General form of a group morphism $\tau : \mathbb{Z}_p \to \operatorname{Aut}(\mathbb{Z}_q)$ such that $\bar{k} \mapsto \tau_{\bar{k}}$

Question

I'm trying to prove the following proposition:

Let $p,q$ be two prime numbers. Then every group morphism $\tau : \mathbb{Z}_p \to Aut(\mathbb{Z}_q)$ such that $\overline{k} \mapsto \tau_{\overline{k}}$ has the form $\tau_{\overline{k}}(\bar{n}) = \overline{r}^k\overline{n}$ $\forall k,n \in \mathbb{Z}$, with $r \in \mathbb{Z}$ such that $r^p \equiv 1 \mod q$.

Here's what I've tried so far:

$\tau_\overline{k}(\overline{n}) = \tau_\overline{k}(\overline{1}+\cdots+\overline{1}) = \overline{n} \tau_\overline{k}(\overline{1})$. I guess now I have to prove that $\tau_\overline{k}(\overline{1}) = \overline{r}^k$, but I don't know how to go on from what I have.

Any help is more than appreciated, thanks in advance!

Answer 1

$\operatorname{Aut}\Bbb Z_q\cong \Bbb Z_q^×$. So necessarily $\tau_{\bar 1}(1)=r$ is relatively prime to $q$.

It follows that $\tau_{\bar k} \bar n=\bar r^k\bar n$.

Also, the homomorphism takes any $\bar k\in\Bbb Z_p$ to an element with order dividing $p$.

Hence the result.