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Math Help - Abstract Algebra: Groups

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Groups

    Hello All,

    I need help on this one. I'm sure it's simple, but I'm coming to the end off an all-nighter and so my brain has shut down.

    Problem: If k and n are positive integers and k is a divisor of n, then \mathbb{Z}_n / \left< [k] \right> \approx \mathbb{Z}_k.

    Thanks,
    Jhevon
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    Super Member Rebesques's Avatar
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    Maybe the statement has an error? If n=dk, then I think

    <br />
\mathbb{Z}_n / \left< [k] \right> \approx \mathbb{Z}_d<br />

    ...Just a thought.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Rebesques View Post
    Maybe the statement has an error? If n=dk, then I think

    <br />
\mathbb{Z}_n / \left< [k] \right> \approx \mathbb{Z}_d<br />

    ...Just a thought.
    i double checked the textbook and i copied it right. so the error may be with the book and not me, but i won't count on that, i've never been so lucky.

    i think a counter-example to your claim would be \mathbb{Z}_n = \mathbb{Z}_6 and \left< [k] \right> = \left< [3] \right> ...or is being awake for 38 hours getting to me?
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    Super Member Rebesques's Avatar
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    Just go to sleep and we 'll figure this out later
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    Super Member Rebesques's Avatar
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    Red face

    Yeap, I am being dumb. If n=dk, then the group generated by [k] has n/gcd(n,k)=d elements, and so <br />
\mathbb{Z}_n / \left< [k] \right> has n/d=k elements.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Rebesques View Post
    Just go to sleep and we 'll figure this out later
    hehe, one of the reasons i did not sleep is that i wanted to get this before then

    are you familiar with the "natural homomorphism" and the "Fundamental homomorphism theorem"? I'm thinking that we have to apply those somehow, but i am having trouble making it all fit
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    Define \mu: \mathbb{Z}_n \mapsto \mathbb{Z}_k as \mu ([x]_n ) = [x]_k. Show this map is well-defined, note, this is were you are going to use the fact that k|n. Now argue that this is a homomorphism. This mapping is onto, and so \mu [ \mathbb{Z}_n ] = \mathbb{Z}_k. The kernel of the homomorphism will be the set (group) \left< [k]_n \right>, I want to mention here another notation, we can write this also as k\mathbb{Z}_n, which looks a little neater. And so by fundamental homomorphism theorem \mathbb{Z}_n/k\mathbb{Z}_n \simeq \mathbb{Z}_k.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Define \mu: \mathbb{Z}_n \mapsto \mathbb{Z}_k as \mu ([x]_n ) = [x]_k. Show this map is well-defined, note, this is were you are going to use the fact that k|n. Now argue that this is a homomorphism. This mapping is onto, and so \mu [ \mathbb{Z}_n ] = \mathbb{Z}_k. The kernel of the homomorphism will be the set (group) \left< [k]_n \right>, I want to mention here another notation, we can write this also as k\mathbb{Z}_n, which looks a little neater. And so by fundamental homomorphism theorem \mathbb{Z}_n/k\mathbb{Z}_n \simeq \mathbb{Z}_k.
    i just want to clarify something before i go storming off to do this. we did a theorem that says a homomorphism is one-to-one if and only if the kernel is the set with only the identity. that does not seem to be happening here, but we need one-to-one for the isomorphism
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    Quote Originally Posted by Jhevon View Post
    i just want to clarify something before i go storming off to do this. we did a theorem that says a homomorphism is one-to-one if and only if the kernel is the set with only the identity. that does not seem to be happening here, but we need one-to-one for the isomorphism
    We need one-to-one to prove that \mathbb{Z}_n\simeq \mathbb{Z}_k. But we are not proving that, we are proving that, \mathbb{Z}_n/k\mathbb{Z}_n\simeq \mathbb{Z}_k.

    The fundamental homormohpsim theorem says that if you got a homomorphism \phi: G\mapsto G' then G/H \simeq \phi [G] where H = \ker \phi. It does not say anything about being one-to-one or onto.
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    Quote Originally Posted by ThePerfectHacker View Post
    We need one-to-one to prove that \mathbb{Z}_n\simeq \mathbb{Z}_k. But we are not proving that, we are proving that, \mathbb{Z}_n/k\mathbb{Z}_n\simeq \mathbb{Z}_k.

