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Let $\mathcal{A}$ be a semisimple $\mathbb{C}$-algebra. By the Artin-Wedderburn theorem, it is isomorphic to a direct product of matrix algebras: $$ \mathcal{A} = \prod_{i=1}^m M_{n_i}(\mathbb{C})$$

Under what conditions on $m,\,n_i$ can we say that $\mathcal{A} \simeq \mathbb{C}[G]$ for some finite group $G$?. This amounts to finding a group $G$ which has exactly $\#\{i \vert \, n_i = k \}$ non-isomorphic complex irreducible representations of dimension $k$, for each $k \in \mathbb{N}$. Note that we must ask for $n_i \vert \sum_{i=1}^m n_i^2$, since the dimensions of irreps divide the order of a group.

Of course, one can also phrase this in terms of "which finite dimensional $C^*$ algebras arise as group $C^*$ algebras?", so one can also consider the question for arbitrary unital $C^*$ algebras, asking under what sufficient conditions does it come from a discrete group?

Any idea or reference will be greatly appreciated.

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    $\begingroup$ See also mathoverflow.net/questions/314502/… $\endgroup$
    – Mare
    Commented Nov 23, 2020 at 18:15
  • $\begingroup$ This indeed seems to make a duplicate (I have the gr.group-theory golden badge which would close the question as duplicate from my only vote, so I don't vote). The other question has no accepted answer, probably because we don't really expect a definite answer. $\endgroup$
    – YCor
    Commented Nov 23, 2020 at 18:18
  • $\begingroup$ @Ycor I see, I only saw that post from Mare's answer, so should I delete this question as a result (I agree it's a duplicate)? $\endgroup$
    – pitariver
    Commented Nov 23, 2020 at 18:19
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    $\begingroup$ Examples of easy necessary conditions: write $n=\sum n_i^2$. Then necessarily $|G|=n$. For many values of $n$ this gives restrictions. For instance, if $n$ is prime, square of prime, or some other values such as $15$, then this is plausible only if all $n_i=1$. If $n\ge 3$ is odd, or of the form $p^aq^b$, $G$ should be solvable, so $n_1$ should be $>1$. $\endgroup$
    – YCor
    Commented Nov 23, 2020 at 18:20
  • $\begingroup$ Maybe. Actually the other question asks several questions, which leaves less focus on this very question, so I have no definite opinion. $\endgroup$
    – YCor
    Commented Nov 23, 2020 at 18:22

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