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Aug 1, 2020 at 18:06 history edited YCor CC BY-SA 4.0
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Aug 1, 2020 at 18:05 comment added Mikhail Borovoi You can find a down-to-earth treatment of nonabelian $H^2$ in my paper :M. Borovoi, Abelianization of the second nonabelian Galois cohomology. Duke Math. J. 72 (1993), 217-239, and also in the references therein (especially in Springer's paper) and in papers referring to my paper and to Springer.
Aug 1, 2020 at 18:01 comment added Mikhail Borovoi There may be more than one neutral class in nonabelian $H^2$.
Aug 1, 2020 at 18:00 comment added Mikhail Borovoi You can lift $\rho$ if and only if the cohomology class is neutral.
Aug 1, 2020 at 17:57 comment added Mikhail Borovoi If yes, then you get a cohomology class in $$H^2(G,{\rm ker}[{\rm Out}(H)\to {\rm Aut}(H^{\rm ab})]).$$
Aug 1, 2020 at 17:53 comment added Mikhail Borovoi First, you should require that $\rho$ lands in the image of ${\rm Out}(H)$.
Aug 1, 2020 at 17:27 history edited curious math guy CC BY-SA 4.0
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Aug 1, 2020 at 17:25 comment added curious math guy You are both absolutely right! We do have a surjection if $H$ is a free group however, which is what I had in mind. I'll change the question accordingly.
Aug 1, 2020 at 17:22 comment added Will Sawin @MarkWildon I don't think your example is the right one, because $H^{\mathrm{ab}} =C_2$. However, $C_5 \rtimes C_4$ works, because its abelianization $C_4$ has an automorphism, which does not lift to $C_5 \rtimes C_4$.
Aug 1, 2020 at 17:20 comment added Mark Wildon The map $\mathrm{Out}(H) \rightarrow \mathrm{Aut}(H^{\mathrm{ab}})$ is not in general surjective. I had an incorrect example and I see Will Sawin has now posted a correct one.
Aug 1, 2020 at 17:20 comment added Will Sawin I am pretty sure this is not a surjection in general. Why should it be?
Aug 1, 2020 at 17:19 history edited LSpice CC BY-SA 4.0
DeclareMathOperator's
Aug 1, 2020 at 17:15 history asked curious math guy CC BY-SA 4.0