Why is $\operatorname{Aut}(\mathbb{Z}_p)\cong \mathbb{Z}_p^*$?

Question

Let $\mathbb{Z}_p$ denote the $p$-adic integers. Why is $\operatorname{Aut}(\mathbb{Z}_p)\cong \mathbb{Z}_p^*$?

I know that $\operatorname{Aut}(\mathbb{Z}/p^n\mathbb{Z})\cong(\mathbb{Z}/p^n\mathbb{Z})^*$, but I'm not sure which result from group theory to apply.

Answer 1

There is an injective map $$ \mathbb{Z}_p^\times \to \text{Aut}_{\text{ab. gp.}}(\mathbb{Z}_p) $$ given by taking $\alpha$ to left-multiplication by $\alpha$ (this is true for any ring).

You wish to know this is surjective. Say $\phi$ is an automorphism of $\mathbb{Z}_p$. Then $\phi$ has to preserve the subgroup $p^n\mathbb{Z}_p$, and therefore $\phi$ has to induce an automorphism of $\mathbb{Z}/p^n\mathbb{Z}$.

This automorphism must be multiplication by a unit $\beta_n$ in $\mathbb{Z}/p^n\mathbb{Z}$, since all automorphisms of this abelian group arise in this way (you seem to already believe this, but proof: take the fact above that was "true for any ring" and observe that injective functions from finite sets to themselves are bijective).

So we have a whole bunch of $\beta_n$. Since for $n > m$ the diagram

$\require{AMScd}$ \begin{CD} \mathbb{Z}_p / p^n \mathbb{Z}_p @>\phi>> \mathbb{Z}_p / p^n \mathbb{Z}_p \\ @V V V @VV V\\ \mathbb{Z}_p / p^m \mathbb{Z}_p @>\phi>> \mathbb{Z}_p / p^m \mathbb{Z}_p \end{CD} commutes, where the vertical arrows are the natural reduction and the horizontal are the maps induced by $\phi$, we must have that $\beta_n$ reduces to $\beta_m$ mod $p^n$.

So the $\beta_n$ form an element $\beta$ of $$ \varprojlim{(\mathbb{Z}_p/p^n\mathbb{Z}_p})^\times = \mathbb{Z}_p^\times, $$ and our automorphism $\phi$ must be multiplication by this $\beta$.