## 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?

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 These are also called malnormal subgroups: en.wikipedia.org/wiki/Malnormal_subgroup – Agol Dec 19 2011 at 23:00 terrytao.wordpress.com/2013/05/24/… – Agol 14 hours ago

## 4 Answers

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.

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Wonderful, thank you very much! – Alex Bartel Apr 27 2011 at 13:10
What Geoff points out also suggests strongly to me that even if someone concocts a character-free proof, it's likely to be much less transparent than the older proof and also probably much longer. – Jim Humphreys Apr 27 2011 at 14:13
@Jim I do not doubt that. On the contrary, the absence of a character-free proof is a very impressive fact for an undergraduate to hear in a rep theory course, so I was very much hoping that it is still true. – Alex Bartel Apr 27 2011 at 14:46
I was not aware of this result, and I will certainly mention it next time I teach representation theory of finite groups (next to the $p^aq^b$ theorem). Are there obvious counter-examples to Frobenius' result for infinite groups? (being a Frobenius complement is usually called "malnormal") – Alain Valette Apr 27 2011 at 16:13
Alex, it's not clear that the lack of a character-free proof would be very impressive to an undergraduate. Perhaps they find that theorem uninteresting! I agree it is good to mention examples of results not directly about representation theory whose only known proofs use characters or whose shortest proofs use characters. But, to take a different example, I never found the p^aq^b theorem to be really exciting, so although I made a mental note that its only short proof uses characters I needed to get my motivation for caring about repn theory from elsewhere (Artin L-functions). – KConrad Apr 27 2011 at 19:15
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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.

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 Dear Geoff, do you happen to have references to other "normal complement" theorems that you refer to? – Alex Bartel Dec 20 2011 at 5:02 There is a book by M.J. Collins which shoud have most of the references, but unfortunately I don't remember its title (it's in CUP). I'll try and dig out some of teh refrences. – Geoff Robinson Dec 20 2011 at 8:46 Dear Geoff, thanks, I have found it. The title is "Representations and characters of finite groups". – Alex Bartel Dec 20 2011 at 9:37 Good. Glad you located it. – Geoff Robinson Dec 20 2011 at 11:08

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!

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You may also be interested in the following references:

K. Corr´adi and E. Horv´ath, 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 Alegbra, 228, 2000, 367-376.

(I hope I didn't screw these up too badly.)

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