# $\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.

$\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.

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Since this isn't a complete answer, I'm answering as a comment. First: find an injection from $\textrm{Aut}(\mathbb{Z}^n)$ into $U_n$. Then consider why every element of $U_n$ acts on $\mathbb{Z}^n$ in a natural way. –  John Stalfos Aug 24 '12 at 15:04

Let $G=\langle a\rangle=\mathbb Z_n$ and get $\phi\in Aut(G)$. Clearly, $$\phi(a)=ta:=\underbrace{a+a+\ldots+a}_t$$ for some $t$. You know that $ta$ is a generator of the group and therefore $(t,n)=1$ necessarily. Here you have $[t]\in U(\mathbb Z_n)$. Now try to show that the following function is an isomorphism: $$\Phi: Aut(G)\longrightarrow U(\mathbb Z_n)$$ $$\Phi(\phi)=[t]$$

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This is a bit misleading notationally, since $\Bbb Z_n$ is an additive group. –  Brian M. Scott Aug 24 '12 at 15:09
@BrianM.Scott: I consider $a^t$ to be $t*a$. –  Babak S. Aug 24 '12 at 15:16
I assumed that you did. I still think that it’s misleading, especially for someone who appears to be having difficulty with the subject: it adds an unnecessary layer of potential confusion. –  Brian M. Scott Aug 24 '12 at 15:20
@BrianM.Scott: Thanks for saying so. Should I delete this answer, dear Brian? I don't know what I should do now. Honestly, if I want to fix every point in my answer, it will be long. Guide me. :) –  Babak S. Aug 24 '12 at 20:00
How about the modification that I just made? –  Brian M. Scott Aug 24 '12 at 20:04

This is a fleshed out version of Brian's hint.

Lemma Let $1\in \mathbb{Z}_n$. If $\varphi\in \operatorname{Aut} \mathbb Z_n$, then $\varphi(1)$ must be a generator of $\mathbb Z_n.$

Proof: Because $\varphi$ is an automorphism (of an abelian group), $\varphi(kx)=k\varphi(x)$ for $k\in \mathbb Z, x\in \mathbb Z_n$. Additionally, $\varphi(0)=0$. Therefore if $k\varphi(1)=0$ for some $k<n$, then applying $\varphi^{-1}$ to both sides yields $k1=0$

By the lemma, we must have $\varphi(1)$ is a generator of $\mathbb Z_n$. The generators are the elements relatively prime to $n$ when the elements of $\mathbb Z_n$ are viewed as a subset of $\mathbb Z$, which are in correspondence with the elements of $U_n$. Moreover, $\varphi(\ell)=\varphi(\ell 1)\equiv \ell\phi(1) \pmod n$, and so the automorphism is given by multiplication by $\varphi(1) \mod n$, and so the map $\Psi:\operatorname{Aut}\mathbb Z_n\to U_n$ sending $\varphi$ to multiplication by $\phi(1)$ is injective. It is not hard to check that it is also surjective. Let us show that it is a group homomorphism.

Let $\varphi,\psi\in \operatorname{Aut}\mathbb Z_n$. Then $(\varphi \circ \psi)(1)=\varphi(\psi(1))=\varphi(1)\psi(1)$ by our calculation above. Therefore $\Psi(\varphi \circ \psi)=\Psi(\varphi)\Psi(\psi)$

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HINT: Suppose that $\varphi\in\operatorname{Aut}\Bbb Z_n$; then $\varphi(1)\in U_n$. (Why?) Consider the map $$h:\operatorname{Aut}\Bbb Z_n\to U_n:\varphi\mapsto\varphi(1)\;.$$

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(If you know about ring theory.) Since $\mathbb Z_n$ is an abelian group, we can consider its endomorphism ring (where addition is component-wise and multiplication is given by composition). This endomorphism ring is simply $\mathbb Z_n$, since the endomorphism is completely determined by its action on a generator, and a generator can go to any element of $\mathbb Z_n$. Therefore, the automorphism group $\mathrm{Aut}(\mathbb Z_n)$ is the group of units in $\mathbb Z_n$, which is $U_n=U(\mathbb Z_n)$.

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Note: This is basically the same proof as Brian, but using Ring theory in a non-essential way. –  M Turgeon Aug 24 '12 at 15:53
It might be using ring theory in a non-essential way, but it is conceptually simpler because the endomorphisms are easier to describe than the automorphisms, and since the invertible elements of $\mathbb Z_n$ are by definition $U_n$, we obtain the result without having to understand what $U_n$ actually looks like. –  Aaron Aug 24 '12 at 17:43