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I have a few related questions. First I would appreciate it if someone could provide me with a reference for the following

"Complex unitary irreducibles of virtually abelian groups have bounded degrees."

I am only concerned with complex unitary representations in this question.

Do we have a quantitative version of the above statement? For example, is the following true?

"If $G$ is a virtually abelian group with a finite index abelian subgroup H, then the degrees of irreducibles of $G$ are bounded above by $[G:H]$"

If H is additionally normal, is it true that the degrees of irreducibles of G divide $[G:H]$?

If these are not true in general, would they be true if $G$ is finitely generated in addition to being virtually abelian?

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    $\begingroup$ What kind of representations are you looking at? What field, etc. $\endgroup$
    – YCor
    Sep 14, 2022 at 14:50
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    $\begingroup$ @user3826143 YCor's point is not that nothing can't or hasn't be done in the "discrete" case, but rather that, on its own, assuming a group is "discrete" means nothing, and that one needs to be more specific. "Discrete" is normally used either with respect to a subgroup which is discrete with respect to a (non-discrete) specified topology on the overgroup, or else is more or less synonymous with "finitely generated", and indeed historically it was used as a counterpoint to continuous (Lie) groups. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Sep 14, 2022 at 16:16
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    $\begingroup$ The Thoma paper you refer to is about the (hard!) converse of your question: namely, he shows that if G is a (discrete) group for which all (continuous) unitary irreps are finite-dimensional then G is virtually abelian. This is much harder than the result you are looking for, which definitely predates Thoma's paper although I am not sure right now of the earliest reference $\endgroup$
    – Yemon Choi
    Sep 14, 2022 at 16:19
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    $\begingroup$ @YCor It seems that the OP's first question is answered by discussions on this old MO question mathoverflow.net/questions/309993/… but I am reading the 1949 paper of Kaplansky that is usually cited, and having some difficulties filling in the details. The paper has a proof that if $A \leq G$ is abelian with $|G:A|=n$, then every fin-dim irreducible unitary rep. of $G$ has degree at most n; but I don't quite follow how he justifies the claim that every topologically irreducible unitary rep. of G is fin-dim $\endgroup$
    – Yemon Choi
    Sep 14, 2022 at 18:40
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    $\begingroup$ @user3826143 I am not completely convinced (i.e. I do not fully understand) why/how Kaplansky's paper shows that every irreducible unitary rep of a VA group is fin-dim, so I am not quite ready to claim that it answers your original question. The paper is called Groups with representations of bounded degree, Canadian J. Math vol. 1 (1949), 105-112 $\endgroup$
    – Yemon Choi
    Sep 14, 2022 at 20:01

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I can at least explain how this bounded degree property works roughly when $G$ is a countable virtually abelian group. Kaplansky shows in his paper Groups with representations of bounded degree, Canadian J. Math vol. 1 (1949), 105-112 that the group ring of a virtually abelian group satisfies a polynomial identity.

If $K$ is any field, then the image of $KG$ under an irreducible representation is a primitive $K$-algebra satisfying this same polynomial identity. In Kaplansky, Rings with a polynomial identity, Bull. Amer. Math. Soc. 54 (1948) 575–580 that a primitive PI algebra is of the form M_k(D) where D is the division algebra which is the commutant of the irreducible representation and that D is finite dimensional over its center. Note that k can be bounded in terms of the degree of the polynomial identity (in Passman's book you can get precise bounds on the degree) because the smallest degree polynomial identity satisfied by $M_r(K)$ goes to infinity as $r$ goes to infinity.

If $G$ is countable, then every simple $\mathbb CG$-module is countable dimensional and hence the commutant of an irreducible representation has countable dimension over $\mathbb C$. But the only division algebra over $\mathbb C$ of countable dimension is $\mathbb C$ itself. Thus any irreducible representation of $\mathbb CG$ has image $M_r(\mathbb C)$ where $r$ is bounded.

In any event once you know that each representation is of finite degree then you can take a normal abelian subgroup of index say $m$ and use Clifford theory to say each irreducible of $G$ has degree at most $m$.

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  • $\begingroup$ In the uncountable case: take a proper extension $K$ of $\mathbf{C}$, e.g. $K=\mathbf{C}(t)$, and then the abelian group $K^*$ acts on $K$. This is an uncountable-dimensional irreducible representation of the abelian group $K^*$. $\endgroup$
    – YCor
    Sep 15, 2022 at 22:33
  • $\begingroup$ @YCor, yes this is true and I even mention this example in a paper a few years ago to show why countable is needed and now I have forgotten. $\endgroup$
    – Benjamin Steinberg
    Sep 15, 2022 at 23:31
  • $\begingroup$ This is fantastic! Thanks so much @BenjaminSteinberg. I love your introductory book on representation theory btw. $\endgroup$
    – user3826143
    Sep 16, 2022 at 15:15
  • $\begingroup$ Thanks, I'm glad you like it. $\endgroup$
    – Benjamin Steinberg
    Sep 16, 2022 at 15:55
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    $\begingroup$ Passman's 1977 book I think also has the more precise result but I could only do from the top of my head the case of a normal abelian subgroup. $\endgroup$
    – Benjamin Steinberg
    Sep 16, 2022 at 17:05

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