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External direct product- homomorphism

  1. Aug 9, 2011 #1
    Theorem 8.3 in Gallian's Contemp. Abst Alg
    says with (s, t) = 1 the group U(st) is isomorphic
    to the external direct product of U(s) and U(t)
    that is, to U(s) (+) U(t)

    U(n) is the group of positive
    integers less than n and relatively prime to n with the
    group operation multiplication mod n

    Defining a map f from U(st) -> U(s)(+) U(t) as f : x -> (x mod s, x mod t)

    I am confused on how to proceed to show that f is HOMOMORPHISM. Please see my working below....

    LET a, b belong to U( st) then

    f ( a x b mod st )=f ( ab mod st)
    = ( ( ab mod st ) mod s , ( ab mod st ) mod t))

    Now, since ab mod st = ab mod s = ab mod t and (ab mod s ) mod s = ab mod s

    = ( ab mod s, ab mod t )
    = ( a mods, a mod t) (b mod s , b mod t)
    = f ( a ) f (b)
     
    Last edited: Aug 9, 2011
  2. jcsd
  3. Aug 9, 2011 #2

    micromass

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    That seems correct.

    Note however that notations like ( ab mod st ) mod s don't make much sense.
    You can write directly f ( ab mod st) = ( ab mod s , ab mod t)
     
  4. Aug 9, 2011 #3

    Hurkyl

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    Actually, it is not unusual to use
    x mod y​
    to refer to the function that selects a canonical representative of the equivalence class of x modulo y. (commonly the one in the interval [0, y-1])

    Also, I don't think it especially surprising to use the postfix operator "- mod f" to refer to the homomorphism G --> G/f. (where G is some appropriate sort of algebraic structure like Abelian group or Ring)
     
  5. Aug 10, 2011 #4
    Why does my idea fail if take this example

    let s= 3 , t = 5. then a = 7 , b= 8 belong to U (st)= U(15)

    f ( 7 x 8 mod 15) = ( ( 7 x 8 mod 15 )mod 5, (7 x 8 mod 15) mod 3) = (4, 0).

    However , if I write

    f ( 7 x 8 mod 15) = ( 56 mod 5, 56 mod 3) = (1, 2).

    which is equal to

    f ( 7) f (8) = ( 7 mod 5, 7 mod 3)( 8 mod 5, 8 mod 3)

    = (2, 1) ( 3, 2)
    = ( 1,2)
     
    Last edited: Aug 10, 2011
  6. Aug 10, 2011 #5

    micromass

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    I don't see how you got this...
     
  7. Aug 10, 2011 #6
    f ( 7 x 8 mod 15) = ( ( 7 x 8 mod 15 )mod 5, (7 x 8 mod 15) mod 3)
    = ( 9 mod 5 , 9 mo d 3)
    = ( 4, 0)
     
  8. Aug 10, 2011 #7

    micromass

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    7x8=11 mod 15, not 9...
     
  9. Aug 10, 2011 #8
    Thanks, then it seems my initial reasoning was correct, that is,


    ( ( ab mod st ) mod s , ( ab mod st ) mod t)) = ( ab mod s, ab mod t ).
     
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