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(Z/10557Z)* as Abelian Groups

  1. May 24, 2012 #1
    If I was to try to work this out I would use the Chinese Remainder Theorem and since 10557 = 3^3 . 17 . 23
    end up with (Z/10557Z)* isomorphic to (Z/27Z)* x (Z/17Z)* x (Z/23Z)* isomorphic to C18 x C16 x C22 where Cn represents the Cyclic group order n.

    How would I then write this as Cn1 x Cn2 x Cn3 s.t. n1 divides n2 divides n3?
  2. jcsd
  3. May 26, 2012 #2


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    this is explained in section 17 of these free notes:

    http://www.math.uga.edu/%7Eroy/844-2.pdf [Broken]

    and probably in many other algebra books.
    Last edited by a moderator: May 6, 2017
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