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Consider $G$ to be a finite group with non-trivial Schur Multipler $H^2(G,U(1))$, where $G$ acts trivially on the circle group $U(1)$.

By Example of a Schur-nontrivial group with no abelian subgroup of the form $H\times H$?, $G$ must have a subgroup of the form $Z_p\times Z_p$ for some prime $p$.

My question is: under what conditions on $G$ can we guarantee that the restriction map $res_{Z_p\times Z_p}:H^2(G,U(1))\to H^2(Z_p\times Z_p,U(1))$ is not the zero map for some $p$ such that $Z_p\times Z_p\subsetneq G$?

Is this satisfied by most $G$, in some quantifiable sense?

The only way I know to approach this is using the fact that, for any Sylow p-subgroup P, $res_P$ is injective from the $p$-primary part of $H^2(G,U(1))$. But here, $Z_p\times Z_p$ is not necessarily Sylow. So I do not know how to use this in general. I would be grateful for some guidance towards other ways to approach the problem.

As a simple counterexample, $G=Z_4\times Z_4$ properly contains only $Z_2\times Z_2$, onto which $res$ is the zero map.

This is likely too broad of a question, so I am also interested in limiting cases. For example, if we restrict $G$ to be abelian this should be easier, making use of the fundamental theorem of finite abelian groups.

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  • $\begingroup$ There are many papers about detecting cohomology of finite group $G$ on its elementary Abelian $p$-subgroups, ( see, eg the work of J.F. Carlson). $\endgroup$ Jul 8, 2016 at 1:11
  • $\begingroup$ If $G$ is abelian, then I think the answer is yes if and only if some Sylow $p$-subgroup of $G$ has $C_p \times C_p$ as a direct factor. $\endgroup$
    – Derek Holt
    Jul 8, 2016 at 8:03
  • $\begingroup$ This was my guess as well, but I wasn't able to rigorously prove both directions. Can you say a bit more about why you think this should be true? $\endgroup$ Jul 8, 2016 at 20:29

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