    The fundamental homormohpsim theorem says that if you got a homomorphism \phi: G\mapsto G' then G/H \simeq \phi [G] where H = \ker \phi. It does not say anything about being one-to-one or onto.
    Ok, thanks. does this work?

    Define \theta : \mathbb{Z}_n \mapsto \mathbb{Z}_k by \theta ( [a]_n) = [a]_k for all [a]_n \in \mathbb{Z}_n. This function is well-defined if and only if k \mid n by homework problem 21.6 (you donít expect me to re-invent the wheel, do you?).

    We claim that this defines a homomorphism. For if [a]_n,[b]_n \in \mathbb{Z}_n, then \theta ([a]_n \oplus [b]_n) = \theta ([a + b]_n) = [a + b]_k = [a]_k \oplus [b]_k = \theta ([a]_n) \theta ([b]_n).

    This mapping is onto, since for any [a]_k \in \mathbb{Z}_k we can find a [a]_n \in \mathbb{Z}_n such that \theta ([a]_n) = [a]_k because k \mid n (should I explain this part more?). So that we have \theta [ \mathbb{Z}_n] = \mathbb{Z}_k. Clearly the kernel of this homomorphism is \left< [k]_n \right> = K.

    Thus we can define a mapping, \phi : \mathbb{Z}_n /K \mapsto \mathbb{Z}_k by \phi (Ka) = \theta (a) for each Ka \in \mathbb{Z}_n /K.

    This mapping is an isomorphism of \mathbb{Z}_n / K onto \mathbb{Z}_k by the Fundamental Homomorphism Theorem. So that \mathbb{Z}_n /K \approx \mathbb{Z}_k

    QED
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    Quote Originally Posted by Jhevon View Post
    This mapping is onto, since for any [a]_k \in \mathbb{Z}_k we can find a [a]_n \in \mathbb{Z}_n such that \theta ([a]_n) = [a]_k because k \mid n (should I explain this part more?)
    You are overkilling it. If [a]_k \in \mathbb{Z}_k then [a]_n \in \mathbb{Z}_n and \theta ([a]_n) = [a]_k.
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    This can also be proved using the third isomorphism theorem. Note that \mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}. And k\mathbb{Z}_n = k\mathbb{Z}/n\mathbb{Z}*.
    Thus, \mathbb{Z}_n / k\mathbb{Z}_n = (\mathbb{Z}/n\mathbb{Z})/(k\mathbb{Z}/n\mathbb{Z}) \simeq \mathbb{Z}/k\mathbb{Z} = \mathbb{Z}_k.




    *)Where m\mathbb{Z} = \{ 0,\pm m,\pm 2 m, ... \}.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    This can also be proved using the third isomorphism theorem. Note that \mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}. And k\mathbb{Z}_n = k\mathbb{Z}/n\mathbb{Z}*.
    Thus, \mathbb{Z}_n / k\mathbb{Z}_n = (\mathbb{Z}/n\mathbb{Z})/(k\mathbb{Z}/n\mathbb{Z}) \simeq \mathbb{Z}/k\mathbb{Z} = \mathbb{Z}_k.




    *)Where m\mathbb{Z} = \{ 0,\pm m,\pm 2 m, ... \}.
    that is very interesting. it seems you can treat it just like a quotient and actually "cancel" the n \mathbb{Z}. very nice.

    thanks for the link as well. i'll read through it when i have the time, or maybe we'll get to this in class and i can use it as supplementary material
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    Quote Originally Posted by Jhevon View Post
    that is very interesting. it seems you can treat it just like a quotient and actually "cancel" the n \mathbb{Z}. very nice.
    Exactly. This is also interesting because it is a quotient group of a quotient group. So you form the cosets. And using the cosets you form a cosets of cosets.

    This is Mine 92th Post!!!
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    It happens to be a good problem to work on if you want to work on your homomorphism.

    Third Theorem: If N\triangleleft H\triangleleft G then H/N\triangleleft G/N and (G/N)/(H/N) \simeq G/H.

    Try to prove it. Hint: Define \theta: G/N\mapsto G/H as \theta (xN) = xH.
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