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Thread: Normal subgroup

  1. #1
    Ant
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    Normal subgroup

    If we take the definition of a normal subgroup to be:

    H is normal if $\displaystyle ghg^{-1} \in H $ for all h in H, g in G.

    So it is clear that $\displaystyle gHg^{-1} \subset H $

    However, I want to show that actually $\displaystyle gHg^{-1} = H $, by showing that $\displaystyle H \subset gHg^{-1} $

    I'm just struggling to find a way to prove the final part? Could anyone offer any help?

    Thanks,
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  2. #2
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    Re: Normal subgroup

    Since $\displaystyle ghg^{-1}\in H$ for all $\displaystyle g\in G,$ then given $\displaystyle g\in G$ you also have $\displaystyle g^{-1}h\left(g^{-1}\right)^{-1}\in H.$ So you also have $\displaystyle g^{-1}Hg\subseteq H$ for all $\displaystyle g\in G,$ i.e. $\displaystyle H\subseteq gHg^{-1}$ for all $\displaystyle g\in G.$
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  3. #3
    Ant
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    Re: Normal subgroup

    Thanks:

    Definition of normal says $\displaystyle ghg^{-1} \in H $ for all g in G and h in H. So we have $\displaystyle ghg^{-1} = h' $ for some h' in H and for all g in G and h in H. This immediately tells us that $\displaystyle gHg^{-1} \subset H $.

    But $\displaystyle ghg^{-1} = h' $ Holds for all g so in particular we can take $\displaystyle g=g^{-1}$ and we find:

    $\displaystyle g^{-1} h g = h'' $ for some h'' in H and for all h in H. This implies $\displaystyle h=gh''g^{-1] $ for any h in H and so we conclude that $\displaystyle gHg^{-1} \subset H$.

    Overall $\displaystyle H \subset gHg^{-1} $ and $\displaystyle gHg^{-1} \subset H $ gives equality.

    (I only really wrote that our for my benefit, thanks for your help!)
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    Re: Normal subgroup

    there are various definitions of normal subgroup floating about out there in the ether. one is:

    a) gHg-1 = H, for all g in G

    another is:

    b) gHg-1 ⊂ H, for all g in G

    and you have just seen that (a) and (b) are equivalent.

    a third one is:

    c) gH = Hg (left cosets are also right cosets), for all g in G

    one could, conceivably, mimic the relationship between (a) and (b) to produce a fourth definition:

    d) gH ⊂ Hg, for all g in G.

    the thing to see is that all 4 definitions produce the same set of subgroups, for any group G. so which one you use is largely a matter of convenience (for sets, it is often easier to show containment, rather than equality).
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