Infinite abelian group counterexample

Question

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 $\mathbb C$.

Answer 1

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 $K\subset\mathbf{C}$.)