I have a semisimple algebra $R$ over a field $k$ that looks like a group algebra $k[G]$ except that it's deformed slightly. That is, instead of a basis $e\_1 ..., e\_n$ closed under multiplication, I have a set of $1$-dimensional subspaces $v\_1,...,v\_n$, such that their direct sum is $R$ (that is, a choice of one generator from each subspace forms a basis) which is also closed under multiplication (that is, for each $i$ and $j$, there is some $k$ such that $v\_i v\_j = v\_k$.)



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This is not a Hopf algebra. There is a natural "diagonal" subalgebra of $R \otimes\_k R$, but it is not necessarily or naturally isomorphic to $R$.


The main reason I can't just view this as a semisimple algebra when I work with it is the "diagonal" subalgebra structure. There is some basic representations theory involving this that I would not like to duplicate.


$R \otimes\_k \bar{k}$ is not necessarily a group algebra. For instance, this is false for $R$ a quaternion algebra.


Other examples include field extensions generated by $n$th roots, and a group algebra modulo the ideal identifying a $2$-torson central element with $-1$. Nonexamples include field extensions not generated by $n$th roots.



What is this deformed group algebra named?
Yes, Dima is right, these are twisted group algebras. They are discussed in [this MathOverflow Question](https://mathoverflow.net/questions/100098/whether-such-an-algebra-has-to-be-the-group-algebra/100101#100101)


