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By a complex irreducible representation of a group $G$, I mean a simple $\mathbb CG$-module. So my representations need not be unitary and we are working in the purely algebraic setting.

It is easy to see that any complex irreducible representation of a virtually abelian group is finite dimensional. (Recall that a group is virtually abelian if it has a finite index abelian subgroup.) I believe it was long ago an open question whether the converse is true and essentially I'm interested in the status of this question:

Is a group virtually abelian if and only if all its complex irreducible representations are finite dimensional?

Here is what I know about the problem:

  1. It was shown by E. Thoma that all irreducible unitary representations of a discrete group are finite dimensional if and only if the group is virtually abelian. This is, of course, a different category than I am looking at, but relevant.
  2. It was shown independently by Snider and Wehrfritz that a solvable group whose complex irreducible representations are all finite dimensional is virtually abelian, cf. here. However, it seems that the paper trail grows cold after this, in the sense that these papers don't seem to have many citations on Mathscinet. I am not sure if this is because they are only cited in old papers that are not recorded in citation part of Mathscinet, of if people stopped working on the problem.
  3. If $G$ is a finitely generated group whose irreducible representations are all finite dimensional, then $G$ is residually finite. This is because $\mathbb CG$ is semiprimitive, which means that it has enough irreducible representations to separate points, and hence $G$ has enough irreducible representations to separate points. But a theorem of Malcev says a finitely generated linear group is residually finite. I don't know if the problem can be reduced to finitely generated groups.
  4. As, @YCor pointed out (and was first observed by Kaplansky) a virtually abelian group has all its complex irreducible representation of bounded degree. Conversely, if $G$ has all its complex irreducible representations of bounded degree, then $G$ is virtually abelian by a result of Isaacs and Passman see here.
  5. B. Hartley proved that the question has a positive answer for locally finite groups.
  6. Passman and Temple showed that the problem reduces to finitely generated groups and also that if all complex irreducible representations of $G$ are finite dimensional, then $G/N$ is virtually abelian for some finitely generated normal subgroup $N$ of $G$, see here.
  7. Snider prove that the question has a positive answer for periodic (=torsion) groups. See here.

I am primarily interested in the case of a countable group.

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    $\begingroup$ Note that a virtually abelian group has all its irreducible representations finite dimensional of bounded dimension. So one can also ask about a (weaker) converse of this (stronger) statement. And one can also wonder whether, for any reason, the existence of irreducible representations of arbitrary large dimension might imply the existence of one of infinite dimension (this is not true in the -distinct- context of unitary representations of compact groups). $\endgroup$ – YCor Sep 6 '18 at 13:09
  • $\begingroup$ @YCor, I did wonder in that direction. I had hoped something like an ultraproduct of finite dimensional simple modules of unbounded dimension would give an infinite dimensional simple, but the problem seems to be that being simple is not first order in the language of modules over $\mathbb CG$ because you are not allowed to quantify over elements of $\mathbb CG$, which are constants, but only over elements of the module. And I guess the same is true for finite dimensionality? $\endgroup$ – Benjamin Steinberg Sep 6 '18 at 13:24
  • $\begingroup$ @YCor, I answer the first part of your comment in the edited question. $\endgroup$ – Benjamin Steinberg Sep 6 '18 at 13:31
  • $\begingroup$ Yes, the ultraproduct is even of uncountable dimension, so is never (for a countable group) a simple module. $\endgroup$ – YCor Sep 6 '18 at 13:42
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    $\begingroup$ Thanks. Actually, the reference to Passman-Temple "Groups with All Irreducible Modules of Finite Degree" yields another important source of examples: the class of groups with only finite-dimensional complex irreps is stable under taking subgroups (Lemma 4(ii)). Thus, if a groups has an infinite-dimensional irrep, so do all its overgroups. In particular, all groups with a subgroup isomorphic to $F_2$ have an $\infty$-di irrep. Combined with the solvable case and the Tits alternative, this shows that linear groups satisfy the conjecture. This paper contains other striking restrictions. $\endgroup$ – YCor Sep 6 '18 at 19:21

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