The question is in the title, but here is some background/reminders:


A subgroup $H\neq\{1\}$ of a finite group $G$ is called a *Frobenius complement* if $H\cap H^g = \{1\}$ for all $g\in G\backslash H$. Given such a Frobenius complement, the corresponding *Frobenius kernel* is defined by
$$
N = \left(G\backslash\bigcup\_{x \in G}H^x\right)\cup\{1\}.
$$
Frobenius proved that $N$ is a normal subgroup of $G$, from which it follows immediately that $G$ is a semidirect product of $N$ and $H$. Frobenius's proof is a little gem of mathematics, using character theory. It is now over 100 years old and, at least at the beginning of this century, no alternative proof was known. My question is just a confirmation request, lest I should say something false in my upcoming representation theory lecture:



> 
> Is there still no proof not using character theory of the fact that a Frobenius kernel is a normal subgroup?
> 
> 
> 



Character-free proof that Frobenius kernel is a normal subgroup?
Nothing much to say here. There is (as of now) no proof of this fact without character theory. Although I think there is a direct counting proof when $H$ has even order, and a transfer argument
tells you that in a minimal counterexample, $H$ must be perfect (since $H$ is a Hall subgroup
of $G$). Hence in a minimal counterexample, $H$ must be a non-trivial perfect group of odd order. There is no such group, but proving that requires a lot more character theory than the proof
of Frobenius.


