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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.

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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.

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  • $\begingroup$ You are correct, I am interested in finite groups. So your argument shows that if a finite G has a non-trivial Schur multiplier, then it must contain a subgroup of the form $H \times H$ for $H$ abelian. This is a very nice result, thank you. $\endgroup$ – David Stephen May 3 '16 at 18:55
  • $\begingroup$ Every group of the form $H\times H$ contains an abelian one of the same form (take a cyclic subgroup of $H$) $\endgroup$ – Lior Silberman Jul 8 '16 at 6:26

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