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Math Help - Subgroup coset

  1. #1
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    [SOLVED]Subgroup coset

    How can i find left and right cosets of subgroup <2> ={0,2,4,6,8,10} of Z12 ???
    thanks
    Last edited by kljoki; June 17th 2012 at 12:03 PM. Reason: [SOLVED]
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  2. #2
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    Re: Subgroup coset

    there is 2 cosets for H={0,2,4,6,8,10}
    0+H = H
    1+H = {1,3,5,7,9,11}
    next is the same with this 2
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  3. #3
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    Re: Subgroup coset

    Z12 is abelian, so the left cosets and right cosets are the same.
    Thanks from kljoki
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  4. #4
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    Re: Subgroup coset

    And there are two such cosets because the given sugroup contains 6= 12/2 members. kljoki, this would have been a good exercise in using the basic definition of "coset" because that is all that is required.
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  5. #5
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    Re: Subgroup coset

    Quote Originally Posted by Deveno View Post
    Z12 is abelian, so the left cosets and right cosets are the same.
    and what if it wasn't abelian, can you give me some example ??
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  6. #6
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    Re: Subgroup coset

    sure. let's use the group of symmetries of a triangle. this is generated by a rotation of 120 degrees, and a reflection about the x-axis, which i shall call a and b. the multiplication operation is composition of functions (we consider the domain of these functions to be the triangle itself; although if you wish, we can consider them as acting on the entire euclidean plane).

    clearly, a3 = e, the identity function ("do nothing" = "rotating 360 degrees")

    also clearly, b2 = e.

    by actual experimentation (you could make a triangle out of straws, for example), we see that:

    ba = a2b.

    thus G = {e,a,a2,b,ab,a2b}.

    we have an obvious subgroup of order 2: H = {e,b}. let's compute the left and right cosets of H. first the left cosets:

    eH = {ee,eb} = {e,b} = H (this one is easy).
    aH = {ae,ab} = {a,ab}
    a2H = {a2e, a2b} = {a2,a2b}

    so far, so good. these 3 cosets are clearly all different.

    bH = {be,bb} = {b,b2} = {b,e} = H (remember, in SETS, order of elements doesn't matter).
    (ab)H = {(ab)e, (ab)b} = {ab,ab2} = {ab,a} = aH (not surprising since (ab)H should equal a(bH) = aH, since bH = H).
    (a2b)H = {(a2b)e, (a2b)b} = {a2b, a2} = a2H

    so we only have 3 "distinct" left cosets, H, aH and a2H.

    now let's look at the right cosets of H:

    He = H (this really should be obvious).
    Ha = {ea,ba} = {a,a2b} (remember that ba = a2b)
    Ha2 = {ea2,ba2}.

    hmm. what *is* ba2? we don't rightly know yet. let's figure it out:

    ba2 = (ba)a = (a2b)a = a2(ba) = a2(a2b) = a4b = (a3)(ab) = (e)(ab) = ab.

    so Ha2 = {a2, ab}.

    note that aH is a different set than Ha, and a2H is a different set than Ha2. this shows that when G is non-abelian, for some subgroups H it might be possible that gH ≠ Hg.

    i leave it to you to verify the the rest of the cosets are:

    Hb = H
    H(ab) = Ha2
    H(a2b) = Ha

    you will discover (shortly) that subgroups for which the left and right cosets are the same, are "special", and have their own name: normal subgroups.
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