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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.

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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.

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  • $\begingroup$ @yuan Thanks for the answer. You gave the perfect answer I was looking for. Just one question: Does the infinite dimension case has been studied or can be derived from some results? $\endgroup$ – Cusp May 14 '15 at 6:02
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    $\begingroup$ @Cusp: group rings of infinite groups seem pretty difficult to understand. There are some big open problems in this area: see, for example, en.wikipedia.org/wiki/Kaplansky%27s_conjecture. $\endgroup$ – Qiaochu Yuan May 14 '15 at 6:05
  • $\begingroup$ It might, however, be interesting to notice that $M_n(k)$ can be realised as a groupoid algebra. $\endgroup$ – Mateusz Wasilewski May 14 '15 at 7:13
  • $\begingroup$ @Wasilewski Can you please elaborate your statement? $\endgroup$ – Cusp May 14 '15 at 16:04
  • $\begingroup$ If you have a groupoid of pairs $(i,j)$, where $i,j \leqslant n$, endowed with the partial composition operation $(i,j)\cdot (j,k) = (i,k)$ and not defined for pairs $(i,j)$ and $(k,l)$ if $j\neq k$, then the groupoid algebra is isomorphic to $M_n(k)$. This is not surprising because $(i,j)$ look like matrix units. $\endgroup$ – Mateusz Wasilewski May 14 '15 at 16:27
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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).

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