow{\kappa^\ast}H^{p+2}(Q;A\_H)$$
>  is (up to sign) cup product with $u$, i.e. $(\kappa^\ast \circ d\_2^{p,1})(x) = - u \cup x$. 
> 
> 
> 


*Proof:* By abuse of notation let $\kappa: A \to A\_H$. Since $\kappa$ is $G$-linear, it induces a map of spectral sequences $\kappa\_r^{pq}: E\_r^{pq}(A) \to E\_r^{pq}(A\_H)$. In particular, $\kappa\_2^{p+2,0}\circ d\_2^{p,1}(A)=d\_2^{p,1}(A\_H)\circ \kappa\_2^{p,1}$. Clearly, $\kappa\_2^{p+2,0}$ is the map $\kappa^\ast$ in the theorem and since $H$ acts trivially on $A\_H$, we obtain $(\kappa^\ast \circ d\_2^{p,1}(A))(x) = d\_2^{p,1}(A\_H)(\kappa\_2^{p,1}x)=-u \cup \kappa\_2^{p,1}x$. Hence, it remains to show 
$u \cup x = u \cup \kappa\_2^{p,1}x$ (note the cup products are w.r.t. different pairings). But this follows immediately by applying $H^\ast(Q;-)$ to the following commutative diagram of pairings: 
$$\begin{array}{ccc}
H\_1(H;\mathbb{Z}) \otimes H^1(H;A) \;\; & \xrightarrow{\cap} & A\_H \newline
\scriptstyle id\otimes \kappa^\ast \displaystyle\downarrow\qquad\quad & & \downarrow\scriptstyle id \newline 
H\_1(H;\mathbb{Z}) \otimes H^1(H;A\_H) & \xrightarrow{\cap} & A\_H. 
\end{array}$$


