Timeline for If $G$ is profinite and $M$ finite abelian, must there exist an open subgroup $H\le G$ with $H^2(H,M) = 0$?
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| Sep 19, 2016 at 15:18 | comment | added | stupid_question_bot | @HJRW I don't think I've even showed that much. I've only showed that for every $z\in H^2(G,M)$, it is zero on some open subgroup $H$. Since $G$ is not assumed to be finitely generated, $H^2(G,M)$ might be infinite. But anyway, this is all I wanted to show after I realized that the question in the OP had a negative answer. | |
| Sep 19, 2016 at 8:57 | comment | added | HJRW | @rtz, your argument seems fine as far as it goes, but it doesn't follow that $H^2(H,M)=0$. I think all you did is show that there's always an open subgroup so that the restriction map $H^2(G,M)\to H^2(H,M)$ is zero. | |
| Sep 16, 2016 at 18:02 | comment | added | stupid_question_bot | @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 $H\le_o G$? | |
| Sep 16, 2016 at 17:48 | comment | added | YCor | No (same example). | |
| Sep 16, 2016 at 17:29 | comment | added | stupid_question_bot | @YCor Okay but it IS true that every element of $H^2(G,M)$ vanishes over some open subgroup $H\le_o G$ right? | |
| Sep 16, 2016 at 16:51 | comment | added | stupid_question_bot | @YCor Ah I suppose that's because $\mathbb{Z}_p^2$ is isomorphic to each of its open subgroups? | |
| Sep 16, 2016 at 7:29 | comment | added | YCor | No ($G=\mathbf{Z}_p^2$, $M=\mathbf{Z}/p\mathbf{Z}$). | |
| Sep 16, 2016 at 5:58 | history | asked | stupid_question_bot | CC BY-SA 3.0 |