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Direct product

  1. Jul 29, 2007 #1
    Let k be a positive integer.

    define G_k = {x| 1<= x <= k with gcd(x,k)=1}

    prove that:
    a)G_k is a group under multiplication modulos k (i can do that).

    b)G_nm = G_n x G_m be defining an isomorphism.
     
  2. jcsd
  3. Jul 29, 2007 #2

    matt grime

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    What have you done for b)? There is only one possible way you can think of to write out a map from G_nm to G_n x G_m, so prove it is an isomorphism. Remember, G_n x G_m looks like pars (x,y)....
     
  4. Jul 29, 2007 #3
    We can use the Chinese Remainder Theorem on this one.

    Define the mapping,
    [tex]\phi: G_{nm}\mapsto G_n\times G_m[/tex]
    As,
    [tex]\phi(x) = (x\bmod{n} , x\bmod{m})[/tex]

    1)The homomorphism part is trivial.
    2)The bijection part is covered by Chinese Remainder Theorem.
     
  5. Jul 30, 2007 #4

    mathwonk

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    but the point is to prove that theorem.
     
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