Enumerating all finite groups is a formidable task and as far as I know an unsolved problem in its full generality. I have thought about what if one approaches this by constructing the groups stepwise by constructing character tables by systematically combining sets of "valid" irreducible representations. That leads me to two basic questions:

  1. Is every group uniquely characterised (up to isomorphism) by the full set of its irreducible representations?

  2. Is every set of "characters" grouped into irreducible representations which fulfills the Schur theorem automatically a "character table" belonging to a group?

Edit: In case the answer to 1. is no: Which information is missing?

  • $\begingroup$ While dwelling on it for a bit longer, I think it must be connected with the associativity of the group elements. I suppose any suffifiently well behaved multiplication table with column- and line-wise unique entries can be reduced to give a "character table" fulfilling Schurs theorem, while many of them will not be associative. $\endgroup$ – Rudi_Birnbaum Feb 13 '18 at 13:00

I think the answer to Question 1 is yes. The irreducible representations of $G$ over the field $K$ are homomorphisms $\rho_i:G \to M_i$ for $1 \le i \le k$, where $M_i = {\rm GL}(n_i,K)$ for some $n_i$. So we can define a homomorphism $\rho:G \to M_1 \times \cdots \times M_k$ by $\rho(g) = (\rho_1(g),\ldots,\rho_k(g))$.

Provided that the characteristic of $K$ does not divide $|G|$, the intersection of the kernels of the $\rho_i$ is trivial, and so $\ker \rho = 1$ and $G \cong {\rm im} \rho$. I am interpreting your assumption as meaning that $\rho_i(g)$ is known for each $i$ and $g$, and hence so is $\rho(g)$.

  • $\begingroup$ OK so far the trivial part (Q1). Up to isomorphism actually means that the map you are referring to is irrelevant, so I suppose thats technically the same as if it is known. $\endgroup$ – Rudi_Birnbaum Feb 14 '18 at 12:55

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