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Thread: Cosets and Lagrange's Theorem

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Cosets and Lagrange's Theorem

    Hello,

    I'm am stuck on this problem (or so I think) XD

    Let $\displaystyle n$ be a positive integer. Let $\displaystyle H=\{0,\pm n,\pm2n,\pm3n,\dots\}$. Find all left cosets of $\displaystyle H$ in $\displaystyle \mathbb{Z}$. How many are there?
    Source: Contemporary Abstract Algebra, 6e, by J. Gallian.

    Since
    $\displaystyle n\in\mathbb{Z}^+$, I thought that it would be safe to say that the left cosets would be

    $\displaystyle \begin{aligned}
    0+H & = \{0,\pm n,\pm2n,\pm 3n,\dots\}= nk+H,\,k\in\mathbb{Z}^+\\
    1+H & = \{\dots,1-2n,1-n,1,1+n,1+2n,\dots\}=nk+1+H,k\in\mathbb{Z}^+\\
    2+H & = \{\dots,2-2n,2-n,2,2+n,2+2n,\dots\}=nk+2+H,k\in\mathbb{Z}^+\\
    & \,\vdots\\
    \left(n-1\right)+H & = \{\dots,-2n-1,-n-1,-1,n-1,2n-1,\dots\}=2nk-1+H,k\in\mathbb{Z}^+
    \end{aligned}$

    From this, it would appear that there are $\displaystyle n$ distinct left cosets. But is there another way to verify that this is the case? Lagrange's Theorem wouldn't apply to this, since $\displaystyle \left|n\mathbb{Z}\right|$ and $\displaystyle \left|\mathbb{Z}\right|$ are infinite.

    Does this look right, or am I way over my head? XD

    I would appreciate any suggestions!
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  2. #2
    Super Member PaulRS's Avatar
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    You had to mention first, that the operation is Addition .

    Okay, let us see: $\displaystyle
    aH = bH \Leftrightarrow H = \left( {a^{ - 1} b} \right)H \Leftrightarrow a^{ - 1} b \in H
    $ or equivalently: $\displaystyle
    n\left| {\left( {b - a} \right)} \right.
    $ ( written in another way: $\displaystyle
    b \equiv a\left( {\bmod .n} \right)
    $ )

    ...

    There are $\displaystyle n$ distinct left-cosets indeed.
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  3. #3
    Super Member Gamma's Avatar
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    Yeah, that is exactly right. $\displaystyle H=n\mathbb{Z}$. It is a normal subgroup as $\displaystyle \mathbb{Z}$ is abelian. You will notice that

    $\displaystyle \mathbb{Z}/H = \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$

    This is precisely how you define the modular numbers. So you can say $\displaystyle [G:H]=n=|\mathbb{Z}_n|$
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