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If a group G has order p^n, show any subgroup of order p^n-1 is normal

  1. Nov 28, 2011 #1
    If a group G has order [itex] p^n [/itex], show that any subgroup of order [itex] p^{n-1} [/itex] is normal in G.

    i have no idea how to start this. i know that to show that a subgroup N is normal in G, i need that [itex] gNg^{-1} = N [/itex]. so i start with any subgroup N of order [itex] p^{n-1} [/itex] but i have no idea how to continue.

    this problem appears in the section before the sylow theorems are introduced so i can't use them. i know that for p-groups, the center is nontrivial and has prime power order. also in this same section, Cauchy's theorem was introduced which says if p divides the order of a group then that group has an element of order p. these concepts were introduced fairly recently to me so this may be why i am having trouble.

    can someone give me a hint or 2 to continue? thanks
     
  2. jcsd
  3. Nov 28, 2011 #2

    micromass

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    Let H be your subgroup of order pn-1. Consider the action of G on the set [itex]X=\{Hg~\vert~g\in G\}[/itex] by multiplication on the right. This determines a group morphism

    [tex]G\rightarrow Sym(X)[/tex]

    with kernel N. The idea is to prove that N=H. Prove this in the following steps:

    1) [itex]N\subseteq H[/itex]
    2) [G:N]=p[H:N]
    3) G/N is a subgroup of Sym(X), so [G:N] divides p!
    4) [H:N]=1.
     
  4. Nov 29, 2011 #3

    Deveno

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    don't you actually want a left-action, so that you have a homomorphism, rather than an anti-homomorphism? i mean, usually mappings are written on the left, so we have x(gH), rather than (Hg)x (or Hg.x, if you prefer). it's a small (minor) issue, but usually actions are taken to be left-actions, because of the way we normally compose functions.
     
  5. Nov 29, 2011 #4

    micromass

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    It would indeed make more sense. But I'm so used to right actions and right cosets that I can't help it :biggrin:
     
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