Results 1 to 3 of 3

Thread: Normalizer and Centralizer

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    Normalizer and Centralizer

    $\displaystyle H\leq G$ and $\displaystyle |H|=2$. Show that $\displaystyle N_G(H)=C_G(H)$. Deduce that if $\displaystyle N_G(H)=G$, then $\displaystyle H\leq Z(G)$

    If $\displaystyle h\in H$, then $\displaystyle |h|=2$ if $\displaystyle h\neq e$.
    $\displaystyle N_G(H)=\{g\in G:gHg^{-1}=H\}$
    $\displaystyle C_G(H)=\{{g\in G:ghg^{-1}=h, \ \forall h\in H\}$.

    How can I get to the point where they are equal with this information?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    Re: Normalizer and Centralizer

    well it's obvious that $\displaystyle C_G(H)$ is a subgroup of $\displaystyle N_G(H)$. so what we need to show is that the normalizer is a subgroup of the centralizer if |H| = 2.

    so let g be in $\displaystyle N_G(H)$. clearly $\displaystyle geg^{-1} = e$, and we know that $\displaystyle ghg^{-1} \in H$, where h is the non-identity element of H.

    but if $\displaystyle ghg^{-1} = e$, then gh = g, and thus h = e, contradiction. hence $\displaystyle ghg^{-1} = h \implies gh = hg$, so $\displaystyle g \in C_G(H)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    Re: Normalizer and Centralizer

    Quote Originally Posted by Deveno View Post
    well it's obvious that $\displaystyle C_G(H)$ is a subgroup of $\displaystyle N_G(H)$. so what we need to show is that the normalizer is a subgroup of the centralizer if |H| = 2.

    so let g be in $\displaystyle N_G(H)$. clearly $\displaystyle geg^{-1} = e$, and we know that $\displaystyle ghg^{-1} \in H$, where h is the non-identity element of H.

    but if $\displaystyle ghg^{-1} = e$, then gh = g, and thus h = e, contradiction. hence $\displaystyle ghg^{-1} = h \implies gh = hg$, so $\displaystyle g \in C_G(H)$
    I am good thanks. I marked the thread solved awhile ago.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Normalizer
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Oct 8th 2011, 01:19 PM
  2. Replies: 4
    Last Post: Mar 2nd 2011, 08:01 PM
  3. One more normalizer q
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 31st 2009, 02:49 PM
  4. Show centralizer of a = centralizer of a^-1
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Sep 25th 2008, 06:47 AM
  5. Normalizer
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Aug 13th 2008, 09:53 PM

/mathhelpforum @mathhelpforum