I am trying to understand Shapiro's lemma for $H^2$ in the concrete language of extensions of finite groups.


Let $H$ be a subgroup of a finite group $G$, and let $A$ be an $H$-module. Let ${\rm Ind}\_G^H(A)$ be the induced module (see Serre, Galois Cohomology, Ch. I, 2.5). Shapiro's lemma says that the inclusion $H\hookrightarrow G$ and the map $\pi\colon {\rm Ind}\_G^H(A)\to A$ of evaluation at $1$ give an isomorphism $H^2(G,{\rm Ind}\_G^H(A))\simeq H^2(H,A)$ (of course, this is true in any degree).


Now let us view elements of $H^2$ as equivalence classes of extensions.
An extension $0\to {\rm Ind}\_G^H(A)\to \hat G\xrightarrow{f}G\to 1$ should correspond under this isomorphism to the extension $0\to A\to f^{-1}(H)/{\rm Ker}(\pi)\to H\to1$.


My question is how to realize the inverse map explicitely:  

That is, starting from an extension $0\to A\to \hat H\to H\to1$, what is the extension of $G$ by ${\rm Ind}\_G^H(A)$ corresponding to it under Shapiro's lemma? 



Shapiro's lemma in the language of group extensions
Shapiro's Lemma boils down to the following isomorphism for a subgroup of finite index: Let us write $G=\bigcup\_{i=1}^t g\_iH$ for left coset representatives of $H$ in $G$. For a $G$-module $P$ we have the following isomorphism $$\Psi:Hom\_H(P,A)\cong Hom\_G(P,Ind\_H^G(A))$$
$$ \Psi(f)(p) = \sum\_{i}g\_i\otimes f(g\_i^{-1}p).$$
If we think now of the extension $$1\to A\to \hat{H}\to H\to 1$$ as given by a two cocycle $\beta:H\times H\to A$, we can proceed in the following way:
We have the Bar Resolutions $B\_H$ and $B\_G$ of the trivial module $\mathbb{Z}$ over $H$ and $G$ respectively. The two cocycle $\beta$ can then be considered as an element in $Hom\_H((B\_H)\_2,A)$. Since $B\_G$ can also be considered as a resolution for $\mathbb{Z}$ over $H$, we will have a lifting of the identity map $B\_G\to B\_H$ as a map of $H$-modules. This will already gives us a two cocycle in $Hom\_H((B\_G)\_2,A)$. Now use the above isomorphism to get a two cocycle in $Hom\_G((B\_G)\_2,Ind\_H^G(A))$. This will give you the desired cocycle and therefore the desired extension. 


The main computational difficulty here (besides calculating the isomorphism $\Psi$) is in lifting the identity map $\mathbb{Z}\to\mathbb{Z}$ to $B\_G\to B\_H$. In degree zero this is rather simple: you choose right coset representatives $g\_i'$ and then map $hg\_i'[]\in (B\_G)\_0$ to $h[]\in (B\_H)\_0$. In degrees one and two it will be more complicated. In many concrete cases you can calculate explicitly projective resolutions for $H$ and for $G$. Also, in case $H$ is normal in $G$ you can write things more concretely, by choosing a two cocycle for $G/H$ with values in $H$ representing the extension $$1\to H\to G\to G/H\to 1$$ (even if $H$ is not abelian).


In the general case I do not think that there is a neater way of writing this down. Of course, I will be more than happy to be proven wrong about this. 


