If $p<q$, $p\nmid q-1$, then “every group of order $pq$ is cyclic” iff “there is no nontrivial homomorphism from $ℤ_p$ into $\mathrm{Aut}(ℤ_q)$”.

Question

Source: Abstract Algebra, $3^{rd}$ edition by Dummit and Foote.

In Section 5.5, there is a paragraph that goes:

“Let $p$ and $q$ be primes with $p<q$ … if $p$ does not divide $q-1$ then every group of order $pq$ is cyclic. This is consistent with the fact that if $p$ does not divide $q-1$, there is no nontrivial homomorphism from $\mathbb{Z}_p$ into $\mathrm{Aut}(\mathbb{Z}_q)$ …"

My Question: How exactly does the boldface texts above follow from “every group of order $pq$ is cyclic”? I am aware of the theorem:

$\phi \colon K \to \mathrm{Aut}(H)$ is the trivial homomorphism if and only if the identity (set) map between $H \rtimes K$ and $H \times K$ is a group homomorphism if and only if $K \lhd H\rtimes K$;

but I struggle to deduce a contradiction from that. Any hint would be greatly appreciated.

Answer 1

The point is that if $H$ and $K$ are abelian, then $\phi:K \to \mathrm{Aut}(H)$ is non-trivial if and only if $H \rtimes K$ is nonabelian.

Answer 2

A group $G$ of order $pq$ has elements of both orders $p$ and $q$ (Cauchy). WLOG, let's assume $q>p$. Then there is one subgroup of order $q$, only, as if they were more, say $H$ and $K$ two of them, then the set $HK\subseteq G$ would have cardinal $q^2>pq$: contradiction. Therefore, said $H$ the (only) subgroup of order $q$ and $L$ any of order $p$, we have: $$G=HL, \space\space H\cap L=\{1\}, \space\space H\unlhd G$$ whence: $$G\cong L\ltimes H\stackrel{(p\nmid q-1)}{=}L\times H\cong C_p\times C_q\cong C_{pq}$$