Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've tried, but I can't solve the question. Please help me prove that:

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

share|cite|improve this question
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: $$\Phi: Aut(G)\longrightarrow U(\mathbb Z_n)$$ $$\Phi(\phi)=[t]$$

share|cite|improve this answer
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 $t*a$. – S. Snape 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. :) – S. Snape 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 $\Psi(\varphi \circ \psi)=\Psi(\varphi)\Psi(\psi)$

share|cite|improve this answer

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

share|cite|improve this answer

(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 $U_n=U(\mathbb Z_n)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.