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All groups in this question are finite, and epimorphism means surjective group homomorphism.

Suppose I have two epimorphisms $f,g\colon G\to H$. This implies that $\ker(f)$ and $\ker(g)$ have the same composition factors, but they need not be isomorphic. I'll say that $f$ and $g$ are compatible if $g=fh$ for some automorphism $h$ of $G$. This would imply that $\ker(g)\simeq\ker(f)$, so it is not always true. I ask: does there always exist $K$ and an epimorphism $p\colon K\to G$ such that $fp$ and $gp$ are compatible?

If $G$ is nilpotent we can reduce to the case where it is a $p$-group, then I think we can take $K$ to be the initial example of an $k$-generator group of exponent $p^n$ and nilpotence class $c$, for sufficiently large $k$, $n$ and $c$. In particular, if $G$ is an abelian $p$-group I think we can take $K=C_{p^n}^k$ for sufficiently large $k$ and $n$. But I am not sure what to do when $G$ is not nilpotent.

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You could start with $K$ being the free group with the elements of $G$ as its free basis, with $p$ the obvious map onto $G$. Then the required automorphism $h$ of $K$ to make the two composite maps compatible is just a permutation of the free basis elements of $K$.

Of course this $K$ is infinite, and you are looking for a finite group. But to get that, we could replace $K$ by $K/N$, for any characteristic subgroup $N$ of finite index in $K$, and with $N \le \ker p$. You could take $N$ to be the intersection of the kernels of all epimorphisms from $K$ to $G$, of which there are finitely many.

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  • $\begingroup$ 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. $\endgroup$
    – Neil Strickland
    Commented Jun 28, 2019 at 13:05
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    $\begingroup$ 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 $gp=fph$. $\endgroup$
    – Neil Strickland
    Commented Jun 28, 2019 at 14:44
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    $\begingroup$ 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 $K/N$. $\endgroup$
    – Derek Holt
    Commented Jun 28, 2019 at 15:16

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