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Let G = D12, given by generators a,b with |a|=6, |b|=2, and ba=a-1b.

  1. Dec 9, 2011 #1
    Let G = D12, given by generators a,b with |a|=6, |b|=2, and ba=a-1b. Let H = { 1, a3, b, a3b }. Find the normalizer of H in G and find the subgroups of G that are conjugate to H.

    The normalizer of H is a subgroup containing H, so since H has index 3, either NG (H) = H or NG (H) = G.

    I'm confused about this part.

    By Lagrange, |G|=|H|[G:H], since normalizer must contain H, then NG(H)≥H. But then its order can be 6. I'm missing something here about [G:NG(H)]
     
  2. jcsd
  3. Dec 9, 2011 #2

    micromass

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    Re: Normalizer

    I honestly don't understand your question :frown:
    Everything you say is true. So what's confusing you?
     
  4. Dec 10, 2011 #3

    Deveno

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    Re: Normalizer

    no the order cannot be 6. NG(H) contains H as a subgroup. H is of order 4, and 4 does not divide 6.
     
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