Timeline for Are $H^3(A,U(1))$ and $\operatorname{Ext}^1(A,A^\vee)$ isomorphic for $A$ finite Abelian?

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Jul 23 at 17:35	history	edited	Dave Benson	CC BY-SA 4.0	added 744 characters in body
Jul 23 at 14:53	comment	added	Dave Benson		The isomorphism $H^3(A,U(1))\cong H^4(A,\mathbb{Z})$ only uses $A$ finite, as does the formula for $\operatorname{\rm Ext}^1(A,A^\vee)$. But the computation of $H^4(A,\mathbb{Z})$ definitely uses $A\cong (\mathbb{Z}/p)^r$ for $p$ odd. If $p=2$ the answer is different, and if $A$ is an odd abelian $p$-group but not elementary abelian the answer is also different.
Jul 23 at 14:49	vote	accept	Andrea Antinucci
Jul 23 at 14:48	comment	added	Andrea Antinucci		Where did you used that $A=\mathbb{Z}_p^r$? Are the results you claimed for $H^3(A,U(1))$ and $\text{Ext}^1(A,A^\vee)$ valid for any finite abelian $A$? If the answer is yes, I suspect the isomorphism I wanted is merely the isomorphism between the symmetric parts $S^2 A^\vee$. Indeed from your answer I realized that the totally antisymmetric part $\Lambda A^\vee$ of $H^3(A,U(1))$ allows to write additional topological actions, simply cubic in $a$.
Jul 23 at 8:02	history	answered	Dave Benson	CC BY-SA 4.0