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Math Help - Cosets and Normal Sub-group: Relation between the two concepts

  1. #1
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    Cosets and Normal Sub-group: Relation between the two concepts

    We know that for any group, G, a subgroup H is normal-subgroup
    IFF every right coset of H is also it's left coset i.e. Ha = aH for all 'a' in G

    What if there is a group where every right coset is a left coset but with some other element
    i.e. Ha = bH where a=/=b (for all the right cosets of H, obviously there is one exception He = eH)

    Basically what I am saying is that there is a one-to-one mapping between set of right cosets and set of left cosets.

    I am not sure if such a structure in groups is relevant / important / logical.

    What does the above tell the relation between bH and aH? Are they same?

    Any thoughts??
    Last edited by aman_cc; August 29th 2009 at 10:23 AM. Reason: wrong typing
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  2. #2
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    A related question - Do right cosets and left cosets partition a Group into same mutually exclusive and exhaustive sets?

    This is obviously true if H is normal sub-group.

    But what if H is not normal is that possible? If yes, what additional property do we have to impose on H?
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  3. #3
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    for any subgroup H of a group G these two statements are equivalent:

    1) Ha=aH, for all a \in G.

    2) for any a \in G, there exists b \in G such that Ha=bH.

    1) implies 2) is trivial. proof of 2) implies 1): so we have b=ha, for some h \in H. then aH=h^{-1}bH=h^{-1}Ha = Ha.

    i'm not sure i understand your second question!
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    A related question - Do right cosets and left cosets partition a Group into same mutually exclusive and exhaustive sets?

    This is obviously true if H is normal sub-group.

    But what if H is not normal is that possible? If yes, what additional property do we have to impose on H?
    I don't think I understand either. Maybe this theorem will help you:

    Let H be a non-trivial subgroup of G. Let k = [G:H], then:

    G = g_1H \cup g_2H \cup ... \cup g_kH where g_1H, g_2H,...,g_kH are the distinct left cosets of H in G. This is also correct for right cosets.
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  5. #5
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    Thanks I follow what you say. My questions are wrong. I did not relaize that Ha = Hb is equivalent to Ha = aH. So if every right coset is also a left coset then Ha = aH for all 'a'.

    Thanks Defunkt and NonCommAlg

    Quote Originally Posted by Defunkt View Post
    I don't think I understand either. Maybe this theorem will help you:

    Let H be a non-trivial subgroup of G. Let k = [G:H], then:

    G = g_1H \cup g_2H \cup ... \cup g_kH where g_1H, g_2H,...,g_kH are the distinct left cosets of H in G. This is also correct for right cosets.
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