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Math Help - Modulo

  1. #1
    Super Member Showcase_22's Avatar
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    Modulo

    I have two questions about modulo n:

    1). In my notes I have \mathbb{Z}_n quoted everywhere! My big problem is: how do I know if this is addition modulo n or multiplication modulo n?

    2). My second question is that I have this:

    \mathbb{Z}_n \oplus \mathbb{Z}_m.

    What does this mean? I know it's the direct sum, so does it mean this:

    For example, \mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2 \} \oplus \{0,1,2,3 \}=\{0,2,4,3\}?

    (I just went through and added the 1st number of each set to the 1st number of the other (and so on)). My big problem was that I didn't have anything to add to the 3 =S

    If someone can clarify these it would help a lot!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    If you're talking about group theory, you probably mean addition modulo n.
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  3. #3
    Super Member Showcase_22's Avatar
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    Thanks!

    I don't suppose you also know how to do the direct product as well? I'm getting stuck on this proof i'm trying to do!!
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post

    1). In my notes I have \mathbb{Z}_n quoted everywhere! My big problem is: how do I know if this is addition modulo n or multiplication modulo n?
    \mathbb{Z}_n is just a number system at this point, there is no operation. You can apply either addition or multiplication to it.


    Quote Originally Posted by Showcase_22 View Post

    2). My second question is that I have this:

    \mathbb{Z}_n \oplus \mathbb{Z}_m.

    What does this mean? I know it's the direct sum, so does it mean this:

    For example, \mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2 \} \oplus \{0,1,2,3 \}=\{0,2,4,3\}?

    (I just went through and added the 1st number of each set to the 1st number of the other (and so on)). My big problem was that I didn't have anything to add to the 3 =S

    If someone can clarify these it would help a lot!
    Not 100% sure here but this is what I would suggest.

    I would take the first element of the first set and add it to each element of the second, then take the second element of the first set and add it to every element of the second and so on.

    \mathbb{Z}_3 \oplus \mathbb{Z}_4 will be the set of distinct solutions.
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  5. #5
    Super Member Showcase_22's Avatar
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    The question says that \mathbb{Z}_n+\mathbb{Z}_m=\mathbb{Z}_{mn} if n and m are relatively prime.

    So I used the example \mathbb{Z}_3 \oplus \mathbb{Z}_4 since 3 and 4 are coprime (and hence relatively prime).

    \mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2\} \oplus \{0,1,2,3 \}=\{0,1,2,3,1,2,3,4,2,3,4,5 \}=\{0,1,2,3,4,5 \} \neq \mathbb{Z}_{12}

    I really can't fathom what this question is going on about =S
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  6. #6
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    Direct Sum

    The symbol

    <br />
\mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2\} \oplus \{0,1,2,3 \}=<br />
    is often used to denote the set of ordered pairs, with the first element from the first set, and the second from the second set. Namely:

    (0,0), (0,1), (0,2), (0,3),
    (1,0), (1,1), (1,2), (1,3),
    (2,0), (2,1), (2,2), (2,3)
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Showcase_22 View Post
    The question says that \mathbb{Z}_n+\mathbb{Z}_m=\mathbb{Z}_{mn} if n and m are relatively prime.

    So I used the example \mathbb{Z}_3 \oplus \mathbb{Z}_4 since 3 and 4 are coprime (and hence relatively prime).

    \mathbb{Z}_3 \oplus \mathbb{Z}_4=\{0,1,2\} \oplus \{0,1,2,3 \}=\{0,1,2,3,1,2,3,4,2,3,4,5 \}=\{0,1,2,3,4,5 \} \neq \mathbb{Z}_{12}

    I really can't fathom what this question is going on about =S
    Firstly. I assume you know what A\times B is? The cartestian product A\times B=\left\{(x,y): x\in A\text{ }y\in B\right\}. Well it is fairly simple to show that if G,G' are groups then G\times G' is a group with (g_1,g_1')(g_2,g_2')=\left(g_1g_2,g_1'g_2'\right). When the two groups are abelian it is customary to write G\oplus G' in lieu of G\times G'


    Secondly, you need to be a tad careful here. Firstly it should be (m,n)=1\implies \mathbb{Z}_n\oplus\mathbb{Z}_m\cong \mathbb{Z}_{nm}. The equals sign is clearly false since the LHS is a set of ordered pairs and the RHS is not. But as is indicated the two groups are "isomorphic" (I believe from previous threads that you are familiar with this concept). To see why the above is true here is a brief outline. For \mathbb{Z}_n\oplus\mathbb{Z}_m to be cyclic we need there to be a generator of order \left|\mathbb{Z}_n\right|\cdot\left|\mathbb{Z}_m\r  ight|=mn. Note that for any z\in\mathbb{Z}_n\oplus\mathbb{Z}_m it is true that z^{\text{lcm}(m,n)}=e (why?). Therefore, to have (and this of course is not a formal proof) a generator of order mn we must have that \text{lcm}(m,n)=mn\implies (m,n)=1. So if (m,n)=1 then \mathbb{Z}_n\oplus\mathbb{Z}_n is a cyclic group of order mn. Also, you should know that any cyclic group of order \ell is isomorphic to \mathbb{Z}_{\ell}. From there the rest is trivial.
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  8. #8
    Super Member Showcase_22's Avatar
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    <br /> <br />
(m,n)=1\implies \mathbb{Z}_n\oplus\mathbb{Z}_m\cong \mathbb{Z}_{nm}<br />
    Sorry, I did mean to write this!

    <br /> <br />
z^{\text{lcm}(m,n)}=e<br />
    What is your notation here? What's e?
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  9. #9
    MHF Contributor chiph588@'s Avatar
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     e is the identity element.
    Last edited by chiph588@; November 25th 2009 at 12:48 PM.
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