Relation between Associative algebra and group algebra Let $A$ be an associative algebra over a filed $k$.
Q) What are the condition we can impose on $A$ such that there exists a $G$ such that $A=k[G]$, the group algebra generated by $G$?
I am particularly interested in the following cases:
1) When $k=\mathbb{Q},\mathbb{R}$ or $\mathbb{C}$. 
2) $A=M_n(k),$ the matrix algebra or a subalgebra of matrix algebra.
PS: I am not sure whether this question is of research level or not. If anybody thinks that this is not proper here please give the references and then vote to close. I have searched it in general but could not find any answer. Thanks in advance.
 A: If $k$ has characteristic zero and $A$ is finite-dimensional, by Maschke's theorem it must be semisimple. If $k$ is in addition algebraically algebraically closed and $A \cong k[G]$, then $A \cong \prod_i M_{n_i}(k)$ where $n_i$ are the dimensions of the irreducible representations of $G$. So in this case the question reduces to the following purely group-theoretic question:

When is a multiset $n_i$ of positive integers the multiset of dimensions of the irreducible representations of a finite group over $k$?

Some miscellaneous necessary conditions are that 


*

*one of the $n_i$ must be equal to $1$ (the trivial representation),

*each $n_i$ must divide $\dim A = |G| = \sum n_i^2$,

*the number of $n_i$ equal to $1$ (the size of the abelianization of $G$) must also divide $\dim A$,

*if $\dim A$ is prime (so $G$ must be cyclic), then each $n_i = 1$. 


The first condition rules out $M_n(k)$ for $n \ge 2$. The second condition rules out lots of examples, such as $k \times M_2(k)$. The third condition rules out lots more examples, such as $k \times k \times M_3(k)$. And so forth. I can't imagine there are any useful sufficient conditions; there are a lot of finite groups and they can be very complicated. 
A: If you take a solvable finite dimensional associative algebra $A$ with 1 having a separable radical factor structure, then, by the Wedderburn-Malcev-Theorem, there exist an abelian subalgebra $T$ with $A=rad(A)\oplus T$. The unit group $E(A)$ is the semidirect product of $1+rad(A)$ and $E(T)$. You can show that $\langle E(A) \rangle _K=A$ holds, if the the underlying field has more than two elements. Therfore you get a surjection from $KE(A)$ to $A=\langle E(A) \rangle _K$. So here we have another connection but not equality. (solvable is defined that $A/rad(A)$ is commutative).
