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Math Help - [SOLVED] If the product of two right cosets of H is a right coset of H, then H is nor

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    [SOLVED] If the product of two right cosets of H is a right coset of H, then H is nor



    I've looked at this a few different ways. I don't really get it.

    I don't see how

    (Ha)(Hb) = Hc \implies gHg^{-1} = H or gH = Hg for all g \in G.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by davismj View Post


    I've looked at this a few different ways. I don't really get it.

    I don't see how

    (Ha)(Hb) = Hc \implies gHg^{-1} = H or gH = Hg for all g \in G.
    It's late, and I don't remember if there is an easier manipulation way..but you'll have to do this eventually might as well start now.

    If HaHb=Hc we may conclude from the partitioning of cosets that c=ab. Thus, we may define a group G/H to be the set \left\{gH:g\in G\right\} with (Hg)(Hg')=H(gg'). Clearly He=H is the identity element, it's associativity is associated from G (why)? It is easy to prove that \left(Hg\right)^{-1}=Hg^{-1} etc. Then, one can prove that \theta:G\to G/H:g\mapsto gH is a well define homomorphism and \ker\theta=H. So?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by davismj View Post

    I've looked at this a few different ways. I don't really get it.

    I don't see how

    (Ha)(Hb) = Hc \implies gHg^{-1} = H or gH = Hg for all g \in G.
    In particular, HgHg^{-1} is some right coset Hc. Now we have 1 = 1g1g^{-1} \in HgHg^{-1}, so 1 \in Hc, i.e. c \in H. Therefore HgHg^{-1}=H and H is normal.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    It's late, and I don't remember if there is an easier manipulation way..but you'll have to do this eventually might as well start now.

    If HaHb=Hc we may conclude from the partitioning of cosets that c=ab.
    No! If HaHb=Hc implies ab=c then, since we also have H(ha)Hb=Hc for any h \in H, it would follow that both ab=c and hab=c...
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    No! If HaHb=Hc implies ab=c then, since we also have H(ha)Hb=Hc for any h \in H, it would follow that both ab=c and hab=c...
    Sorry. Too late at night for group theory. What I meant to say is that HaHb=Hc then Hc=Hab since evidently e\in H\implies a\in aH,b\in bH and thus ab\in HaHb from where it follows by the partitioning that HaHb=Hab. What I said is clearly wrong since given any subgroup of cardinality greater than two we have that Hh_1=Hh_2 but h_1\ne h_2
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