TITLE: $2$-cohomology group of semi-direct products
QUESTION [5 upvotes]: Let $G=N\rtimes T$ and let $A$ be a $G$-module with a trivial $G$-action.
The action of $T$ on induce a natural action of $T$ on the second cohomology group of $N$. Denote by $H^2(N,A)^T$ the $T$-invariant cohomology classes in $H^2(N,A)$. 
For $A=\mathbb{C}^*$ and $(|N|,|A|)=1$ it is shown (e.g. in "The Schur Multiplier" by G. Karpilovsky) that 
$$H^2(G,\mathbb{C}^*)=H^2(N,\mathbb{C}^*)^T\times H^2(T,\mathbb{C}^*).$$
My question is: is this result is true when we replace $\mathbb{C}^*$ with $A$ and if not what is the generalization?
The proof given in Karpilovsky does not work.
However, in http://link.springer.com/article/10.1007%2FBF01181625#page-1, Tahara deals with similar questions but I couldn't derive the desired result from there.   
I will appreciate any help.

REPLY [2 votes]: From the spectral sequence mentioned by Anton one can derive an exact sequence
in low degrees 
$$ 0\to E_2^{1,0}\to E^1\to E_2^{0,1}\to E_2^{2,0}
    \to \mathrm{ker}\left[E^2\to E_2^{0,2}\right]\to E_2^{1,1}\to E_2^{3,0}\ , $$
where $E^n=H^n(G,V)$ for $n=1,2$. I have found this exact sequence on page 38 of Sansuc's paper   (without proof). Maybe this is what you need.
Note also that in order to construct the spectral sequence and the exact sequence in low degrees, you need only the short exact sequence
$$ 1\to N\to G\to T\to 1.$$
From the splitting of this short exact sequence you can extract additional information about the exact sequence in low degrees.