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 $n$ , $q=n$ .