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Jul 23, 2018 at 14:15 vote accept Abhishodh
Jul 22, 2018 at 22:08 answer added S. Carnahan timeline score: 4
Jul 22, 2018 at 19:10 comment added Abhishodh Consider $G= \mathbb{Z}_2$ which has $H^2(G,U(1)) =0$. In this case, the covering group is $\mathbb{Z}_2$ itself. All irreducible representations of $\mathbb{Z}_2$ are labeled by the trivial element of $H^2(G,U(1))$.
Jul 22, 2018 at 18:50 comment added abx I do not understand your question at all. Could you explain it in the case $H^2(G,U(1))=0$? There are a number of finite groups with that property.
Jul 22, 2018 at 18:19 history asked Abhishodh CC BY-SA 4.0