Abelian groups cannot have characters of degree 2

Question

I was attempting the following exercise:

Assume that $G$ is a simple group. Let $\chi$ be an irreducible character of degree $2$, and $g \in G$ be an element of order $2$. Prove that $\chi (G) \neq 2$.

After spending a while on it, I didn't get far with it, so decided to look at a skeleton solution. This solution relies on the following fact:

Abelian groups cannot have irreducible characters of degree $2$.

I can't figure out why this is true though! It doesn't seem particularly obvious to me, and I don't see how I would go about verifying this fact for myself. Would anyone be able to give me any ideas?

Thanks in advance!

Answer 1

If you haven't done the relation between the number of irreducible characters and conjugacy classes yet, but you have covered Schur's lemma, you can follow up on Adam's hint. Let $V$ be an irreducible $G$ module with $G$ a finite abelian group.

1. The module action $f_g:x\to g\cdot x, x\in V,$ by a (fixed) element $g\in G$ is a homomorphism of $G$-modules, because for all $h\in G$ $$f_g(h\cdot x)=g\cdot(h\cdot x)=(gh)\cdot x=(hg)\cdot x=h\cdot f_g(x).$$
2. By Schur's lemma the mapping $f_g$ is multiplication by a scalar $\lambda_g$, i.e. $$g\cdot x=\lambda_g x$$ for all $x\in V$.
3. The above applies to all the elements of $G$, so they all act as scalars.
4. Therefore every subspace of $V$ is also a submodule.
5. So for $V$ to be irreducible it must be 1-dimensional.

Answer 2

As $G$ is abelian therefore $G$ has $o(G)$ conjugacy classes which also implies $G$ has $o(G)$ irreducible characters.Setting,$o(G)=n$. Suppose $v_1,v_2,...,v_n$ be dimensions of these charactes then taking the regular $CG$ module one obtains that $v_1^2+v_2^2+...+v_n^2$ =n.Now as each $v_i$ is a nonnegative integer therefore $v_i=1$ for each $i$