# Examples of isomorphic non-equivalent twisted group algebras

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.

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 $${\mathbb Z}/3$$.
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 $$F^{\times 4}$$.
• 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. Commented Apr 8, 2023 at 14:22