Which groups have undetectable third U(1)-cohomology?

Let $$G$$ be a finite group. A categorical Schur detector for $$G$$ is a set $$\mathcal{S}$$ of proper subgroups $$S \subsetneq G$$ such that the total restriction map $$\mathrm{rest}_{\mathcal{S}} : \mathrm{H}^3(G; \mathrm{U}(1)) \to \prod_{S \in \mathcal{S}} \mathrm{H}^3(S; \mathrm{U}(1))$$ is an injection. The name is a riff on the suggestion of Epa and Ganter to call $$\mathrm{H}_3(G;\mathbb{Z})$$ the categorical Schur multiplier of $$G$$ (Platonic and alternating 2-groups, Higher Structures 1(1):122–146, 2017).

I know of two classes of groups which do not admit categorical Schur detectors: the cyclic and the binary dihedral (aka dicyclic) groups of prime power order. (Of course, a binary dihedral group of prime power order is necessarily of order $$2^n$$.) This follows from the fact that for any finite subgroup $$G \subset \mathrm{SU}(2)$$, there is an isomorphism $$\mathrm{H}^3(G;\mathrm{U}(1)) \cong \mathbb{Z}/\lvert G\rvert$$.

Question: Are there any other finite groups which do not admit categorical Schur detectors?

Of course, one could ask the same question about cohomology of other degrees, or with other coefficients. $$\mathrm{H}^3(-;\mathrm{U}(1))$$ is of particular interest in "moonshine", because it is home to global symmetry anomalies of 2D QFTs.

• I record the trivial remark that a group of non-prime-power order always admits a categorical Schur detector, namely the set of its Sylow subgroups. Note that my definition of "categorical Schur detector" requires that each subgroup in the set be proper. – Theo Johnson-Freyd Dec 5 '20 at 16:25
• A variant of your question has been studied in the literature: the case when the coefficients are $\mathbb Z/p$, and the common kernel of restrictions to all proper subgroups is called essential cohomology. Adem and Karagueuzian show that a p-group has essential cohomology if all its elements of order p are central. (I gave a different proof, giving a degree in which this happens.) – Nicholas Kuhn Dec 5 '20 at 17:42
• @NicholasKuhn Thank you. I did not know the term "essential cohomology". Is the paper of you mention arxiv.org/abs/math/0612133? – Theo Johnson-Freyd Dec 5 '20 at 20:18
• Yes. And it appeared in Advances in 2007. – Nicholas Kuhn Dec 6 '20 at 2:52
• @NicholasKuhn Thanks! – Theo Johnson-Freyd Dec 7 '20 at 3:18

Inspired by YCor's remark I realized that every finite abelian $$p$$-group $$G$$ of rank three will have essential elements in $$H^3(G;U(1))$$. I think that this works for $$p=2$$ as well, but here's an argument for $$p$$ odd. Let $$y_1,y_2,y_3$$ be linearly independent elements of $$H^1(G;\mathbb{F}_p)\cong \mathrm{Hom}(G,\mathbb{F}_p)$$. If $$S$$ is any proper subgroup of $$G$$, then the image of $$H^1(G;\mathbb{F}_p)\rightarrow H^1(S;\mathbb{F}_p)$$ has rank two. Thus the images of $$y_1,y_2,y_3$$ in $$H^1(S;\mathbb{F}_p)$$ are no longer linearly independent, and so the product $$y_1y_2y_3$$ is an element of $$H^3(G;\mathbb{F}_p)$$ that is essential.
It remains to show that $$y_1y_2y_3$$ has non-zero image under the map $$H^3(G;\mathbb{F}_p)\rightarrow H^3(G;U(1))$$ induced by the inclusion of additive groups $$\mathbb{F}_p\rightarrow U(1)$$. For this, look at the long exact sequence in cohomology coming from the short exact coefficient sequence $$0\rightarrow \mathbb{F}_p\rightarrow U(1)\rightarrow U(1)\rightarrow 0$$. It suffices to show that the image of $$H^2(G;U(1))\rightarrow H^3(G;\mathbb{F}_p)$$ does not contain $$y_1y_2y_3$$. I struggle with $$U(1)$$ coefficients, so I would instead show the equivalent fact that $$y_1y_2y_3$$ is not in the image of $$H^3(G;\mathbb{Z})\rightarrow H^3(G;\mathbb{F}_p)$$. This is because $$H^3(G;\mathbb{Z})$$ is detected by proper subgroups. Roughly speaking this is because for a finite cyclic group $$C$$, $$H^1(C;\mathbb{Z})=0$$. In terms of the Kunneth theorem, the only way to get elements in degree three for a product of cyclic groups is as a `Tor' term coming from a pair of elements of $$H^2(C;\mathbb{Z})$$ and $$H^2(C';\mathbb{Z})$$ for cyclic subgroups $$C$$, $$C'$$. But such an element is detected on the subgroup $$C\times C'$$.
This should generalize: if $$G$$ is a finite abelian $$p$$-group then every element of $$H^{k+1}(G;\mathbb{Z})$$ will be detected on subgroups of rank $$k$$. If so, then for $$G$$ an abelian $$p$$-group $$H^3(G;U(1))$$ will contain essential elements if and only if $$G$$ has rank 1 or 3.