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Let $G$ be a profinite topological group, $M$ a discrete $G$-module.

If $M$ is "P", is every $H^i_{\rm cont}(G,M)$ also "P"? or at least is it a subgroup/subquotient of an abelian group that is "P"? for "P" one of the following properties:

  • torsion divisible.
  • finitely generated as an abelian group.

I would benefit from a reference.

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  • $\begingroup$ 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. $\endgroup$
    – dorebell
    Commented Feb 19, 2018 at 9:02
  • $\begingroup$ 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. $\endgroup$
    – dorebell
    Commented Feb 19, 2018 at 9:08

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