Automorphism group of ${\bf Z}_p$

Question

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 $a_0$ ?

Answer 1

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.

Answer 2

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.