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Let $G$ be a finite group, $K$ a characteristic-$p$ algebraically closed field (say $p$ divides $|G|$), and let $M$ be a finite-dimensional $KG$-module.

What hypotheses are needed on $G$, $M$ to ensure that $H^{1}(G,M) \cong H^{1}(G,M^{*})$ where $M^{*}$ denotes the dual module Hom$(M,K)$?

In fact, in the situation I'm faced with, $G$ is a finite simple group and calculation suggests the stronger result that $Ext^1(M,N) \cong Ext^1(N,M)$ when $M$ and $N$ are irreducible, and at least one of $M$,$N$ is self-dual. There are counterexamples when the latter condition fails, e.g. Alt$_9$ has non-self-dual modules of dimension 8 and 20 with $Ext^1(M,N) \neq Ext^1(N,M)$.

I feel like this result should be well-known but I've been unable to find a reference for it yet.

A weaker result that would also satisfy me is any criterion to ensure $H^1(G,M) \neq 0 \Rightarrow H^1(G,M^{*}) \neq 0$.

Thanks in advance!

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There is a cup-product pairing $H^1(G,M) \times H^1(G,M^*) \to H^2(G,K)$. I don't think this will typically be a perfect pairing, but in some situations it may be. (This is the kind of thing that came to mind when I read your question.) –  Matt E Apr 12 '11 at 0:08
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If it helps, it is not generally true that $H^1(G,M) \neq 0 \implies H^1(G,M^*) \neq 0$. One can take G to be a smallish symmetric group and M a Specht (or dual Specht, depending on your definition). I found this particularly irritating since M and M* lifted to isomorphic QG modules, but had different first cohomology mod 2. In this example, neither G nor M is simple, but I believe it should be easy to find examples where G is simple (and M is not). One should be able to find examples where G has a cyclic Sylow p-subgroup. –  Jack Schmidt May 19 '11 at 16:28
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