1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sylow theory problem

  1. Mar 4, 2008 #1
    [SOLVED] sylow theory problem

    1. The problem statement, all variables and given/known data
    Let G be a finite group and let p be a prime dividing |G|. Let P be a Sylow p-subgroup of G. Show that N[N[P]]=N[P].

    2. Relevant equations

    3. The attempt at a solution
    Please confirm that this proof is correct.

    The Second Sylow Theorem tells us that any two Sylow p-subgroups of a finite group must be conjugate.
    The normalizer of P is the subgroup N[P]={[itex]g \in G[/itex] : [itex]gPg^{-1} = P[/itex]}. It is obvious that N[P] is the largest subgroup of G in which P is normal.
    It follows immediately that P is the only Sylow p-subgroup in N[P].
    The normalizer of the normalizer of P is the subgroup N[N[P]] = {[itex]g \in G[/itex] : [itex]gN[P]g^{-1} = N[P][/itex]}. I claim that P is normal in N[N[P]]. If g in N[N[P]], then [itex] gPg^{-1} \subset N[P] [/itex] and since P is the only Sylow p-subgroup in N[P], it follows immediately that gPg^{-1} = P. Thus g is contained in N[P] and N[N[P]] is contained in N[P].

    Why did they tell us that p is a prime that divides |G|? Doesn't that follow from the fact that there exists a Sylow-p-subgroup?
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Looks kosher.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Sylow theory problem Date
Subnormal p-Sylow Subgroup of Finite Group Dec 28, 2014
Sylow's Theorems and Simple Groups Nov 17, 2012
Sylow's First Theorem Oct 6, 2012
Group Theory (Sylow) Feb 25, 2009
Sylow theory Mar 4, 2008