# Is the solvable radical of a finite perfect group contained in the Schur multiplier of the quotient of the group modulo the solvable radical?

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?

• "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.
– YCor
May 20, 2020 at 11:10
• Thanks a lot! -- And can you perhaps also tell what would be the answer if the assume $N = {\rm Z}(G)$? May 20, 2020 at 19:45
• 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.
– YCor
May 20, 2020 at 19:46

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.