If $G$ is an abelian group, then every irreducible character has dimension one (i.e. is linear), for finite group we also have a converse. Do we have a converse for infinite groups? Or:
Does there exists an infinite group all whose irreducible characters are linear, which is not abelian?
I append a proof that $G$ abelian implies all irreducible characters are linear: If $G$ is an abelian group and $V$ an irreduble $\mathbb C[G]$-module, then by Schur's lemma we have $\operatorname{dim}\operatorname{Hom}_{\mathbb C[G]}(V, V) = 1$. As $G$ is abelian, every $g$ corresponds to a $\mathbb C[G]$-homomorphism. Hence in its action on $V$ could be identified with some homomorphism of the form $\lambda \cdot I_n$ with $n = \operatorname{dim}V$. If $n > 1$ then such a map would leave every space invariant, also the ones of dimension $< n$, hence $n = 1$. $\square$
Yes. Take an infinite simple group $G$, such as $PSL_3(K)$ for a sufficiently large field $K$ of sufficiently large characteristic, with cardinality strictly larger than $\mathbb{C}$. Then every finite-dimensional representation of $G$ over $\mathbb{C}$ is trivial, so its only irreducible character is the trivial one.
There are more complicated examples of finitely generated or even finitely presented groups all of whose finite-dimensional representations are trivial.