# Automorphism group of ${\bf Z}_p$

### Theorem

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

### Proof

$G= \bf Z_p$ 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 $$f_a\circ f_b = f_{ab\ (p)}$$

### Question

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 $a_0$ ?

• 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. – mrs Jan 3 '14 at 8:07
• $U({\bf Z}_n)$ ? What is $U$ ? – HK Lee Jan 3 '14 at 8:08
• If $R$ is a ring then $U(R)$ denotes its group of units. – mrs Jan 3 '14 at 8:10
• 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 $1 <a < n-1$. – 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.
• 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. – 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.
• 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 $\langle a\rangle={\bf Z}_p$. – HK Lee Jan 3 '14 at 9:14
• 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) – HK Lee Jan 5 '14 at 3:24