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


