Find non trivial homomorphism $\mathbb{Z}/q\mathbb{Z} \rightarrow \text{Aut}(\mathbb{Z}/p\mathbb{Z})$ Let $H= \mathbb{Z}/p\mathbb{Z}$ and $F=\mathbb{Z}/q\mathbb{Z}$ with $p,q$ primes. As indication, I got that $${\rm Aut}(H) \cong (\mathbb{Z}/p\mathbb{Z})^\times \cong \mathbb{Z}/(p-1)\mathbb{Z}.$$
I have to show that if $q$ divides $p-1$ there exist a non-trivial homomorphism $\psi$: $F \rightarrow{\rm Aut}(H)$.
I understand if $q$ divides $p-1$ then there is an element of order $q$ in $\mathbb{Z}/(p-1)\mathbb{Z}$. But, the problem I have, is that I don't really understand how to construct homomorphisms so I'd like to have a step-by-step explanation if possible, please. For me, this non-trivial homomorphism is simply of form $\psi([x]_q): y \rightarrow y \cdot g^x$, where $g$ is generator of $(\mathbb{Z}/p\mathbb{Z})^\times$. But I was told that it is wrong.
Thanks in advance and have a nice day.
 A: The map in your problem could represent the morphism induced by a semidirect product of a $p$-Sylow and a $q$-Sylow $\mathbb Z_p\rtimes_{\psi}\mathbb Z_q$.
Consider the homomorhism $\psi: \mathbb Z_q\to \operatorname{Aut}(\mathbb Z_p)$ you have that $|\psi(\mathbb Z_q)|$ divides $|\mathbb Z_q|$ and $|\operatorname{Aut}(\mathbb Z_p)|$, where $|\operatorname{Aut}(\mathbb Z_p)|=\varphi(p)=p-1$ (for example the two groups could be $\mathbb Z_2$ and $\mathbb Z_3$).
Note: Since $q$ divides $|\operatorname{Aut}(\mathbb Z_p)|$, which is a cyclic group, you could take the homomorphism that maps a generator of $\mathbb Z_q$ to a generator of the (single) $q$-group in $\operatorname{Aut}(\mathbb Z_p)$ (in a $p$-cyclic group $P$ every element of $P$ has the same order of the group and generate it).
If $q|p-1$, we can assume $|\psi(\mathbb Z_q)|=q$ and for the first theorem of isomorphism, we know that $$|\psi(\mathbb Z_q)|=q=|F:Ker(\psi)|\implies|Ker(\psi)|=\dfrac{|\mathbb Z_q|}{q}=1 \implies \psi\ne id_{\mathbb Z_q}$$
Since the order of the Kernel of the morphism is not equal to the order of the group, this shows that $$\exists\psi\in \operatorname{Hom}(\mathbb Z_q,\operatorname{Aut}\mathbb Z_q)\text{, where }\psi\text{ is non-trivial and it is also injective}.$$
