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I already asked this on Math Stackexchange, but got no answers, so I am trying here. https://math.stackexchange.com/questions/1914836/when-is-first-group-cohomology-isomorphic-to-conjugacy-classes-of-sections

In chapter 1, exercise $\S$2.1 of Neukirch's Cohomology of Number Fields page 24, there is the following situation: $G$ is a profinite group, $A$ is a (possibly nonabelian) discrete group with a continuous $G$-action, $\hat{G} := A \rtimes G$, and $\mathrm{SEC}(\hat{G}\to G)$ is defined as the $A$-conjugacy classes of homomorphic sections of the natural exact sequence $$1 \to A \to \hat{G} \xrightarrow{\pi} G \to 1.$$ The result in question is that if either $G$ or $A$ is finite, then there is a canonical bijection of pointed sets $H^1(G,A) \cong \mathrm{SEC}(\hat{G}\to G)$.

My questions are:

(1) Why is it necessary for either $G$ or $A$ to be finite?

(2) Would the result be true if $A$ were not discrete but profinite?

(3) If $A$ were profinite, would the finiteness condition be necessary?

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    $\begingroup$ It's certainly true whenever $A$ is abelian, see Chapter IV of Brown. $\endgroup$
    – Mark Grant
    Commented Sep 22, 2016 at 6:05
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    $\begingroup$ (1) The semidirect product of profinite groups doesn't need to be profinite (cf. mathoverflow.net/questions/98778/…). But if one of the groups is finite, $\hat{G}$ is always profinite. I guess that's why the authors required this condition. (2), (3) Note that at the stage of chap. 1, in the definition of the cohomology of a profinite group the coefficients are required to be discrete. $\endgroup$
    – tj_
    Commented Sep 22, 2016 at 8:03
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    $\begingroup$ @Mark Grant: Brown uses a different cohomology theory: Brown uses abstact maps in the bar resolution, while in the cohomology of profinite groups, continuous maps are used. Therefore, one has be be a bit cautious when transporting such results from one theory to the other. $\endgroup$
    – tj_
    Commented Sep 22, 2016 at 8:11

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