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A finite group $G$ is abelian iff all its irreducible representation $\rho$ have dimension 1.

I'm looking for a counter-example when $G$ is an infinite group. Are there any?

EDIT We're dealing with finite representations over $\mathbb C$.

Asaf Karagila
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draks ...
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    See http://math.stackexchange.com/questions/358104/nonabelian-group-with-all-irreducible-representations-one-dimensional – anon Jun 03 '14 at 09:46
  • The correct statement you're thinking of is "a finite group is abelian iff all its irreducible representations have dimension $1$," which is not the subject of your link. – anon Jun 03 '14 at 09:46
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    @draks... [...] The answer is then no: for instance if $G$ is an infinite finitely generated simple group, then its only finite-dimensional rep is the trivial one. On the other hand it's true that for an abelian group, any irreducible finite-dimensional complex representation is 1-dimensional. (ed ajf) – YCor Jun 03 '14 at 12:28
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    additional example: if $K$ is an algebraically closed field of characteristic 0, then every finite-dimensional irreducible complex representation of $K^*\ltimes K$ is 1-dimensional. The difference with the previous example is that this group admits a faithful finite-dimensional complex representation if $K$ is not too big. – YCor Jun 03 '14 at 12:31
  • @YCor: Since the title states "Infinite Abelian group counterexample", I would guess the intended meaning of the question is "find an Abelian ininite group with irreducible representation of (finite) dimension $>1$" - that is, a counterexample to the opposite implication. (I just mention it because I am curious whether such a thing exists). – Pavel Čoupek Jun 03 '14 at 13:08
  • @PavelC: I answered this in the comment: there's no such example (trivial consequence of Burnside's density theorem). – YCor Jun 03 '14 at 13:58
  • @YCor would you mind posting an answer? – draks ... Jun 03 '14 at 13:59

1 Answers1

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1) for any abelian group $G$, every irreducible finite-dimensional complex representation $\rho$ is 1-dimensional. Indeed, let $A$ be the $\mathbf{C}$-linear subspace generated by $\rho(G)$; this equals the $\mathbf{C}$-algebra generated by $\rho(G)$, and is abelian because $G$ is abelian. By Burnside's theorem, $A=M_d(\mathbf{C})$. Therefore $d=1$. (The argument works over any algebraically closed field.)

2) on the other hand there are many non-abelian groups for which every finite-dimensional irreducible complex representation is 1-dimensional. Here are various sources of examples:

a) groups with no non-trivial finite-dimensional representation. For finitely generated groups, these are exactly those with no nontrivial finite quotient, e.g. infinite finitely generated simple groups. Actually it's already fine if every finite-dimensional linear representation has an abelian image (e.g., Thompson's group $F$)

b) solvable groups with no nontrivial finite quotients. One example is the affine "$ax+b$" group $K^*\ltimes K$ when $K$ is an algebraically closed field. (unlike groups in (a) these groups have faithful finite-dimensional linear representations, over $\mathbf{C}$ when $K\subset\mathbf{C}$.)

YCor
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  • What happens if we allow infinite-dimensional, say, unitary, representations? Are there nonabelian (locally compact) groups such that every continuous irreducible unitary representation is 1-dimensional? – Moishe Kohan Jun 04 '14 at 00:38
  • No: it's clear that unitary representations of $G$ (locally compact) separate the points (e.g. take $L^2(G)$); then irreducible unitary representations also separate the points (this can be reduced to a convexity argument). But 1-dimensional representations can only separate the points in abelian groups. – YCor Jun 04 '14 at 08:25