$p,q$ primes, $p\mid q-1$. Weaker assumption in the proof of the existence of non-trivial $C_p\ltimes C_q$?

Question

Motivated by the fact that the non-existence of non-trivial $C_p\ltimes C_q$, for $p\nmid q-1$, can be proven without any piece of information on the structure of $\operatorname{Aut}(C_q)$, not even its order, I wonder whether to proving the existence of a non-trivial semidirect product (then necessarily for $p\mid q-1$) really requires the result $\operatorname{Aut}(C_q)\cong C_{q-1}$ (as, e.g., this and this seem to indicate). Namely: is there any proof of such existence, which uses solely the (weaker) result $\left|\operatorname{Aut}(C_q)\right|=q-1$, or is there any theoretic reason why this is not possible and the "full result" $\operatorname{Aut}(C_q)\cong C_{q-1}$ is then required?

Answer 1

If $p \mid q-1$, by Cauchy's theorem there is an element of order $p$ in a group of order $|Aut(C_q)| = q-1$. Hence a nontrivial homomorphism $C_p \rightarrow Aut(C_q)$.