Timeline for p-groups as Sylow subgroups
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2011 at 4:54 | comment | added | Jack Schmidt | The abelian group 4×2 is not a Sylow 2-subgroup of GL(n,q), SL(n,q), Sym(n), or Alt(n) for the reasons stated in this answer: it is 2-nilpotent forcing. However, one takes advantage of [GL,GL]=SL and [Sym,Sym]=Alt (in virtually all cases) to see an abelian Sylow 2-subgroup of a 2-nilpotent group from these families has to be cyclic. | |
| Feb 7, 2011 at 18:47 | comment | added | Frieder Ladisch | A somewhat silly comment: The cyclic group of order $2^n$ is a Sylow $2$-subgroup of $GL(1,p)=\mathbb{F}_p^{*}$, if $p$ is a prime of the form $p=1+2^n+r2^{n+1}$. (By Dirichlet, there are such primes.) | |
| Feb 7, 2011 at 17:22 | comment | added | Steve D | A simpler proof is to look at the action of an element of order 2 via conjugation, it gives a homomorphism to the symmetric group S_|G| which is not contained in A_|G|, then proceed by induction. | |
| Feb 7, 2011 at 15:03 | comment | added | Frieder Ladisch | The first assertion follows from Burnside's normal $p$-complement theorem: if a Sylow $p$-subgroup of $G$ is contained in the center of its normalizer, then $G$ has a normal $p$-complement. This theorem and its proof are contained in nearly any group theory book. | |
| Feb 7, 2011 at 14:30 | history | answered | Chris Godsil | CC BY-SA 2.5 |