Let G be a group and Aut G the set of all automorphisms of G. What is Aut(adsbygoogle = window.adsbygoogle || []).push({}); Z[itex]_{n}[/itex] for arbitrary n[itex]\in[/itex]N?

Sol

Let f[itex]\in[/itex]AutZ[itex]_{n}[/itex]. SinceZ[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 generateZ[itex]_{n}[/itex], that is,Z[itex]_{n}[/itex]=<f(1)>.

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

Now at this point I'm certain that AutZ[itex]_{n}[/itex][itex]\cong[/itex]H, but I just can't find a homomorphism from AutZ[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:AutZ[itex]_{n}\rightarrow[/itex]H by c(f) = f(1). Choose f,g[itex]\in[/itex]AutZ[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 inZ[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!

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# Help with classification of the set of automorphisms on Z/n

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