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Math Help - Given a subgroup A in G, consider the normalizer. Prove A is normal iff normalizer=G.

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    Given a subgroup A in G, consider the normalizer. Prove A is normal iff normalizer=G.

    Given a subgroup A \subseteq G, consider the set N_G(A) = \{g \in G | gAg^{-1} = A\}.

    Prove that A is normal if and only if N_G(A) = G.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JJMC89 View Post
    Given a subgroup A \subseteq G, consider the set N_G(A) = \{g \in G | gAg^{-1} = A\}.

    Prove that A is normal if and only if N_G(A) = G.
    This is by definition isn't it......namely N_G(A) is the largest subgroup H (and yes, it is always a subgroup) such that A\unlhd H
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    Quote Originally Posted by Drexel28 View Post
    This is by definition isn't it......namely N_G(A) is the largest subgroup H (and yes, it is always a subgroup) such that A\unlhd H
    The only definition I have for the normalizer is that it is the set N_G(A) = \{g \in G | gAg^{-1} = A\}.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by JJMC89 View Post
    The only definition I have for the normalizer is that it is the set N_G(A) = \{g \in G | gAg^{-1} = A\}.
    I mean for normality. One has that \displaystyle N\unlhd G if and only if gNg^{-1}=N for all g\in G which is precisely saying that N_g(N)=N.
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    Quote Originally Posted by Drexel28 View Post
    I mean for normality. One has that \displaystyle N\unlhd G if and only if gNg^{-1}=N for all g\in G which is precisely saying that N_g(N)=N.
    Oh right.
    I've got it now.
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