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Thread: Normalizer of a subgroup of a p - group

  1. #1
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    Normalizer of a subgroup of a p - group

    Let $\displaystyle G $ be a group of order $\displaystyle p^n $ where $\displaystyle p $ is prime, and let $\displaystyle H $ be a proper subgroup of $\displaystyle G $. Prove that $\displaystyle H = N(H) $ where $\displaystyle N(H) $ is the normalizer of $\displaystyle H $.

    I have not yet studied theorems on p - groups. All I have done is theorems and concepts of subgroups, cosets and the Lagrange's Theorem. Can this question be solved using these concepts?
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  2. #2
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    I don't think I believe that this is true. If G is abelian, then the normalizer of any subgroup is equal to G.
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  3. #3
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    Quote Originally Posted by sashikanth View Post
    Let $\displaystyle G $ be a group of order $\displaystyle p^n $ where $\displaystyle p $ is prime, and let $\displaystyle H $ be a proper subgroup of $\displaystyle G $. Prove that $\displaystyle H = N(H) $ where $\displaystyle N(H) $ is the normalizer of $\displaystyle H $.

    I have not yet studied theorems on p - groups. All I have done is theorems and concepts of subgroups, cosets and the Lagrange's Theorem. Can this question be solved using these concepts?

    The claim is false: any finite p-group is nilpotent and thus fulfills the normalizer condition: $\displaystyle H<N_G(H)$ for any proper subgroup $\displaystyle H\,\,of\,\,G$

    Tonio
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