Character free proof that Frobenius kernel is a normal subgroup? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:00:11Z http://mathoverflow.net/feeds/question/63142 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63142/character-free-proof-that-frobenius-kernel-is-a-normal-subgroup Character free proof that Frobenius kernel is a normal subgroup? Alex Bartel 2011-04-27T09:52:10Z 2011-12-19T23:10:21Z <p>The question is in the title, but here is some background/reminders:</p> <p>A subgroup <code>$H\neq\{1\}$</code> of a finite group $G$ is called a <em>Frobenius complement</em> if <code>$H\cap H^g = \{1\}$</code> for all $g\in G\backslash H$. Given such a Frobenius complement, the corresponding <em>Frobenius kernel</em> is defined by <code>$$N = \left(G\backslash\bigcup_{x \in G}H^x\right)\cup\{1\}.$$</code> 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:</p> <blockquote> <p>Is there still no proof not using character theory of the fact that a Frobenius kernel is a normal subgroup?</p> </blockquote> http://mathoverflow.net/questions/63142/character-free-proof-that-frobenius-kernel-is-a-normal-subgroup/63156#63156 Answer by Geoff Robinson for Character free proof that Frobenius kernel is a normal subgroup? Geoff Robinson 2011-04-27T12:24:01Z 2011-04-27T12:46:38Z <p>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.</p> http://mathoverflow.net/questions/63142/character-free-proof-that-frobenius-kernel-is-a-normal-subgroup/63333#63333 Answer by DavidLHarden for Character free proof that Frobenius kernel is a normal subgroup? DavidLHarden 2011-04-28T20:38:40Z 2011-04-28T20:38:40Z <p>You may also be interested in the following references:</p> <p>K. Corr´adi and E. Horv´ath, Steps towards an elementary proof of Frobenius’ Theorem, Comm. in Algebra, 24, No. 7 1996, 2285-2292.</p> <p>Paul Flavell, A Note on Frobenius Groups, Journal of Alegbra, 228, 2000, 367-376. </p> <p>(I hope I didn't screw these up too badly.)</p> http://mathoverflow.net/questions/63142/character-free-proof-that-frobenius-kernel-is-a-normal-subgroup/63361#63361 Answer by F Zaldivar for Character free proof that Frobenius kernel is a normal subgroup? F Zaldivar 2011-04-29T00:11:35Z 2011-04-29T00:11:35Z <p>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!</p> http://mathoverflow.net/questions/63142/character-free-proof-that-frobenius-kernel-is-a-normal-subgroup/83908#83908 Answer by Geoff Robinson for Character free proof that Frobenius kernel is a normal subgroup? Geoff Robinson 2011-12-19T22:42:32Z 2011-12-19T23:10:21Z <p>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 <code>${\rm Res}^{X}_{E}(\theta)$</code> 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 <code>${\tilde \mu}$</code> of $G$ by setting <code>${\tilde \mu}(x) = \mu(1)$ whenever the order of $x$ is coprime to $|H|$ and</code>${\tilde \mu}(x) = \mu(h)$` 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.</p>