solvable groups have at least two irreducible representations of dim 1

Question

Let G be a finite group and $n$ the number of irreducible characters of dimension $1$ of G.

Prove that $n>1$.

The hint I have is not convincing or somewhat unclear: "$G^{ab}:=G/[G, G]$ is non trivial because G is solvable" This is okay.

Then it goes on: "Hence it has a non trivial irreducible representation of dimension 1". I don't see how this proves that n>1, since it states that $G^{ab}$ has an irreducible representation of dimension $1$ but we're dealing with $G$ here. Did I miss something?

Thank you for your help.

Answer 1

Use the following three facts:

1. If $N$ is a normal subgroup then the irreducible representations of $G/N$ correspond to the irreducible representations of $G$ on which $N$ acts trivially.
2. The one-dimensional irreducible representations of $G$ correspond to the irreducible representations of $G^{ab} = G/[G,G]$.
3. If $G$ is finite abelian then there are $|G|$ irreducible (necessarily one-dimensional) isomorphism classes of representations of $G$.