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](https://arxiv.org/abs/0812.4291) the [largest Mathieu group](https://en.wikipedia.org/wiki/Mathieu_group_M24) $M\_{24}$ 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](https://mathoverflow.net/questions/227033/restriction-of-h3m-24-u1-to-m-12-rtimes-mathbbz-2).



Is the map $\mathrm H^4(S_{24}) \to \mathrm H^4(M_{24})$ surjective?
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 $\mathbb Z/6 \subset \mathbb Z/12$.


