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Math Help - index of subgroup

  1. #1
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    index of subgroup

    Let [G:H]=2 and a, b not in H, then show that ab is in H.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: index of subgroup

    Hi again srv123!

    Suppose that ab is not in H.
    Then ab must be in the other coset.

    Which other coset is there?

    Can you draw any conclusions from there (that lead to a contradiction)?
    Thanks from srv123
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  3. #3
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    Re: index of subgroup

    hello,
    It has something to do with normality of H ryt?
    i need the nxt clue:-(
    ab is in abH if ab is not in H..
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: index of subgroup

    Well, you have [G:H]=2.
    This means there are exactly 2 cosets of H.
    The first coset is simply H
    The other coset contains both a and b (and through our assumption also ab).

    Let's call the other coset aH.

    So what can you conclude about b (and through our assumption about ab)?
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  5. #5
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    Re: index of subgroup

    then b will be in H, ryt?
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: index of subgroup

    Huh?

    Can you expand a little?
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  7. #7
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    Re: index of subgroup

    starting from ur point, i was thinking lyk this...
    b and ab both in aH; so b.(ab)^-1 is in H
    hence a is in H and we reach a contradiction...

    again if we suppose ab not in H, then it is in aH

    so, ab=ah

    now b= a^-1.a.b= a^-1.a.h = h belonging to H...contradiction...
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  8. #8
    Super Member ILikeSerena's Avatar
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    Re: index of subgroup

    Hmm, it seems you're on the right road.... but I don't completely get what you're doing.

    If b and ab are (indeed) both in aH, why would b.(ab)^-1 be in H?
    Are you applying a rule there?
    Which rule?

    And once you have your first contradiction, it follows that ab in not in aH.
    Therefore it has to be in the only other coset, which is H.

    What's the other contradiction for?
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  9. #9
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    Re: index of subgroup

    No no, the last part was just another alternate way I found to reach the cntradiction....

    and about the first part,
    i thought like this....

    the cosets partition the group...ryt? the equivalence relation(R) on the parent group G is aRb iff ab^-1 is in H. Then we can get equivalnce classes of the right cosets; and two
    members a and b of G are in same class iff ab^-1 is in H. ryt?

    Now in our problm if we take H and Ha are the two cosets; then b and ab are in same class so b.(ab)^-1 should be in H...isn't it?
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  10. #10
    Super Member ILikeSerena's Avatar
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    Re: index of subgroup

    Ah okay.
    Yep. That's right:
    Two members a and b of G are in same class iff ab^-1 is in H.
    That's an equivalent axiom for a subgroup!

    It helps if you mention something like that to avoid leaving people (like me) in the dark.

    And yes, if b and ab are in same class then b.(ab)^-1 should be in H.
    Last edited by ILikeSerena; June 4th 2012 at 01:16 PM.
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