# Shapiro's lemma in the language of group extensions

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?

• For the interpretation with central extensions the action on the modules needs to be trivial. Dec 2 '16 at 8:32
• My main source for this is the book by Neukirch, Schmidt, and Wingberg, Cohomology of Number Fields, Springer 2008. There the interpretation of elements of the second cohomology as central extensions (Theorem 1.2.4) does not seem to assume that the module is trivial. Am I missing something? Dec 2 '16 at 9:27
• $H^2(H,A)$ classifies extension of $H$ by $A$ inducing the given action of $H$ on $A$. By definition, such an extension is central iff $H$ acts trivially on $A$.
– js21
Dec 2 '16 at 9:39
• I now see the point, thanks. I edited the question by replacing "central extensions" by "extensions". Indeed, the book by Neukirch et al does not say "central". I hope that now the question makes sense. Dec 2 '16 at 10:28

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).
• I spent a while trying to make the lift of $2$-coycles $H \times H \rightarrow A$ to $2$-cocycles $G \times G \rightarrow A$ explicit. (In the post above, this comes from the comparison theorem, lifting the identity map $\mathbb{Z} \rightarrow \mathbb{Z}$ to a map on resolutions $B_G \rightarrow B_H$.) Even in simple cases, for example when $H$ has index $2$ in $G$, and there are reasonable maps $\mathbb{Z}(G^i) \rightarrow \mathbb{Z}(H^i)$ that (after switching to the bar resolution) induce the required lifting, it still looks like a mess to me. Dec 4 '16 at 14:47