Find non trivial homomorphism $\mathbb{Z}/q\mathbb{Z} \rightarrow \text{Aut}(\mathbb{Z}/p\mathbb{Z})$

Solution 1:

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 $$\exists\psi\in \operatorname{Hom}(\mathbb Z_q,\operatorname{Aut}\mathbb Z_q)\text{, where }\psi\text{ is non-trivial and it is also injective}.$$