Background: For a projective representation of $G$ on a Hilbert space there is a 2-cocycle $c:G\times G \to \mathbb T$ where the cocycle condition $\delta c=0$ reads $c(f,g)c(fg,k) =c(f,hk)c(h,k)$ and comes from associativity. The proj. representation lifts to a true representation of the central extension $\tilde G$ of $G$ $$ 1 \to \mathbb T \to \tilde G \to G \to 1$$

Two projective representations are equivalent if the cocycles representent the same element in $H^2(G,\mathbb T)$, i.e. they differ by a cobundary $$\delta b(g,h)=\frac{b(g)b(h)}{b(gh)}$$. For a 2-cocycle the commutator map or antisymmetric part is $$\hat c(f,g) =\frac {c(f,g)}{c(g,f)}$$.

The following theorem is well-known. If $G\cong \mathbb Z^n$ then $c\mapsto \hat c$ is an isomorphism of $H^2(G,\mathbb T)$ to a subgroup of $Z^2(G,\mathbb T)$.

From this follows that two proj. repres. are equivalent iff they have the same commuator map.

I was wondering how much this theorem generalizes.

Question: Does this theorem generalize to arbitrary Abelian groups? Or to what kind of groups?