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Math Help - Left cosets

  1. #1
    MHF Contributor Amer's Avatar
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    Left cosets

    find all left coset taking the subgroup generated by <(1,2)>

    the group is \mathbb{Z}_2\times \mathbb{Z}_4

    we have the subgroup h={(0,0),(1,2)}

    \mathbb{Z}_2\times \mathbb{Z}_4 =[(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)]

    I take all elements not in h and add them to h to get the left coset but I have repeated elements so how I can chose elements such that I do not have repeated ones

    the book said

    h+(0,0) = h

    h+(1,0)

    h+(0,1)

    h+(1,1)

    how the book take these elements can you help
    is there a way to chose these elements
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    find all left coset taking the subgroup generated by <(1,2)>

    the group is \mathbb{Z}_2\times \mathbb{Z}_4

    we have the subgroup h={(0,0),(1,2)}

    \mathbb{Z}_2\times \mathbb{Z}_4 =[(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)]

    I take all elements not in h and add them to h to get the left coset but I have repeated elements so how I can chose elements such that I do not have repeated ones

    the book said

    h+(0,0) = h

    h+(1,0)

    h+(0,1)

    h+(1,1)

    how the book take these elements can you help
    is there a way to chose these elements
    Do you know Lagrange's theorem? Since \left|\mathbb{Z}_2\oplus \mathbb{Z}_4\right|=8 and \text{ord }\left(1,2\right)=2\implies\left|\left \langle \left(1,2\right)\right \rangle\right|=2 we mus only find \frac{8}{2}=4 subgroups. And, since the subgroup itself is always a coset we must only find three more subgroup. So how about the four cosets \left\langle \left(1,2\right)\right \rangle=H,\left(1,1\right)+H,\left(0,2\right)+H,\l  eft(0,3\right)+H?

    There is no diehard way, that I know to pick the elements. In this case, and some others, you can eye-ball the subgroup to see what will give you distinct cosets.
    Last edited by Drexel28; January 11th 2010 at 10:47 AM.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Do you know Lagrange's theorem? Since \left|\mathbb{Z}_2\oplus \mathbb{Z}_4\right|=8 and \text{ord }\left(1,2\right)=4\implies\left|\left \langle \left(1,2\right)\right \rangle\right|=2 we mus only find \frac{8}{2}=4 subgroups. And, since the subgroup itself is always a coset we must only find one more subgroup. So how about the four cosets \left\langle \left(1,2\right)\right \rangle=H,\left(1,1\right)+H,\left(0,2\right)+H,\l  eft(0,3\right)+H?

    There is no diehard way, that I know to pick the elements. In this case, and some others, you can eye-ball the subgroup to see what will give you distinct cosets.
    I know it, and I know I have 4 subgroups in the left coset

    it is clear now thank you very much
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Amer View Post
    I know it, and I know I have 4 subgroups in the left coset

    it is clear now thank you very much
    I made a typo! I'm sure you paid no mind to it, but I though I should point it out.
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