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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Out{Out}\newcommand\ab{^{\text{ab}}}$Let $G$ and $H$ be a groups and assume that $H$ is non-abelian. Then we have a morphism $$\Out(H)\rightarrow \Aut(H\ab).$$ For example if $H=F_2$, the free group with two generators, this map is an isomorphism, for $F_n$ with $n\geq 3$ this is only a surjection and for more general groups this is not a surjection, as was pointed out in the comments by Will Sawin and Mark Wildon.

My question thus is if we are given a morphism $$\rho:G\rightarrow \Aut(H\ab)$$ does there exist a way to tell if this comes from a morphism $$G\rightarrow \Out(H)?$$

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    $\begingroup$ I am pretty sure this is not a surjection in general. Why should it be? $\endgroup$
    – Will Sawin
    Commented Aug 1, 2020 at 17:20
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    $\begingroup$ 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. $\endgroup$
    – Mark Wildon
    Commented Aug 1, 2020 at 17:20
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    $\begingroup$ @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 $C_5 \rtimes C_4$. $\endgroup$
    – Will Sawin
    Commented Aug 1, 2020 at 17:22
  • $\begingroup$ 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. $\endgroup$
    – curious math guy
    Commented Aug 1, 2020 at 17:25
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    $\begingroup$ 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. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 1, 2020 at 18:05

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