# When is the restriction map $res:H^2(G,U(1))\to H^2(Z_p\times Z_p,U(1))$ not the zero map?

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.

## Answer

Attribution
Source : Link , Question Author : David Stephen , Answer Author : Community