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Abstract->p-sylow groups

  1. Dec 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Let P be a p-sylow sbgrp of a finite group G.
    N(P) will be the normalizer of P in G. The quotient group N(P)/P is cyclic from order n.

    PROVE that there is an element a in N(P) from order n and that every element such as a represnts a generator of the quotient group N(P)/P

    2. Relevant equations
    3. The attempt at a solution

    Welll.... there is mP in N(P)/P such as (mP)^n = P -> m^n*P=P -> m^n = 1 ...
    If m has order that is less the n, we'll get a contradiction to the fact that mP is from order n.
    It's pretty obvious that every element of this kind is a generator of this group...But I realy feel I'm missing something... It's a 20 points question and the answer will take me 2 lines...

    Where is my mistake?

    TNX in advance
  2. jcsd
  3. Dec 21, 2009 #2
    You cannot deduce that [tex]m^n = 1[/tex] from [tex]m^n P = P[/tex]; that tells you only that [tex]m^n \in P[/tex]. However, there is a simple way to produce the element of order [tex]n[/tex] you need, using [tex]m^n[/tex].
  4. Dec 21, 2009 #3
    Hmmmm...Yep, you're right...
    So we have m^n is in P... We need to produce an element k of order n that is in N(P)... Hmmmm we have m^n*P=P, and m^n is in P, m is in N(P). We know something more about this m? We know nothing about its order but we know that mP=Pm, which means m^n*P=P*m^n, and it's still gives us nothing...

    Can you please give more detailed directions?
    I'll appreciate any kind of further help...

    TNX a lot
  5. Dec 21, 2009 #4
    Wait a sec! We know that m^n is in P and that P has order p^r for some r in N...So m^n must have an order that is a power of p, say p^k... so we know that m^(p^k) has order n in N(P)...
    But how will we discover what k is? And why it's a generator of N(P)/P?

    Will you help me please?

    Last edited: Dec 22, 2009
  6. Dec 22, 2009 #5
    bouncing this message
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