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Jul 9, 2016 at 20:32 comment added Nicolas Thanks very much once again. It is very clear and very helpful :-)
Jul 9, 2016 at 20:08 comment added Geoff Robinson It would be easier to look up the standard proof as I only gave a sketch: Sylow $7$ $S$ is Abelian of exponent (at most) $7$ ( the exponent $7$ comes form looking at reduction (mod $7$)). If $S$ contains an elementary Abelian subgroup $A$ of order $49$, then $A$ may be assumed to consist of all diagonal matrices of order $7$ determinant $1$ so contains an element of order $7$ with eigenvalues as stated, which contradicts a result of Blichfeldt ( see L. Dornhoff, Part A) for example).
Jul 9, 2016 at 19:28 comment added Nicolas Hi Geoff, I may ask again for your help as I am really stuck on this one: Now $G$ has cyclic Sylow $7$-subgroups for otherwise, $G$ contains a element of order $7$ with eigenvalues $1,\omega, \omega^{-1}$, where $\omega = e^{\frac{2 \pi i}{7}}$, I could show that the $7-$Sylow is abelian but that's all.
Jul 8, 2016 at 21:11 comment added Geoff Robinson An indication of the power of modular representation theory is given by the fact that finite primitive subgroups of ${\rm GL}(4,\mathbb{C})$ were determined by Blichfeldt in about 1917, while the primitive subgroups of ${\rm GL}(5,\mathbb{C})$ were determined by Brauer in 1965.
Jul 8, 2016 at 21:05 comment added Nicolas Thanks very much Geoff for the explanation. I would not have found this one by myself :-) This glimpse at modular representation theory shows it is very powerful !
Jul 8, 2016 at 18:17 comment added Geoff Robinson As for reduction $(mod $p$)$, I like Curtis and Reiner 1962 ( Representation Theory of Finite Groups and Associative Algebras) but all books on modular representation theory cover it.
Jul 8, 2016 at 18:16 comment added Geoff Robinson The Sylow $7$-subgroup is self-centralizing by a subtle argument which I think originates with Brauer. Take a prime $q \neq p$ and suppose $x \in C_{G}(P)$ is an element of order $q$. Reduce the representation (mod $q$), and note that $P$ is diagonalizable, and since a generator of $P$ has three different eigenvalues, anything which centralizes $p$ is daigonalizable too. Hence $x$ acts trivially (mod $q$), so $x \in O_{q}(G) = 1$. Alsao $N_{G}(P)$ can't be a Frobenius group of order $42$. I have still left out a few details.
Jul 8, 2016 at 15:10 comment added Nicolas @ Geoff. Do you have a preferred reference for some examples of reduction? Also, how do you prove that a Sylow $-$normalizer must have order 21. All I have is $C_G(P<N)G(P)$ by Burnside and $[N_G(P):C_G(P)]\in \{2,3,6\}$ by the N/C lemma.
Jul 8, 2016 at 12:55 vote accept Nicolas
Jul 7, 2016 at 22:06 comment added Geoff Robinson Yes, reduction mod p refers to realising the representation over a local ring and looking at images mod the unique maximal of that ring in the standard manner.
Jul 7, 2016 at 21:06 comment added Nicolas Thanks vm Geoff. It will take me some time to go through your proof but it is extremely precious ! May I just ask: when you mention reduction, do you refer to the use of modular representations or something else ?
Jul 7, 2016 at 20:49 history edited Geoff Robinson CC BY-SA 3.0
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Jul 7, 2016 at 20:32 history answered Geoff Robinson CC BY-SA 3.0