Let $F$ be a field, $G$ be a finite group and $\alpha \in Z^2(G, F^*)$ . The twisted group algebra $F^{\alpha}G$ is a $F$-algebra with $F$ vector basis, $\{\bar g : g \in G \},$ and multiplication defined by $\bar x \bar y = \alpha(x,y) \overline{xy}$ for $x,y \in G$ and extended distributively.
Let $F^{\alpha} G$ and $F^{\beta}G$ be twisted group algebras with bases $\{\bar g : g \in G\}$ and $\{\tilde{g} : g \in G\}$ respectively. We say that $F^{\alpha}G$ and $F^{\beta}G$ are said to be equivalent if there exist an $F$-algebra isomorphism $\psi : F^{\alpha}G \to F^{\beta} G$ and a map $t : G \to F^*$ such that $\psi(\bar g) = t(g) \tilde{g}.$
Are there are examples twisted group algebras which are isomorphic as $F$ algebras but not equivalent?
There is a equivalent criterion which says that $F^{\alpha} G$ and $F^{\beta}G$ are equivalent iff $\alpha$ and $\beta$ are in same cohomology class. So we need to find examples where $H^2(G,F^*)$ is non trivial. For example cyclic groups, certain abelian groups such as $\mathbb{Z}_n\times \mathbb{Z}_m$ with $(m,n)=1$ would not work when $F$ is algebraically closed field. I am also not aware of any result which gives the wedderburn decomposition of $F^{\alpha} G$ when $G$ is abelian . The case when $G$ is abelian, there might be potential counter examples.
