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Groupy Theory: Cosets

  1. Nov 19, 2003 #1
    Hello,

    It should be common knowledge now that I have trouble with Group Theory. I would like to go back and start from the beginning but I haven't the luxury of time at this point. So for the present time I am resigned to just keeping up with the class the best I can. For anyone has the time and patience, I would appreciate it if someone can look over my work for the following 2 questions. Advice on how to approach the question, hints, interpretations of concepts, and expansions on concepts are welcome.

    Question1:

    Let G = D6 = {u, y, y2, x, xy, xy2} where x2 = u, y3 = u, and yx = xy-1. Let H = {u,x}. (u = the identity element).

    i) Write down the elements of the right cosets A = Hy and B = Hy2.

    ii) Calculate the product AB = (Hy)(Hy2) of the cosets Hy and Hy2 (ie., write down and simplify every possible product ab, where a is an element of A and b is an element of B).

    iii) Is AB a coset of H in G?

    iv) Is AB a coset of any subgroup of G? (Hint: Use Lagrange's Theorem).


    i)

    A = Hy = {uy, xy} = {y, xy}
    B = Hy2 = {uy2, xy2} = {y2, xy2}.

    ii)

    y*y2 = y3 = u
    y*xy2 = xy2

    xy*y2 = xy3= xu = x
    xy*xy2 = x*xy-1*y2 = u*y = y

    Therefore AB = {u, y, x, xy2}

    iii)

    I am not sure about this part of question 1, but I would think that AB is not a coset of H in G since AB has 4 elements while H has only two.

    iv)

    I am also not sure about this part of question 1. However, I think that since the order of AB is 4 and that the order of G is 6, 4 is not a divisor of 6 hence AB cannot be a subgroup of G. If there cannot be a subgroup of order 4, AB cannot be a coset of any subgroup since there are no subgroup of order 4. (?)


    Question2:

    i) Let G be a group, and let H be a subgroup of G. What condition tells you that H is a normal subgroup of G?

    ii) Prove the following: H is normal in G iff g-1Hg = H for every g which is an element of G.


    i)

    A subgroup H of a group G is a normal subgroup of G if the following is true:

    Condition: gH = Hg for every g which is an element of G. That is, the right coset Hg of H in G, generated by g, is equal to the left coset gH of H in G, generated by g (where g is an element of G).

    ii)

    (Still to come).
     
    Last edited by a moderator: Nov 19, 2003
  2. jcsd
  3. Nov 19, 2003 #2

    HallsofIvy

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    Why is y*xy2= xy2?

    I get y*xy2= (yx)(y2
    = (xy-1(y2)
    = xy.

    (iii) and (iv) are correct.
     
  4. Nov 19, 2003 #3
    Yes. My mistake. I was doing this late last night. Thanks for pointing that out to me HallsofIvy.

    As well, thankyou for looking over my work. That is appreciated very much.

    I'm not so sure on how to start question 2, part ii however.


    I know that a subgroup H of a group G is a normal subgroup of G if gH = Hg for every g which is an element of G.

    So am I to prove

    If gH = Hg then g-1Hg = H

    as well as

    If g-1Hg = H then gH = Hg

    for every g which is an element of G?


    If so, I wasn't sure how to proceed. For the first one I let x = g-1 which is an element of G. Then

    x * gH = x * Hg

    Since x = g-1 then

    g-1 * gH = g-1 * Hg

    And since g-1 * g = u it follows that

    u * H = g-1 * H * g = H.

    Was this the correct way to proceed? I am unsure what to do.


    For

    g-1Hg = H then gH = Hg

    I let x = g and followed much the same steps as above. I don't think I am doing this correctly.

    Any further input is appreciated. Thankyou.
     
  5. Nov 19, 2003 #4

    NateTG

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    Homework Helper

    Question 1 Part ii:
    You should get the answer that
    AB=H

    Question 1 Part iii:
    Your reasoning is correct, but this depends on part ii.

    Question 1 Part iv:
    Your reasoning is correct but the results from Part ii have changed

    Question 2 Part i:
    Looks good

    Question 2 Part iii

    What you have is probably acceptable. It's unnecessary to introduce x, since you already have g-1.

    So you have, for example
    H=g-1Hg (by hypothesis)
    then multiply both sides by g on the left
    g*H=g*g-1Hg
    gH=uHg
    gH=Hg

    If you want to be more formal, H,Hg, gH and g-1Hg are all sets, so you can consider operating on the elements in the sets, but there's not a whole lot of extra insight to be gained:
    e.g
    by hypothesis we have that
    H=g-1Hg
    then for every
    h in H there is h' in g-1Hg such that
    h=g-1h'g
    multiply both sides by g on the left
    gh=gg-1h'g
    so
    gh=h'g
    This is true for every h in H, so gH is a subset of Hg.
    Hg and gH also have the same number of elements.
    Therefore Hg=gH
     
  6. Nov 19, 2003 #5
    I tried this over and over and I cannot get AB = H.

    What am I missing?

    Given that A = {y, xy} and B = {y^2, xy^2}

    yy^2 = u
    yxy^2 = xy^-1y^2 =xy

    xyy^2 = x
    xyxy^2 = xxy^-1y^2 = u u y = y

    I still get AB = {u,xy,x,y} which is not equal to H.

    [?]
     
  7. Nov 19, 2003 #6

    NateTG

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    Homework Helper

    Dorf. I need glasses today.

    You're right.
     
  8. Nov 19, 2003 #7
    Thank god. I spent an hour and a half trying to see how AB = H. I thought I was regressing once again. Need more confdence in my skills I guess.

    Thanks for spending the time looking over my work HallsofIvy and NateTG. I appreciate your efforts greatly.

    Cheers.
     
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