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Let $G$ be a finite perfect group, and let $N$ be the solvable radical of $G$. If $G/N$ is a non-abelian simple group, then is it true that $N$ is contained in the Schur multiplier of $G/N$?

If this is not true in general, then does it hold at least in case $G/N$ is either of type ${\rm PSL}(2,2^p)$ ($p$ prime) or isomorphic to one of ${\rm PSL}(2,7)$ or ${\rm Sz}(8)$?

Furthermore, can the finite perfect groups whose quotient modulo their solvable radical is isomorphic to one of the simple groups mentioned in the previous paragraph reasonably be classified?

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    $\begingroup$ "Is contained", you mean "is isomorphic to a subgroup of"? Anyway, in many cases $N$ is non-abelian, so this sounds hopeless. Even with $N$ abelian, you have plenty of perfect semidirect products $G=N\rtimes S$ with $S$ simple non-abelian (for given $S$ you have such $G$ with $N$ arbitrary large). Classifying them for $N$ abelian of prime exponent is essentially classifying reps of $S$ in arbitrary finite fields. Classifying them for $N$ nilpotent is even much harder. So this sounds hopeless. $\endgroup$
    – YCor
    Commented May 20, 2020 at 11:10
  • $\begingroup$ Thanks a lot! -- And can you perhaps also tell what would be the answer if the assume $N = {\rm Z}(G)$? $\endgroup$
    – Leyli Jafari
    Commented May 20, 2020 at 19:45
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    $\begingroup$ The answer is positive if $N=Z(G)$, but this is somewhat immediate from the definition. I think that the question belongs on MathSE rather than here. $\endgroup$
    – YCor
    Commented May 20, 2020 at 19:46

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The answer to the question as asked is definitely "no" in general. For any value of $n>1 $ and any odd prime $p$, we may take a perfect group $G$ which is a semidirect product of the form $E.{\rm Sp}(2n,p),$ where $E$ is extra special of order $p^{2n+1}$ and the action of the given symplectic group on $E$ is the natural one. Then $E$ is the solvable radical of $G$, and is non-Abelian, so is not (isomorphic to) a subgroup of any Schur multiplier.

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