Aut(Z_n): Automorphism Group of Z_n

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We prove that Aut(\mathbb{Z}_n)\cong (\mathbb{Z}/n\mathbb{Z})^*, also known as U(n) (easier to type).

Define \Psi: Aut(\mathbb{Z}_n)\to U(n) by \Psi(\phi)=\phi (1).

First we show that it is a homomorphism:

\begin{aligned}\Psi(\phi_1 \circ \phi_2)&=\phi_1(\phi_2(1))\\    &=\phi_1 (1+1+\cdots +1)\ \ \ (\phi_2 (1) \text{ times})\\    &=\phi_1 (1)+\phi_1 (1)+\cdots +\phi_1 (1)\ \ \ (\phi_2 (1) \text{ times})\\    &=\phi_2 (1) \cdot \phi_1 (1)\\    &=\Psi (\phi_2)\cdot \Psi (\phi_1)\\    &=\Psi (\phi_1) \cdot \Psi (\phi_2)\ \ \ \text{since} (\mathbb{Z}/n\mathbb{Z})^* \text{ is abelian.}    \end{aligned}

Next we show that it is injective:

\Psi (\phi) =1

Thus, \phi (1)=1.

Let x\in \mathbb{Z}_n.

\phi (x)=x\phi (1)=x\cdot 1=x.

Thus, the only automorphism that maps to 1 is the identity.

Thus, \ker \Psi is trivial.

Finally, we show that it is surjective.

Let x\in (\mathbb{Z}/n\mathbb{Z})^*. Consider \phi such that \phi (0)=0, \phi (1)=x, \phi (i)=ix, …, \phi (n-1)=(n-1)x.

We claim that \phi is an automorphism of \mathbb{Z}_n.

Firstly, we need to show that \{0,1,2,\cdots, n-1\}=x\{0, 1, 2, \cdots, n-1\}. This is because \gcd (x,n)=1. Hence if q is the order of x, i.e. qx\equiv 0 \pmod n, then n\vert qx, which implies that n\vert q which implies that q is at least n. Since the order of x is also at most n, q=n.

Finally, we have \phi(a+b)=(a+b)x=ax+bx=\phi (a)+\phi (b) and thus we may take \phi as the preimage of x.

Hence \Psi is surjective.

This is a detailed explanation of the proof, it can be made more concise to fit in a few paragraphs!

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