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Gcd(m, n)=1 implies Zmn isomph to Zm x Zn

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data
    show that if m, n are relatively prime, that is, greatest common divisor of m and n is 1, then [itex] \mathbb{Z} _mn \approx \mathbb{Z} _m \times \mathbb{Z} _n [/itex]


    2. Relevant equations

    I need to show that [itex] \theta [/itex] is operation preserving, and I need to show that it is one to one and onto.

    3. The attempt at a solution

    For theta, [itex] \theta ([a]_{mn} + _{mn}) = \theta ([a+b]_{mn})=([a+b]_m,[a+b]_n)= [/itex]
    [itex] ([a]_m+_m,[a]_n+_n)=([a]_m,[a]_n)+([b)_m,_n)= \theta ([a]_{mn}) + \theta (_{mn}) [/itex]
    Did I assume anything I shouldn't have there?
    I'm going to consult my notes about proving 1-1. going to try the kernel thing.
    As for onto, how do I show that?
    I'm concerned that I haven't used the fact that m, n are relatively prime.
     
    Last edited: Nov 11, 2011
  2. jcsd
  3. Nov 11, 2011 #2

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    Hi Arcana!

    You assumed your definition of theta, which I have to deduce from your work now.
    Beyond that your proof of the preservation of the operation is correct.

    So what's left is to proof that theta is 1-1, for which you need that m and n are relatively prime.
     
  4. Nov 11, 2011 #3
    The definition of theta was given in a previous problem. I'm just tired of typing latex. Guess I need more practice so I'm not so slow at it.
    going to consult notes about kernel thingy now for 1-1, but what about onto?
     
  5. Nov 11, 2011 #4

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    How many elements does each set have?
    Are they the same?

    I'm not sure what your kernel thingy is, but if you can say each element is mapped onto a different element, you can deduce "onto" by the fact that each set has the same number of elements.
     
    Last edited: Nov 11, 2011
  6. Nov 11, 2011 #5
    difficulty:
    theta is 1-1 iff ker(theta)= {e_Z_mn}
    [tex] \theta ([a]_{mn})=([0]_m,[0]_n) \Rightarrow [/tex]
    [tex] m \mid a \wedge n \mid a \Rightarrow [/tex]
    [tex] a=km \wedge a=jn [/tex]

    I know that being relatively prime is the reason a must =mn but I'm not sure why, so I don't know what to say next.
     
  7. Nov 11, 2011 #6

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    You have n|a.

    So from a=km, you know that n|(km).
    Since n does not have any common factors with m, it must be that n|k.
     
  8. Nov 11, 2011 #7

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    This means that k=i n.
    And in turn that:
    a=km=(in)m=i(mn)
     
  9. Nov 11, 2011 #8
    What is i? 1? And why does k=in? Sorry I'm being dense here.
     
  10. Nov 11, 2011 #9

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    Since n|k, there must be a number i such that k=in.
     
  11. Nov 11, 2011 #10

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    Not dense, I know how hard it is to follow other people's reasoning in algebra.
    The trick is to set up your own reasoning.
    I just hope my comments can help you in your reasoning.
     
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