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Thread: 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 $\displaystyle A \subseteq G$, consider the set $\displaystyle N_G(A) = \{g \in G | gAg^{-1} = A\}$.

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

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