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Let $G_1$ and $G_2$ be two groups (of some kind, e.g. finite groups).

Let $M_1, N_1$ be $G_1$-modules, and $M_2, N_2$ be $G_2$ modules, always with coefficients in $\mathbb{C}$.

Write $G = G_1 \times G_2$, $M = M_1 \otimes M_2$ and $N = N_1 \otimes N_2$.

Is $Ext_G(M, N)$ related in some way to $Ext_{G_1}(M_1, N_1)$ and $Ext_{G_2}(M_2, N_2)$ (eventually with some extra conditions)?

In a more vague way, is there a way to "break down" $Ext_G(M, N)$?

Here I don't precise the degree of $Ext$, but $Ext^1$ will be of primary interest.

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    $\begingroup$ That's in general non-trivial even for Hom and for the trivial groups, isn't it? $\endgroup$ – Fernando Muro May 31 '16 at 18:24
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    $\begingroup$ 1. If the groups are finite, and all modules have coefficients in $\mathbb C$, don't higher ext groups vanish, since all modules are projective? 2. If $M_1$ and $M_2$ are finitely generated, then it looks like there ought to be a Kunneth formula. $\endgroup$ – Gregory Arone May 31 '16 at 19:22
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The answer is given by Kunneth formula. In your case for each pair of modules $N_1$, $N_2$ we can consider a functor from the category of representations of $G_1\times G_2$ to $\mathbb C$-vector spaces, sending $M$ to $\mathrm{Hom}_{\mathbb C[G_1\times G_2]}(M, N_1\otimes N_2)$. $\mathrm{Ext}^\bullet_{\mathbb C[G_1\times G_2]}(M_1\otimes M_2, N_1\otimes N_2)$ is its right derived functor computed on $M=M_1\otimes M_2$ and the computation can be done replacing $M_1\otimes M_2$ by a projective resolution. It is easy to see that if $P^\bullet$ and $Q^\bullet$ are projective resolutions of $M_1$ and $M_2$, then $P^\bullet\otimes Q^\bullet$ (this is bicomplex and I mean its totalisation) is a projective resolution of $M_1\otimes M_2$ and when you compute $\mathrm{Hom}_{\mathbb C[G_1\times G_2]}(P^\bullet\otimes Q^\bullet, N_1\otimes N_2)$ you can check in particular that $\mathrm{Ext}^1_{\mathbb C[G_1\times G_2]}(M_1\otimes M_2, N_1\otimes N_2)=\mathrm{Ext}^1_{\mathbb C[G_1]}(M_1, N_1)\otimes \mathrm{Hom}_{\mathbb C[G_2]}(M_2, N_2)\oplus \mathrm{Hom}_{\mathbb C[G_1]}(M_1, N_1)\otimes \mathrm{Ext}^1_{\mathbb C[G_2]}(M_2, N_2)$.

In the case of finite groups categories of representations over $\mathbb C$ are semisimple and both sides are zero.

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