Group of order $pq$ (both primes) where $p\nmid q-1,~q\nmid p-1$

Question

This little number theory problem came from an application of Sylow theorem which said for groups of order $pq$, where $p,~q$ are primes and $p\nmid q-1,~q\nmid p-1$. Then $G$ is cyclic. By Sylow III Thm, $n_p=pk+1\mid pq$, but why this implies $n_p=1$? I can't see the clear reason.

Answer 1

There is a stronger consequence of the Sylow theorems. If $n = p^Rm$ with $n$ the order of $G$ and $p \not | m$, then the amount of $p$-Sylows of $G$ which I will note $n_p$ is $1$ modulo $p$ and $n_p | m$. In this case, we get that $n_p | q$ and so in particular either $n_p = q$ or $n_p = 1$. The former is impossible by hypothesis: if $n_p = pk + 1 = q$, then $q-1 = pk$ and so $p | q-1$.

Hence there is a unique $p$-Sylow. By symmetry there is a unique $q$-Sylow. Both are normal, disjoint, and their product contains at least $pq$ elements: it must be the whole group. If $S_p,S_q$ are such Sylow groups, this tells us that $(s,t) \in S_p \times S_q \mapsto st \in G$ is an isomorphism. Consequently, $G \simeq S_p \times S_q \simeq \mathbb{Z}_p \oplus \mathbb{Z}_q \stackrel{(CRT)}{\simeq} \mathbb{Z}_{pq}.$