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I want to show that $\text{Aut}(\mathbb{Z}_n)$ is an abelian group of order $\phi (n)$.

The automorphism group of the group $\mathbb{Z}_n$, $\text{Aut}(\mathbb{Z}_n)$, is the group of isomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_n$.

How do the elements of $\text{Aut}(\mathbb{Z}_n)$ look like?


marked as duplicate by Dietrich Burde, Clément Guérin, David Hill, user147263, Silvia Ghinassi Feb 18 '16 at 0:04

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  • $\begingroup$ Multiplication by an invertible element of the ring $Z_n$. $\endgroup$ – David Feb 17 '16 at 19:52
  • $\begingroup$ Generalize what we talked about in your other question: how many elements of order $\;n\;$ are there in $\;\Bbb Z_n\;?$ . Exactly $\;\phi(n)\;$ , and since automorphisms map generators to generators... $\endgroup$ – DonAntonio Feb 17 '16 at 19:52
  • $\begingroup$ Also answered here. $\endgroup$ – Dietrich Burde Feb 17 '16 at 19:59
  • $\begingroup$ There are $\phi(n)$ generators for $\Bbb Z/n\Bbb Z$. A map from $\Bbb Z/n\Bbb Z$ is determined completely by where $\overline 1$ goes since $h(\overline m)=mh(\overline 1)$. Now $\overline 1$ can be mapped to any element and extended to a homomorphism. In order for the map to be surjective it has to map $\overline 1$ to a generator. And since there are $\phi(n)$ generators there are $\phi(n)$ elements in Aut$(\Bbb Z/n\Bbb Z)$. $\endgroup$ – Gregory Grant Feb 17 '16 at 20:17
  • $\begingroup$ How do we know that there are $\phi (n)$ elements in $\mathbb{Z}_n$ that have order $n$ ? @Joanpemo $\endgroup$ – Mary Star Feb 22 '16 at 16:48

Suppose $\phi\in\mathrm{Aut}(\mathbb{Z}_n)$ and suppose $\phi([1])=[m]$. Then, for $0\leq r<n$ we have \begin{align*} \phi([r])=&\phi([\underbrace{1+\cdots+1}_r])\\ &=\phi(\underbrace{[1]+\cdots+[1]}_r)\\ &=\underbrace{\phi([1])+\cdots+\phi([1])}_r=\underbrace{[m]+\cdots+[m]}_r=[mr] \end{align*} Hence, $\phi$ is multiplication by $m$ and we may write $\phi=\phi_m$. But, $\mathrm{Aut}(\mathbb{Z}_n)$ is closed under inverses and we have $\phi_m^{-1}=\phi_k$ for some $k$. Now, $$ [1]=\phi_m^{-1}\phi_m([1])=\phi_k\phi_m([1])=\phi_k([m])=[mk]=[m][k] $$ Hence, $[k]=[m]^{-1}$ and $[m]\in\mathbb{Z}_n^{\times}$.

Now, a computation similar to the one above also shows that $\phi_r\phi_s=\phi_{rs}$. Hence, the map $$\Phi:\mathrm{Aut}(\mathbb{Z}_n)\to\mathbb{Z}_n^{\times}$$ given by $\Phi(\phi_m)=[m]$ is a homomorphism. It is now straighforward to check that this map is bijective. Hence, $\mathrm{Aut}(\mathbb{Z}_n)$ is an abelian group of order $|\mathbb{Z}_n^\times|=\varphi(n)$.

  • $\begingroup$ What do the brackets mean at $\phi([1])=[m]$ ? Knowing that $\Phi (\phi_m)=[m]$ is a homomorphism and bijective, how do we conclude that $\mathrm{Aut}(\mathbb{Z}_n)$ is an abelian group of order $|\mathbb{Z}_n^\times|=\varphi(n)$ ? $\endgroup$ – Mary Star Feb 18 '16 at 18:12
  • $\begingroup$ I am using the notation $[x]$ for the congruence class of $x$ in $\mathbb{Z}_n$. You may also have seen $\overline{x}=[x]$. For your second question, you have an isomorphism $$\mathrm{Aut}(\mathbb{Z}_n)\to\mathbb{Z}_n^\times=\{[x]\in\mathbb{Z}_n\mid \gcd(x,n)=1\}$$ Since $\mathbb{Z}_n^\times$ is an abelian group, so is $\mathrm{Aut}(\mathbb{Z}_n)$. Also, the two groups have the same size. . . $\endgroup$ – David Hill Feb 18 '16 at 18:43
  • $\begingroup$ Could you explain to me why we take the function $\Phi (\phi_m)=[m]$ ? Why do we take as input the multiplication of $m$ ? $\endgroup$ – Mary Star Feb 18 '16 at 20:55
  • $\begingroup$ Because it gives the desired isomorphism. I think you should work out all the details. $\endgroup$ – David Hill Feb 18 '16 at 21:16

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