# (Infinite) Non-abelian group with only linear characters

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 $n = 1$. $\square$

• For the form of Schur's lemma you're using you need the assumption that $V$ is finite-dimensional. – Qiaochu Yuan Jan 17 '16 at 23:06
• @QiaochuYuan Thanks for pointing out! So this direction is also not working in general? – StefanH Jan 17 '16 at 23:06
• Well, it depends on whether you care about irreducible characters or irreducible representations. If the latter, sometimes irreducible representations are infinite-dimensional, so don't have characters in any obvious sense. – Qiaochu Yuan Jan 17 '16 at 23:07
• @QiaochuYuan Okay, think to answer that satisfactory I need to learn more representation theory, I just started, so maybe I will come back for more specific questions later when I understand the theory better. – StefanH Jan 17 '16 at 23:15

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.