It is a fairly well-known fact, often used as an exercise in textbooks/lecture courses on complex representation theory, that if $G$ is a non-abelian simple group, and $\chi$ is a non-trivial irreducible complex character of $G$, then $\chi(1)>2$.
The "standard" approach (at least the one I've seen) is the following.
- if $G$ is non-abelian simple, then $G'=G$ (here $G'$ is the derived subgroup). Therefore, $G$ has only one linear character, namely the trivial character. So $\chi(1)>1$.
- Using the fact that the degree of an irreducible complex character must divide the group order, we have that if $\chi(1)=2$ then $G$ has an element $g$ of order 2, and if $\rho$ is the representation affording $\chi$ then we can consider $\det\rho(g)$ (which is a linear character) and we quickly arrive at the result.
The analogous result in modular representation theory is the following.
- If $G$ is a non-abelian simple group, $p$ is an odd prime number, and $\chi$ is a non-trivial irreducible Brauer character of $G$, then $\chi(1)>2$.
This can be proved analogously as in the ordinary case: the first step works equally well and in order to find an element of order $2$ in $G$ we can use the Feit-Thompson theorem. However, I would prefer not to have to appeal to a theorem whose proof I don't know, and I can't think of any other way to do this. So my question is:
- Is there a proof of the above fact, which does not use the Feit-Thompson theorem?
The more elementary the solution, the better. Apologies if this has been asked on the website before.