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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.

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Let $G={\mathbb Z}/3 \times {\mathbb Z}/3$ and $\alpha$ be the $2$-cocycle corresponding to the extraspecial group of exponent $3$ and order $27$. Then the twisted group algebras defined using $\alpha$ and $-\alpha$ are isomorphic, by swapping the two copies of ${\mathbb Z}/3$.

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Another answer already supplied an answer to the OP’s question, by pointing out that automorphisms of a group $G$ can act nontrivially on $H^2(G; F^\times)$. An alternate interpretation of the question is to find isomorphic twisted group algebras not related by an isomorphism of their grading groups. This is also easy: untwisted group algebras are special cases of twisted group algebras; if $F$ is algebraically closed, then both $F[\mathbb{Z}/4]$ and $F[(\mathbb{Z}/2)^2]$ are isomorphic to $F^{\times 4}$.

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  • $\begingroup$ I took the question to be asking for isomorphic but non-equivalent group algebras on the same group $G$, which obviously cannot happen without some twisting. $\endgroup$
    – LSpice
    Commented Apr 8, 2023 at 14:22
  • $\begingroup$ @LSpice and indeed Benson’s answer was accepted. $\endgroup$ Commented Apr 8, 2023 at 14:24

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