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Math Help - Cosets on S3

  1. #1
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    Cosets on S3

    First the problem:
    Let H be the cyclic subgroup (of order 2) of S_3 generated by \binom {1~2~3}{2~1~3}. Then no left coset of H (except H itself) is also a right coset. There exist an element a in S_3 such that aH \cap Ha = \{ a \}.
    If I am correct the problem is simply asking about a specific example, one that does not necessarily generalize. I can easily show that, for a in H, aH = Ha = H for any subgroup, due to closure of H. And, as it happens I have showed (by construction) that for any element a of S_3 that is not in H, aH \neq Ha. (I note that this property is specific to the subgroup H. If I take a new cyclic subgroup K = <(123)> of S_3 that aK = Ka for all a in S_3.) (Note: please forgive me if my cyclic notation is off. I rarely use it.)

    The last part is bugging me. aH \cap Ha = \{ a \} is a true statement for all a in S_3! So why are they commenting on it??

    -Dan
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  2. #2
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    because if for every g in S3, gH = Hg, then gH∩Hg = Hg (=gH).

    the fact that aH ∩ Ha = {a} ≠ aH indicates a certain defect in H.

    your last statement isn't true. if h is in H, then hH = Hh = H, so hH ∩ Hh = H ∩ H = H, NOT {h}.

    in particular, eH ∩ He = {e, (1 2)} (i am just writing the generator of H in cycle form), which is not {e},

    and (1 2)H ∩ H(1 2) = {(1 2), (1 2)^2} ∩ {(1 2), (1 2)^2} = {(1 2), e} ∩ {(1 2),e} = {(1,2), e} which is not {(1 2)}.
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