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If I know that a certain group has only finitely many irreducible representations (let's say over the complex numbers), is that group necessarily finite?

In the following cases, you can see that there will be an infinite number of irreducible representations(irreps).

  1. If the group is a direct product of infinitely many (non-trivial)groups.
  2. If the group is finitely generated, infinite, and abelian.
  3. If there exists a normal subgroup N such that G/N satisfies conditions 1 or 2.

I'm stuck. Any ideas? Given an arbitrary infinite group, can you construct infinitely many irreps?

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For 2., you can drop "finitely generated". – user2035 Jun 20 '12 at 7:55
Are you talking about unitary (possibly infinite-dimensional) Hilbert space representations? Clearly, there are groups which do not have any non-trival finite-dimensional dimensional representation. – Andreas Thom Jun 20 '12 at 8:13
Seconding Andreas's comment (cf. Bugs Bunny's answer). Do you want only finitely many fin-dim irreps, or groups with genuinely only finitely many irreps? – Yemon Choi Jun 20 '12 at 8:22

This is not an answer, but it's a bit too long for a comment.

If I'm not mistaken, any group $G$ which embeds densely into a compact Hausdorff group $H$ has the property that it is finite if and only if it has finitely many irreducible representations. This includes the residually finite groups, which embed densely into their profinite completions. To see this, use the fact that by Peter-Weyl $L^2(H)$ decomposes as a direct sum $\bigoplus_i n_i V_i$ where $V_i$ are the finite-dimensional unitary irreducible representations of $H$ and $n_i = \dim V_i$. It follows that there are infinitely many $V_i$ if and only if $H$ is infinite if and only if $G$ is infinite; moreover, by density the $V_i$ are irreducible representations of $G$, any two which are inequivalent as $H$-representations are inequivalent as $G$-representations.

So to find a counterexample we should look for groups that are not residually finite. Infinite simple groups certainly have this property, but I don't know any of them well enough to analyze their irreducible representations.

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What is about $\mathbb Z$, doc? Last time I saw it, it was sitting densely in the torus but it has infinitely many irreducible complex reps! – Bugs Bunny Jun 20 '12 at 8:54
It is even in ${\mathbb C}/{\mathbb Z}$ instead of torus... – Bugs Bunny Jun 20 '12 at 8:56
@Bugs: yes, and it is infinite... – Qiaochu Yuan Jun 20 '12 at 9:07
Sure! Thanks for pointing this out :-)) – Bugs Bunny Jun 20 '12 at 9:14

No way, doc! Take $PSL_2$ over a field of cardinality alef-2012. It has exactly one complex irreducible rep up to an equivalence...

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Certainly it has exactly one finite-dimensional complex irreducible rep, but for any group $G$ the group algebra $\mathbb{C}[G]$ has a maximal left ideal by Zorn's lemma and the quotient with respect to this ideal is necessarily a nontrivial complex irreducible representation of $G$, possibly of dimension $|G|$. – Qiaochu Yuan Jun 20 '12 at 7:49
You also seem to have an assumption like $\aleph_{2012}>2^{\aleph_0}$. – user2035 Jun 20 '12 at 7:54
thank you! now...are there any nice characterizations of such groups (with finitely many irreps)? – Amudhan Jun 20 '12 at 7:54
I doubt it. Course', you can say that it is equivalent to ${\mathbb C}G^\circ$ being finite dimensional but it is not useful... – Bugs Bunny Jun 20 '12 at 9:19

What happens for infinite simple groups?

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Evidently any non-trivial representation of a simple group is faithful, so the group is linear. But if a linear group is finitely generated (and there are many such examples of finitely generated simple groups) then it is residually finite, which is impossible for an infinite simple group. – HJRW Jun 20 '12 at 8:55
@HW: I understand "linear" to refer to faithful finite-dimensional representations (and Mal'cev's theorem requires this hypothesis). Every group is linear in the sense that it admits a faithful representation (the regular representation). – Qiaochu Yuan Jun 20 '12 at 16:08

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