If $p$ is odd prime then $\DeclareMathOperator{Aut}{Aut}\Aut(\bf Z_p) = (\bf Z_p)^\times =\bf Z_{p-1}$


$ G= \bf Z_p$ Let $$f_a \colon G \to G,\; f_a(x)=ax,\; 0<a<p $$

Then since $(a,p)=1$, $$ f_a(\bf Z_p) = \{0, a, 2a, \dots, (p-1)a \} = \langle a \rangle = \bf Z_p. $$ So $f_a\in \Aut (\bf Z_p)$, and $$f_a\circ f_b = f_{ab\ (p)}$$


Can you finish the proof ? That is, a group $H=\{ f_a \mid 0< a < p\}$ has order $p-1$.

But I cannot show that it is cyclic

That is we must show that there exists $0< a_0 <p$ s.t. $$ \{ a_0, a_0^2, \dots, a_0^{p-1}\} = \bf Z_p^\times $$

For example, $$a_0(p=3)=2,\; a_0(5)=2,\; a_0(7)=3,\; \ldots $$

How can we show that the existence of $a_0$ ?

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    $\begingroup$ Have you ever tried to show that if $G$ is cyclic then $Aut(G)\cong U(\mathbb Z_n)$? I think this ways is easier to walk. $\endgroup$ – mrs Jan 3 '14 at 8:07
  • $\begingroup$ $U({\bf Z}_n)$ ? What is $U$ ? $\endgroup$ – HK Lee Jan 3 '14 at 8:08
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    $\begingroup$ If $R$ is a ring then $U(R)$ denotes its group of units. $\endgroup$ – mrs Jan 3 '14 at 8:10
  • $\begingroup$ I agree that ${\rm Aut}\ ({\bf Z}_n) = U({\bf Z}_n)$. But I think that there still exists a doubt of cyclicity. My try is ${\rm Aut}\ ({\bf Z}_n) = \langle a\rangle $ where $(a,n-1)=1$ and $1 <a < n-1$. $\endgroup$ – HK Lee Jan 3 '14 at 8:52

I think this way is easier. Have look at it and check if it is helpful for you. Let $G=\langle a\rangle$ of order $n$. If we pick $\phi \in \operatorname{Aut}(G)$, then $\phi(a)=a^k$ such that $\gcd(n,k)=1$. What is $U(\mathbb Z_n)$? It is $$U(\mathbb Z_n)=\{r\in\mathbb Z_n\mid\text{for an element}~~s\in\mathbb Z_n, rs=sr=1\}$$ Here, for this $k$ we have $[k]\in U(\mathbb Z_n)$. Now try to show that $\Phi: \operatorname{Aut}(G)\to U(\mathbb Z_n),~~\phi\mapsto[k]$ is an isomorphism.

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    $\begingroup$ Thank you. This way is good since we do not need to restrict to the case where $p$ is prime : (1) $\Phi$ is well-defined : $(n,k)=1$ implies $an+bk=1$. So $bk=kb=1$. (2) $\Phi$ is a homomorphsim : $\Phi (\phi_k \phi_m)=\Phi(\phi_{km})=[km]=[k][m]=\Phi(\phi_k)\Phi(\phi_m)$ (3) $bk=kb=1$ implies $(n,k)=1$. So $\Phi$ is onto. (4) For $1\leq k <n$, $[k]$ are distinct in $U({\bf Z}_n)$ So $\Phi$ is one-to-one. $\endgroup$ – HK Lee Jan 3 '14 at 9:37

As this is about algebra, let's use the power of structure preserving maps.$\DeclareMathOperator{Aut}{Aut}$

What you have done tacitly is defining a map $$(\bf Z_p )^\times \rightarrow \Aut(\bf Z_p) \\ a \mapsto f_a$$ You have shown that this map is a homomorphism and it is clear that the map is one-to-one. You only have to show that it is onto to conclude the desired isomorphism. You can do so by defining an inverse map. I think this is easier than showing that $\Aut(\bf Z_p)$ is cyclic.

  • $\begingroup$ I read again my posting. I agree your opinion. Since ${\bf Z}_p = \langle 1 \rangle $, if $f\in {\rm Aut}\ ({\bf Z}_p)$, then $f(1)=a$ determines $f$, where $\langle a\rangle={\bf Z}_p$. $\endgroup$ – HK Lee Jan 3 '14 at 9:14
  • $\begingroup$ @HeeKwonLee exactly, that's what I was referring to with the inverse map. $\endgroup$ – benh Jan 3 '14 at 9:16
  • $\begingroup$ Facts for Proof (${\rm Aut}\ ({\bf Z}_p)$ is cyclic) : (1) ${\rm Aut}\ ({\bf Z}_p)$ is abelian so that we have a decomposition ${\bf Z}_{n_1}\times \cdots \times {\bf Z}_{n_k}$ with $n_k|n_{k-1}|\cdots |n_1$. (2) ${\bf Z}_p[x]$ is UFD so that $x^{n_k}-1$ has at most $n_k$ roots in the field ${\bf Z}_p$. (reference : chapter - polynomial ring in Dummit and Foote's book) $\endgroup$ – HK Lee Jan 5 '14 at 3:24

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