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Homework Help: Help with classification of the set of automorphisms on Z/n

  1. Oct 2, 2011 #1
    Let G be a group and Aut G the set of all automorphisms of G. What is Aut Z[itex]_{n}[/itex] for arbitrary n[itex]\in[/itex]N?

    Sol

    Let f[itex]\in[/itex]Aut Z[itex]_{n}[/itex]. Since Z[itex]_{n}[/itex]=<1>, f can be completely characterized by f(1), i.e., f(k) = f(1)[itex]^{m}[/itex], for k[itex]\in[/itex]Z[itex]_{n}[/itex] and m[itex]\in[/itex]N. By these facts, it follows that f(1) must generate Z[itex]_{n}[/itex], that is, Z[itex]_{n}[/itex]=<f(1)>.

    Define H as the group of all generators of Z[itex]_{n}[/itex] under multiplication. Then f(1)[itex]\in[/itex]H.

    Now at this point I'm certain that Aut Z[itex]_{n}[/itex][itex]\cong[/itex]H, but I just can't find a homomorphism from Aut Z[itex]_{n}[/itex][itex]\rightarrow[/itex]H. I can't concoct an epimorphism in the opposite direction either. The closest I got is the below:

    Also define the map c:Aut Z[itex]_{n}\rightarrow[/itex]H by c(f) = f(1). Choose f,g[itex]\in[/itex]Aut Z[itex]_{n}[/itex]. Then:
    c(g[itex]\circ[/itex]f) = g[itex]\circ[/itex]f(1) = g(f(1)) = g(1)[itex]^{l}[/itex]. But any l'[itex]\equiv[/itex]l (mod n) [itex]\Rightarrow[/itex]g(f(1)) = g(1)[itex]^{l'}[/itex], so we can consider l' an equivalence class in Z[itex]_{n}[/itex]. That implies l = f(1)[itex]^{p}[/itex] for some p[itex]\in[/itex]N, so that c(g[itex]\circ[/itex]f) = f(1)[itex]^{p}[/itex]*g(1) = c(g)c(f)[itex]^{p}[/itex].

    I can't see what I'm overlooking, so I'd appreciate if someone can help point me in the right direction. Thanks!
     
  2. jcsd
  3. Oct 2, 2011 #2
    Well, H is just Z/nZ isn't it? Find an isomorphism from Aut(Z_n) to Z/nZ. You know that if if is in Aut(Z_n) then it is uniquely determined by what it does to 1. For example, f(1) = a*1 for some a in Z. Now try to use this to construct an isomorphism from Aut(Z_n) to Z/nZ.
     
  4. Oct 2, 2011 #3
    Not quite. H[itex]\subset[/itex]Z[itex]_{n}[/itex], but not a subgroup since Z[itex]_{n}[/itex] is additive and H is multiplicative. In particular, H is always a proper subset since 0[itex]\in[/itex]Z[itex]_{n}[/itex] but 0[itex]\notin[/itex]H. So I have to find an isomorphism from Aut Z[itex]_{n}[/itex] directly to H.
     
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