Does a perfect $4^{11}\cdot M_{24}$ exist?

Question

Is there any perfect group which could be notated as $4^{11}\cdot M_{24}$ (a non-split extension of the largest Mathieu group by a homocyclic group of type $4^{11}$)?

Answer 1

I think the answer is no. In fact there are no perfect split extensions with structure $4^{11}:M_{24}$. I deduced this from some cohomology calculations in Magma.

Note that there are two (mutually dual) irreducible $11$-dimensional modules for $M_{24}$ over ${\mathbb F}_2$, and the extension $4^{11}:M_{24}$ doesn't exist for either of these. But the nonsplit extension $2^{11} \cdot M_{24}$ exists for only one of these modules, say $M$, so that must be the one you are referring to.

To do the calculation, I started with the split extension $E := 2^{11}:M_{24}$,with action from module $M$, as a permutation group of degree $2048$. The module $M$ can also be regarded as a module for $E$ using inflation, and I computed the cohomology group $H^2(E,M)$.

This had dimension $1$, so there is a unique nonsplit extension, but that is the extension with structure $2^{11+11} \cdot M_{24}$, coming from the nonsplit extension $2^{11} \cdot M_{24}$. So there can be no nonsplit extension of $M$ by $E$ with structure $4^{11}:M_{24}$.