# 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?

• These are also called malnormal subgroups: en.wikipedia.org/wiki/Malnormal_subgroup Dec 19, 2011 at 23:00
• terrytao.wordpress.com/2013/05/24/… May 25, 2013 at 12:58
• 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. May 29, 2013 at 2:16

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.

• @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. 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") 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). Apr 27, 2011 at 19:15
• We are looking at transitive permutation groups where non-identity elements fix at most one point. All nontrivial finite permutation groups in which a non-identity element fixes a point has a subgroup of this form (if n is the max number of points fixed by a nonidentity element, let H be the pointwise stabilizer of n-1-points each fixed by a non-identity element, and take an H-orbit on which an element $\neq 1$ has n-1 fixed points. Also see literature on Zassenhaus groups -(consider linear groups with free action on the non-zero elements of the vector space, which are Frobenius complements). May 29, 2011 at 13:56
• @AlainValette, better late than never, there are very easy counterexample in the infinite case. The theorem says that every malnormal subgroup of a finite group is a retract. This fails, for instance, for any subgroup of a free group generated by a commutator.
– HJRW
Dec 8, 2016 at 22:31

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!

• The even order case was done by Burnside (1898), ¶105 page 143 (or ¶134 page 172 in the 2nd ed), not Bender. Jan 31, 2014 at 19:39
• Thanks, Jack, indeed it is Burnsides's. Somehow I knew that, but when writing the answer I came across a reference that said Bender proved that part (my recollection was a book on character theory by B. Huppert, but now I am not that sure). The reference for Grun is: Beitrage zur Gruppentheorie II. Uber einen Satz von Frobenius. J. fur die Reine und Angew. Math. 186 (1945), 165-169. The reference for Shaw is: Remark on a theorem of Frobenius. Proc. A. M. S. 3 (1952), 970-972. Dec 9, 2016 at 0:03

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.)

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.

• Dear Geoff, do you happen to have references to other "normal complement" theorems that you refer to? 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. Dec 20, 2011 at 8:46
• Dear Geoff, thanks, I have found it. The title is "Representations and characters of finite groups". Dec 20, 2011 at 9:37
• Good. Glad you located it. Dec 20, 2011 at 11:08
• You do indeed need to know need to know that $\tilde{\mu}$ is actually a character of $G$. But we know this because: it is a generalized character by the characterization of characters, as indicated above. Furthermore, we do have $\langle \tilde{\mu}, \tilde{\mu} \rangle = 1$ (because of the embedding of $H$). Hence $\tilde{\mu}$ is $\pm$ an irreducible. But since $\tilde{\mu}(1) >0$, $\tilde{\mu}$ is indeed a genuine (irreducible) character. The last two facts are implicit in the "as before" in the answer as written. May 21, 2021 at 9:36

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.

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?

• It is not true that all odd-order Frobenius complements are cyclic, and there are other possibilities for those with even order - for example the group ${\rm SL}_2(5)$ of order 120. This is all "well-known". Apr 18, 2020 at 16:29
• As Derek Holt says, the perfect $G = {\rm SL}(2,5)$ does occur as a Frobenius complement (though it is the only perfect group which does). H. Zassenhaus knew the structure of Frobenius complements long ago. There is a Frobenius group $H = V.G$, where $V$ is elementary Abelian of order $121$, for example ( and $G \cong {\rm SL}(2,5)$.) Apr 18, 2020 at 17:36
• As a specialist in combinatorics rather than algebra, I don't know the structure of SL(2,5), but I have constructed the Frobenius group of order (121)(120) and discovered an element of order 120, and I stand by my answer. Apr 30, 2020 at 13:05
• There is such a Frobenius group. In fact, there is a Frobenius group of order $q^{n}(q^{n}-1)$ whenever $q$ is a prime power, with a cyclic Frobenius complement of order $q^{n}-1$, soming from a "Singer cycle in ${\rm GL}(n,q)$. In the case of $q^{n} = 121,$ there is asl a Frobenius group of order $121 \times 120$, where the Frobenius complement is the non-Abelian group ${\rm SL}(2,5)$ of order $120$. Apr 30, 2020 at 13:41