Finding a homomorphism from Aut(Z/n) to the group of generators

[TopCat]

Let G be a group and Aut G the set of all automorphisms of G. What is Aut Z$_{n}$ for arbitrary n$\in$N?

Sol

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

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

Now at this point I'm certain that Aut Z$_{n}$$\cong$H, but I just can't find a homomorphism from Aut Z$_{n}$$\rightarrow$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$_{n}\rightarrow$H by c(f) = f(1). Choose f,g$\in$Aut Z$_{n}$. Then:
c(g$\circ$f) = g$\circ$f(1) = g(f(1)) = g(1)$^{l}$. But any l'$\equiv$l (mod n) $\Rightarrow$g(f(1)) = g(1)$^{l'}$, so we can consider l' an equivalence class in Z$_{n}$. That implies l = f(1)$^{p}$ for some p$\in$N, so that c(g$\circ$f) = f(1)$^{p}$*g(1) = c(g)c(f)$^{p}$.

I can't see what I'm overlooking, so I'd appreciate if someone can help point me in the right direction. Thanks!

 

[Robert1986]

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.

 

[TopCat]

Not quite. H$\subset$Z$_{n}$, but not a subgroup since Z$_{n}$ is additive and H is multiplicative. In particular, H is always a proper subset since 0$\in$Z$_{n}$ but 0$\notin$H. So I have to find an isomorphism from Aut Z$_{n}$ directly to H.