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

1. Feb 18, 2009

### zcdfhn

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 $$\in$$ 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.

$$f \in Aut (Z)$$ implies $$f:Z \rightarrow Z$$ and $$f^{-1}:Z \rightarrow Z$$ should be injective. Thus, both the kernels of $$f, f^{-1}$$ 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 $$Aut(Z_n)$$ is $$\phi(n)$$ (Euler totient function).
Once you find an automorphism, you need to make sure that the order of each element is same with $$x \in Z_n$$ and $$f(x) \in Z_n$$, where f is an automorphism of $$Z_n$$.