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Let $G$ be a finite group and let $S$ be a subgroup of $G$. Let $\alpha$ be a one-cocycle $G\to M$ for some $G$-module $M$, and suppose that $Res_{S}^{G}\alpha=0$. For $g\in G$, and $S^{g}=gSg^{-1}$, $Res_{S^{g}}^{G}\alpha$ is a coboundary. But according to Weiss, Cohomology of Groups, if $\tau=gsg^{-1}\in S^{g}$ then $\alpha_{\tau}=g\alpha_{s}$

which would mean that $\alpha_{\tau}=0$. I must be misunderstanding something.

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  • $\begingroup$ What is $\alpha_s$ resp. $\alpha_\tau$ ? $\endgroup$
    – Mark Opitz
    Commented Oct 29, 2013 at 19:39
  • $\begingroup$ Which page in Weiss? $\endgroup$
    – Boris Novikov
    Commented Oct 29, 2013 at 20:58
  • $\begingroup$ $\alpha$_s = $\alpha$(s) $\endgroup$
    – Gabrielle
    Commented Oct 31, 2013 at 15:44
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    $\begingroup$ A 1-cocycle is just a derivation. Hence $\alpha(\tau)=g\alpha(s) + (1-\tau)\alpha(g)$. Here $\tau \mapsto (1-\tau)\alpha(g)$ is an inner derivation (=1-coboundary). Perhaps Weiss' equation has to be understood "modulo a coboundary" ? $\endgroup$
    – Mark Opitz
    Commented Oct 31, 2013 at 21:39
  • $\begingroup$ I think you are right, thanks very much. $\endgroup$
    – Gabrielle
    Commented Oct 31, 2013 at 21:45

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I trhink you mixed $\alpha$ and ${\rm con}\alpha$. In fact, ${\rm con_g}\alpha (\tau) =g\alpha (g^{-1}sg)$, and I don't see any contradiction here.

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