There is a result in Cartan Eilenberg (XII 6.4) that says that if $G$ is a finite group and $D$ a divisible abelian group with trivial $G$-action then for any $G$-module $M$ the cup product $$\widehat{H}^{r-1}(G,Hom(M,D)) \times \widehat{H}^{-r}(G,M) \longrightarrow \widehat{H}^{-1}(G,D)$$ induces an isomorphism $$\widehat{H}^{r-1}(G,Hom(M,D)) \longrightarrow Hom(\widehat{H}^{-r}(G,M),D)$$ for all $r \in \mathbb{Z}$


I wanted to know if this result would hold for non-Tate cohomology and homology groups, i.e.,
if instead of having a finite group $G$, we used an arbitrary group $G$ and instead of using cup products we used cap products, would the result still hold for all $r \in \mathbb{Z}$ or maybe for some $r$?


Thank you



cap products and injective abelian groups
The isomorphism, at least, exists. Take $M\leftarrow Y$ and $\mathbb Z\leftarrow X$ projective resolutions of $M$ and of $\mathbb Z$ as $G$-modules. The double complex $\hom\_G(X,\hom\_{\mathbb Z}(Y,D))$ gives rise, as usual, to two spectral sequences. Computing first cohomology with respect to the differential induced by that of $Y$ and then by that of $X$, we get $H^\bullet(G,\hom(M,D))$. On the other hand, rewriting the complex as $\hom\_{\mathbb Z}(Y\otimes\_G X,D)$ using standard adjunctions, and then computing homology first with respect to the differential induced by that of $X$ and then by that of $Y$ gives the other side of your isomorphism. Convergence then does the trick.


That the isorphism is given by cap products should follow from general nonsense... 


