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Thread: Help with a proof: Normal subgroup

  1. #1
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    Help with a proof: Normal subgroup

    Hi.
    I need to prove:

    $\displaystyle H\triangleleft G \Leftrightarrow \forall g\in G,\hspace{5} \forall h\in H:\hspace{5} g^{-1}hg\in H$

    So, one direction is quite easy:

    $\displaystyle \Rightarrow$ :
    If $\displaystyle H$ is a subgroup of $\displaystyle G$, it follows that $\displaystyle \forall g\in G$: $\displaystyle gH=Hg$.
    in particular, for every $\displaystyle h\in H$, $\displaystyle gh=hg$ , so $\displaystyle h=g^{-1}hg$, $\displaystyle \forall g\in G, \forall h\in H$.

    but the second direction is not quite clear for me:
    $\displaystyle \Leftarrow$ :
    Since $\displaystyle \forall g\in G$, $\displaystyle \forall h\in H$: $\displaystyle g^{-1}hg\in H$, there exist some $\displaystyle h_0\in H$, such that $\displaystyle g^{-1}hg=h_0$
    and then:
    $\displaystyle gh_0=hg$

    but how do I know that $\displaystyle h_0=h$?

    Thanks in advance.
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  2. #2
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    Re: Help with a proof: Normal subgroup

    Quote Originally Posted by Stormey View Post
    $\displaystyle H\triangleleft G \Leftrightarrow \forall g\in G,\hspace{5} \forall h\in H:\hspace{5} g^{-1}hg\in H$

    So, one direction is quite easy:

    $\displaystyle \Rightarrow$ :
    If $\displaystyle H$ is a subgroup of $\displaystyle G$, it follows that $\displaystyle \forall g\in G$: $\displaystyle gH=Hg$.
    You mean, if H is a normal subgroup.

    Quote Originally Posted by Stormey View Post
    in particular, for every $\displaystyle h\in H$, $\displaystyle gh=hg$
    $\displaystyle gH=Hg$ does not mean this. It means that sets gH and Hg are equal. Now, gH = {gh | h ∈ H} and Hg = {h'g | h' ∈ H}. For an arbitrary h ∈ H, the fact that gh ∈ {h'g | h' ∈ H} means that there exists some h' ∈ H such that gh = h'g. Similarly, for every h' ∈ H there exists an h such that gh = h'g.
    Thanks from topsquark
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    Re: Help with a proof: Normal subgroup

    Hi, emakarov.
    thanks for the help.

    Quote Originally Posted by emakarov View Post
    You mean, if H is a normal subgroup.
    Yes, sorry. I meant normal.

    Quote Originally Posted by emakarov View Post
    $\displaystyle gH=Hg$ does not mean this. It means that sets gH and Hg are equal. Now, gH = {gh | h ∈ H} and Hg = {h'g | h' ∈ H}. For an arbitrary h ∈ H, the fact that gh ∈ {h'g | h' ∈ H} means that there exists some h' ∈ H such that gh = h'g. Similarly, for every h' ∈ H there exists an h such that gh = h'g.
    OK, I get now why it's wrong to say that $\displaystyle gH=Hg$ means gh=hg, so then how do I go from "there exists an h such that gh = h'g" to gh=hg for every g in G, and for every h in H?

    **edit

    scrap that last post.
    I got it.
    for some reason I thought that every normal subgroup is abelian...
    Last edited by Stormey; Nov 27th 2013 at 10:14 AM.
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    Re: Help with a proof: Normal subgroup

    Quote Originally Posted by Stormey View Post
    how do I go from "there exists an h such that gh = h'g" to gh=hg for every g in G, and for every h in H?
    You don't because gH = Hg does not imply gh = hg for every h ∈ H and g ∈ G. But gh = hg is not required for the proof you need.
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  5. #5
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    Re: Help with a proof: Normal subgroup

    Thanks.
    edited my last post.
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