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Math Help - Abstract Algebra

  1. #1
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    Abstract Algebra

    Let G be a finite group. and let H be a subgroup of G, and call N(H) the normalizer of H in G defined as, {x|xHx^-1 = H}.
    Prove that N(H) is a subgroup of G
    and prove that |{xHx^-1|x in G}| = |G/N(H)|
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  2. #2
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    To show N(H) is a subgroup take a,b \in N(H) and show ab^{-1}\in N(H). This is fairy simple if you first note that if xHx^{-1}=H then x^{-1}Hx=H.
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  3. #3
    Super Member Gamma's Avatar
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    The second part is a result of the counting formula which states that the size of the orbit is equal to the index of the stabilizer . N(H)\unlhd G and N(H)=stab(H) under the group action conjugation. The LHS is the orbit of H under conjugation, so these sets are equal in cardinality.

    Plus there is a really easy bijection to establish this fact. \phi : xHx^{-1} \rightarrow G/N(H) by \phi (xHx^{-1})=xN(H). Surjectivity is clear.
    Injectivity and well-defined is also pretty obvious.
    xHx^{-1}=yHy^{-1} \Leftrightarrow y^{-1}xHx^{-1}y=H\Leftrightarrow (y^{-1}x)H(y^{-1}x)^{-1}=H  \Leftrightarrow y^{-1}x \in N(H)\Leftrightarrow xN(H)=yN(H)
    Last edited by Gamma; December 7th 2008 at 04:05 AM. Reason: added well definition
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