Let $G$ be a profinite group, and $M$ a finite abelian group with trivial $G$-action. Must there exist an open subgroup $H\le G$ with $H^2(H,M) = 0$?
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asked Sep 16, 2016 at 5:58
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5$\begingroup$ No ($G=\mathbf{Z}_p^2$, $M=\mathbf{Z}/p\mathbf{Z}$). $\endgroup$– YCorCommented Sep 16, 2016 at 7:29
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$\begingroup$ @YCor Ah I suppose that's because $\mathbb{Z}_p^2$ is isomorphic to each of its open subgroups? $\endgroup$– stupid_question_botCommented Sep 16, 2016 at 16:51
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$\begingroup$ @YCor Okay but it IS true that every element of $H^2(G,M)$ vanishes over some open subgroup $H\le_o G$ right? $\endgroup$– stupid_question_botCommented Sep 16, 2016 at 17:29
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$\begingroup$ No (same example). $\endgroup$– YCorCommented Sep 16, 2016 at 17:48
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$\begingroup$ @YCor Isn't every class in $H^2(G,M)$ represented by a continuous cocycle $z : G^2\rightarrow M$ such that $z(1,G) = z(G,1) = 0$? By continuity $z^{-1}(0)$ is open, and contains $(1,1)$, and hence by the definition of the topology on $G\times G$, $z^{-1}(0)$ must contain an open subgroup of $G\times G$, which must in turn contain $H\times H$ for some $H\le_o G$? $\endgroup$– stupid_question_botCommented Sep 16, 2016 at 18:02
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