The group $\mathrm{Co}_0$ has a 24-dimensional module. This induces a map $\mathrm H^4(O(24),\mathbb Z) \to \mathrm H^4(\mathrm{Co}_0,\mathbb Z)$. Has this map been computed? Has the right hand side been computed? Where can I learn how to do these types of computations?
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asked Aug 31, 2016 at 17:50
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$\begingroup$ This paper might have some relevance. Geoffrey Mason, Reed–Muller codes, the fourth cohomology group of a finite group, and the β-invariant,J. Algebra 312, Issue 1, 1 June 2007, Pages 218-227. $\endgroup$– John MurrayCommented Sep 1, 2016 at 8:20
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$\begingroup$ The domain is ZxZ/2, with first factor generated by p1 and second factor by the Bockstein of w1w2. If you think the same question for H^{1,2,3} is easy, we can focus on p1. I bet the codomain is unknown, I don't know why. But you can start by restricting the rep. to easy subgroups. In general there's a split injection of H4(G;Z_p) into H4(p-Sylow;Z_p), which is Z/p for p=11,13,23. For p=7 the Sylow is Z/pxZ/p, likely easy. The 2-Sylow is complicated, but it's the same as the 2-Sylow of the normalizer of a Z/2^12-subgroup -- with known H4? I don't know how far this train goes. $\endgroup$– David TreumannCommented Sep 6, 2016 at 0:53
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$\begingroup$ @DavidTreumann great. I really only want to know the image of p1. Benson has worked out all details for p=2, I think, but I don't know much and so didn't try to understand his answer. I'll look again. $\endgroup$– Theo Johnson-FreydCommented Sep 7, 2016 at 13:59
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$\begingroup$ @DavidTreumann H^1(Co0) = H^2(Co0) = 0, so that should handle the easier factors $\endgroup$– Theo Johnson-FreydCommented Sep 9, 2016 at 13:50
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