Smallest subgroup of $A_{24}$ that strictly contains the Mathieu group $M_{24}$ $M_{24}$ has many interesting subgroups living in it that one wouldn't think would all fit together in $A_{24}$ without generating all of $A_{24}$. It's really quite amazing to me.
Is there any subgroup strictly contained in $A_{24}$ which strictly contains $M_{24}$, i.e. something sandwiched in the middle? It seems unlikely to me, as adding another new permutation to $M_{24}$ makes it seem like it might "burst". If there do exist such intermediate groups, I'd be curious about what the structure of the lattice of subgroups is, or at least what the minimal groups strictly containing $M_{24}$ are.
 A: $M_{24}$ is a maximal subgroup of $A_{24}$. A citable reference is e.g.
Liebeck, Praeger, Saxl, A Classification of the Maximal Subgroups of the Finite Alternating and Symmetric Groups, J.Alg. 111, 365-383 (1987),
though the result is older. (What GAP does -- and I write this as the person who implemented the routine -- is to use pre-stored tables of maximal subgroups for certain simple groups, the calculation thus is really just a lookup.)
One reason for $M_{24}$ being maximal in $A_{24}$ is that $M_{24}$ is already quintuply transitive, and any hypothetically larger group would also have to be so. But such groups do not exist, unless they contain the alteranting group of the appropriate degree. (This however in itself is not an easy proof.)
I am not aware of any self-contained elementary proof of the maximality that could fit in an answer here.
A: GAP says the answer is no, with this sequence of commands.
b11 := (1,4)(2,7)(3,17)(5,13)(6,9)(8,15)(10,19)(11,18)(12,21)(14,16)(20,24)(22,23);
b21 := (1,4,6)(2,21,14)(3,9,15)(5,18,10)(13,17,16)(19,24,23);
m24 := Group(b11, b21);
a24 := AlternatingGroup(24);

IntermediateGroup(a24, m24);

returns fail.
