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If Ha=Hb then b belongs to Ha

  1. Oct 28, 2007 #1
    this is how i set up the problem, am i thinking correctly? for all h belong H b,a belong to G where Ha and Hb are right cosets.

    ha= some hb
    a=b.
    then hb belongs Ha.

    since identity e belongs to H.

    eb belongs to Ha. so b belongs to Ha.
     
  2. jcsd
  3. Oct 28, 2007 #2

    morphism

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    Are you saying a=b? Because this isn't necessarily true. We have that h_1 a = h_2 b, where h_1 and h_2 need not be the same element.
     
  4. Oct 28, 2007 #3

    JasonRox

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    Try to understand what morphism and not just accept it. That's a mistake I see students make often and I myself. Once you understand why, you'll never confuse yourself again.
     
  5. Oct 31, 2007 #4

    HallsofIvy

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    What exactly are you trying to prove? If it is what you gave as the title, "if Ha= Hb, then b belongs to Ha", what you have done is at least a start: Since e is in H, eb= b is in Hb and then since Ha= Hb, b is in Ha. You don't need to write, nor is it true that "a= b".
    What you wrote, I don't know why, was "if ha= hb for some h, then a= b". If you meant h as some member of a group rather than meaning the subgroup H, then, since you could "multiply" on the left on both sides by h-1, yes, a= b.

    (Edited after learningphysics correction.)
     
    Last edited: Nov 1, 2007
  6. Oct 31, 2007 #5

    learningphysics

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    I think you meant "Since e is in H, eb = b is in Hb" etc...
     
  7. Nov 1, 2007 #6
    new question

    okay, i am beginning to understand better now. it's obvious now that if Ha=Hb then ha need not equal hb. however, i only understand this by the argument that different elements in the subgroup H can correspond to ha=hb.i.e. h(1)a=h(2)b. but the difficulty for me lies in conceptualizing what exactly a coset is. we are told (however this is not proved to us in class) that do to the equivalence relation that gives us the coset of H, G is paritioned into a disjoint union of these cosets. then if a,b belong to G and h belongs to H then just because Ha=Hb this does not necessarily imply that a=b because these a and b do not represent single elements in the group, but rather several elements say a(1) a(2) and b(1), b(2). then h(1)a(1) belongs to Ha and h(2)b(2) belongs to Ha and h(2)a(1) belongs to Ha, etc... then these need not necessarily be equal on the individual level. but in the end the collection of these elements must be the same. so some of them must equal others, but we don't necessarily have h(1)a(n)=h(1)b(n). is this the general idea here?

    i guess what i am asking here is the following: what does the statement a=b mean? does it mean that a and b represent the same collection of elements from the group G? can then a(1)=b(2) because the numbering convention is meaningless at this level? so we are just saying that some a belonging to G is equal to some b belonging to G?

    sorry i know this question must seem a little silly, but i'm having some difficulty picturing what is going on in this process of dividing the group into cosets.
     
  8. Nov 1, 2007 #7

    learningphysics

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    a and b are single elements... H is a collection of elements... (H is a collection of the type of elements of the type that a and b are... a and b themselves may be collections of other types of elements, but we don't need to worry about that.)

    Ha is a collection of elements... the set of all elements h*a such that h is an element of H.

    Hb is a collection of elements... the set of all elements h*b such that h is an element of H.

    Ha = Hb means that Ha is the same collection of elements as Hb.

    a(1), a(2) etc... doesn't make sense... a is just an element... there is no a(1) or a(2)...

    H is a collection... so you can say h(1), h(2) etc... representing the different elements of H.

    Ha = Hb

    means every element of Ha is an element of Hb and every element of Hb is an element of Ha. so for example, h(1)*a is an element of Hb... that means that h(1)*a = h(k)*b... for some k. now if a and b are not equal then k is not equal to 1...
     
  9. Nov 1, 2007 #8

    matt grime

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    I\m going to put in an example.

    a,b and lower case letters denote elements of a goup. H,G etc are goups.

    Example. G=S_3 symmetric group on 3 elements. H={e,(123), (132)} a cyclic subgroup of order 3.

    Ha means the set of elements {ha : h in H}. It is a subset of G. Work out what H(12) is for the above example. In fact work out Ha for all a in S_3. Verify that cosets are either identical or disjoint.
     
  10. Nov 1, 2007 #9
    wow, that example unlocked it for me thanks matt grime.
     
  11. Nov 1, 2007 #10
    quick question

    just wanted to verify that (123)=e for S_3.
     
  12. Nov 1, 2007 #11

    learningphysics

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    No. e is just the identity permutation... ie no elements are permuted... e = (1)(2)(3).
     
  13. Nov 1, 2007 #12

    matt grime

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    No. In fact I said {e,(123),(132)} was a subgroup of order 3, so e must not equal (123). Sorry if I used an unfamiliar notation - I'm using the cycle representation, so (123) sends 1 to 2, 2 to 3 and 3 to 1.
     
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