Character-free proof that Frobenius kernel is a normal subgroup? 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?

 A: You may also be interested in the following references:
K. Corrádi and E. Horváth, Steps towards an elementary proof of Frobenius’ Theorem,
Comm. in Algebra, 24, No. 7 1996, 2285-2292.
Paul Flavell, A Note on Frobenius Groups, Journal of Algebra, 228, 2000, 367-376. 
(I hope I didn't screw these up too badly.)
A: Perhaps it is not too late to elaborate on Geoff answer. For the case when the subgroup $H$ has even order, H. Bender has a character-free proof, actually quite short. Next, when $H$ is solvable, O. Grun has a character-free proof essentially based on a transfer argument (this proof seems to be quite similar to one by R. H. Shaw). Now, by the Feit-Thompson odd-order theorem these two cases exhaust all possibilities for $H$; but alas, the odd-order theorem runs deeper and in its proof there is a lot of character theory!
A: It did occur to me that allowing even more character theory, namely Brauer's characterization of characters, there is a way to prove this theorem of Frobenius which is more amenable to generalization. Recall that Brauer's characterization of characters states that a class function $\theta$ of a finite group $X$ is a generalized character of and only if ${\rm Res}^{X}_{E}(\theta)$ is a generalized character for each Brauer elementary subgroup $E$ of $X$, where a Brauer  elementary subgroup of $X$ is a subgroup which is a direct product of a $p$-group and a cyclic group for a prime $p$ (which is not fixed in this definition). It is easy to see under the hypotheses of Frobenius' theorem that every Brauer elementary subgroup of $G$ is either conjugate to a subgroup of $H$ or else has order coprime to $|H|$. It follows, then, that whenever $\mu$ is an irreducible character of $H$, we may extend $\mu$ to a well-defined generalized character ${\tilde \mu}$ of $G$ by setting ${\tilde \mu}(x) = \mu(1)$ whenever the order of $x$ is coprime to $|H|$ and ${\tilde \mu}(x) = \mu(h)$ whenever $x$ is $G$-conjugate to $h \in H.$ Once this is done, the existence of the complement $K$ follows as before. There are many other "normal complement" theorems which can be proved by similar methods, by authors such as Brauer, Suzuki, Dade and Reynolds. Indeed, the use of "tamely imbedded" subsets to produce isometries in character rings occurs in the proof of the Feit-Thompson odd order theorem, and was used to eliminate some difficult residual group-theoretic configurations.
A: 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.
A: CW answer since this has only been posted in comments. After this question was posted in 2011, Terry Tao has (in 2013) posted a proof reducing to character theory only of abelian groups.
From Tao's post (May 24, 2013):

I didn’t succeed in obtaining a completely elementary proof, but I did find an argument which replaces character theory (which can be viewed as coming from the representation theory of the non-commutative group algebra ${{\bf C} G \equiv L^2(G)})$ with the Fourier analysis of class functions (i.e. the representation theory of the centre ${Z({\bf C} G) \equiv L^2(G)^G}$ of the group algebra), thus replacing non-commutative representation theory by commutative representation theory.

Copy of Geoff Robinson's comment (May 29, 2013):

The new proof of Terry Tao is an alternative proof. It reduces the problem in an ingenious way to one of character theory of commutative semisimple algebras, though it is close in spirit to the original character-theoretic proof. It had not been devised at the time when the accepted post below was written. It is probably still fair to say that there is no purely group-theoretic proof of Frobenius's theorem.

A: I'm pretty certain I have a purely group theoretic/combinatorial proof of Frobenius' Theorem only a few pages long. According to me, all odd-order Frobenius complements are cyclic, and I believe the only non-cyclic complements are the ordinary quaternions (Q8?) (K = Z3 x Z3 or Z5 x z5) and a group of order 24 with quaternion subgroup (Z5 x Z5). I believe as well that the same methods produce a fairly easy proof of the Feit-Thompson theorem, though I haven't got every detail down. Anyone interested in having a look? 
