If a finite group has only 1D irreducible representations, is it abelian?

Question

I know abelian groups have only 1D representations. Is the converse proposition true? i.e. If a finite group has only 1D irreducible representations, is it abelian?

Answer 1

The number of irreducible representations is the number of conjugacy classes. Since the sum of the squares of the dimensions of the irreducible representations is the size of the group, if they're all one-dimensional, there are as many conjugacy classes as group elements, i.e. each group element is in a conjugacy class of its own. Hence the group is abelian.

Answer 2

Yes. The group algebra ${\bf C}[G]$ is the direct sum of matrix algebras, one for each irreducible representation: a $d$-dimensional representation contributes a copy of the algebra of $d \times d$ matrices. If each representation has $d=1$, then each of these matrix algebras is commutative, whence so is ${\bf C}[G]$, whence so is $G$. $\Box$