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Math Help - 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  ghg^{-1} \in H for all h in H, g in G.

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

    However, I want to show that actually gHg^{-1} = H , by showing that  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
    Member Sylvia104's Avatar
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    Re: Normal subgroup

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

    Thanks:

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

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

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

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

    (I only really wrote that our for my benefit, thanks for your help!)
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  4. #4
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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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