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Math Help - Help with finite p-group

  1. #1
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    Exclamation Help with finite p-group

    Let G be a finite p-group and H<G. Prove that H<NG(H).
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  2. #2
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    I think the proof goes as follows (though I'm used to asking questions here, not helping ).

    G is finite and with a prime order. Therefore it's cyclic (have you studied that? It's pretty basic).
    A cyclic group is of course commutative (abelian).
    Therefore, for every sub-group H in R, the normalizer in G N(H) is actually G itself...
    So obviously H is a subgroup of N(H)=G...
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  3. #3
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    Quote Originally Posted by bogazichili View Post
    Let G be a finite p-group and H<G. Prove that H<NG(H).
    suppose the claim is false for some p-groups. choose G to be one of those groups but with smallest possible order. so there exists a subgroup H of G such that N_G(H)=H.

    we know that in a p-group |Z(G)| > 1. clearly Z(G) \subseteq N_G(H)=H. let G_1=\frac{G}{Z(G)}. then |G_1| < |G|. now draw a contradiction!


    Remark: the above result holds in a much larger family of groups, e.g. any finitely generated nilpotent group.
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