Timeline for A question about continuous group cohomology

5 events

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Feb 19, 2018 at 9:08	comment	added	dorebell		I'm not sure about being a subquotient of a torsion divisible group (perhaps one can always embed a torsion group into a torsion injective, hence divisible group?), but $M$ torsion divisible does not imply $H^i_{\mathrm{cont}}(G, M)$ is torsion divisible. For example, take $G$ as the absolute Galois group of $\mathbf{F}_p$ and $M = \overline{\mathbf{F}_p}^\times$. This is torsion divisible since every element algebraic over $\mathbf{F}_p$ is a root of unity but $\overline{\mathbf{F}_p}$ is algebraically closed. Then $H^0_{\mathrm{cont}}(G, M) = \mathbf{F}_p^\times$, which is not divisible.
Feb 19, 2018 at 9:02	comment	added	dorebell		I would not expect this to be true for the latter property: let $G$ be a free pro-p group on infinitely many generators, and let $M$ be an abelian pro-p group with trivial $G$-action (e.g. $M = \mathbf{Z}[p^{-1}]/\mathbf{Z}$. Then $H^1_{\mathrm{cont}}(G, M) = \mathrm{Hom}(G, M)$, and the universal property of free pro-p groups should imply that this is infinitely generated. Of course, subquotients of finitely generated abelian groups are finitely generated, so no help by relaxing to that.
Feb 19, 2018 at 8:46	history	edited	YCor		edited tags
Feb 19, 2018 at 8:37	history	edited	user95222		edited tags
Feb 19, 2018 at 8:30	history	asked	user95222	CC BY-SA 3.0