TITLE: Is the map $\mathrm H^4(S_{24}) \to \mathrm H^4(M_{24})$ surjective?
QUESTION [19 upvotes]: The group $S_{24}$ of permutations of $24$ things has fourth integral cohomology $\mathrm H^4(S_{24};\mathbb Z) \cong \mathbb Z/2 \oplus \mathbb Z/2 \oplus \mathbb Z/12$.  According to Sikiric and Ellis the largest Mathieu group $M_{24}$ has $\mathrm H^4(M_{24};\mathbb Z) \cong \mathbb Z/12$.  The Mathieu group is defined in terms of a permutation representation on $24$ things (namely, the coordinate vectors in the extended binary Golay code), and so there is a restriction map $\mathrm H^4(S_{24};\mathbb Z) \to \mathrm H^4(M_{24};\mathbb Z)$.
The number $12$ being somewhat magical, I expect that this map is a surjection.  Is it?  Is the answer to this question known?  My impression of the literature is that $M_{24}$ is just beyond where current technology can fully work out its (2-local, or even $\mathbb F_2$) cohomology.  Compare this older MO question.

REPLY [11 votes]: The answer to my question is No.  The generator of the $\mathbb Z/12$ part of $H^4(S_{24})$ is $p_1$ of the permutation representation.  That representation restricts to $M_{24}$ to a Spin representation, i.e. one with $w_1=w_2=0$.  For any such representation, $p_1$ is automatically even.
Thus the map $H^4(S_{24}) \to H^4(M_{24})$ has image within the $\mathbb Z/6 \subset \mathbb Z/12$.