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Jun 28, 2019 at 15:16 comment added Derek Holt Note also that $K/N$ is a subdirect product of (an unfortunately large number of) copies of $G$, so if $G$ is abelian, nilpotent, solvable, etc, then so is $K/N$.
Jun 28, 2019 at 14:48 vote accept Neil Strickland
Jun 28, 2019 at 14:44 comment added Neil Strickland To spell it out, the fibers of $f$ and $g$ will all have the same size (namely $|G|/|H|$), so we can choose a bijection $h_1\colon G \to G$ (not a homomorphism) with $g=fh_1$. Then $h_1$ induces an automorphism of $K$, which preserves the characteristic subgroup $N$ and so induces an automorphism $h$ of $K/N$. This will satisfy $gp=fph$.
Jun 28, 2019 at 13:05 comment added Neil Strickland I thought about this construction before. I could not see how to prove that it had the required property, but perhaps I do see that now. I will try to sort out the details.
Jun 28, 2019 at 12:25 history answered Derek Holt CC BY-SA 4.0