A group $G$ is Schur-nontrivial if the Schur multipler $H^2(G,U(1))$ is not the trivial group.


I am trying to find an example of a Schur-nontrivial group which does not contain a subgroup of the form $H\times H$ where $H$ is a finite abelian group (thus $H\times H$ is a group of central type). Every example I have seen (browsing Groupprops, for example) does contain such a subgroup.


I have not yet looked through all of the finite simple groups, as I am not very familiar with them. I am particularly interested in nice groups which could perhaps represent a physical symmetry. 



Example of a Schur-nontrivial group with no abelian subgroup of the form $H\times H$?
I presume that since you use the definition $H^2\left(G,U(1)\right)$ for the Schur multiplier, you're interested in finite groups? (For infinite groups this is not in general equivalent to the standard definition of $H\_2(G,\mathbb{Z})$, and in any case, for infinite groups there are easy examples, such as $G=\mathbb{Z}\times\mathbb{Z}$.)


There are no finite examples.


If $G$ is finite, then your condition is equivalent to $G$ having $p$-rank at most one for every prime $p$ (i.e., not having a subgroup isomorphic to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$), which is in turn equivalent to all the Sylow $p$-subgroups being cyclic or (for $p=2$) generalized quaternion.


Cyclic and generalized quaternion groups have trivial Schur multiplier, and this implies that $G$ has trivial Schur multiplier, since if $P$ is a Sylow $p$-subgroup of $G$ then the restriction map $H^2\left(G,U(1)\right)\to H^2\left(P,U(1)\right)$ is injective on the $p$-primary part of $H^2\left(G,U(1)\right)$.


In fact, for finite groups your condition is equivalent to all subgroups of $G$ having trivial Schur multiplier.


