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Find Aut(Z n ) for cases n = 2,3,4, and 5

  1. Feb 18, 2009 #1
    Find Aut(Zn) for cases n = 2,3,4, and 5

    My question is to find Aut(Zn) for cases n = 2,3,4, and 5

    First, I am a little confused about the case Aut(Z), the integers. I know that the only functions that satisfy Aut(Z) are f(x) = x, and f(x) = -x, but what is wrong with f(x) = x+1?

    It would be great if you explain to me how to find automorphic functions, and why the previous example works and doesn't work.

    My attempts at Aut(Z2): I think that f(x) = kx and f(x) = x+k, k [tex]\in[/tex] Z are the only functions that apply in this case.

    The rest, I believe, are similar, but I just need help before I commit to an answer.

    Thank you for your time.
     
  2. jcsd
  3. Feb 18, 2009 #2
    Re: Automorphisms

    [tex] f \in Aut (Z)[/tex] implies [tex]f:Z \rightarrow Z[/tex] and [tex]f^{-1}:Z \rightarrow Z[/tex] should be injective. Thus, both the kernels of [tex]f, f^{-1}[/tex] should be identity. That implies the identity is mapped to an identity in Z.

    Generally speaking, the set of automorphisms forms a group and the order of [tex]Aut(Z_n)[/tex] is [tex]\phi(n)[/tex] (Euler totient function).
    Once you find an automorphism, you need to make sure that the order of each element is same with [tex]x \in Z_n[/tex] and [tex]f(x) \in Z_n[/tex], where f is an automorphism of [tex]Z_n[/tex].
     
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