Suppose $G$ is a group of order $pq$ with $p<q, p \nmid q-1$ and $p,q$ are primes. Then $G$ is cyclic.
The way our instructor has proved this theorem is as follows $:$
He first proved that $G$ has a unique Sylow-$p$-subgroup $S_p$ of order $p$ and a unique Sylow-$q$-subgroup of order $q$ by using Sylow's second theorem. Since conjugation of a Sylow-$p$-subgroup by the elements of $G$ is another Sylow-$p$-subgroup of $G$ so by uniqueness it follows that $S_p$ and $S_q$ are normal subgroups of $G$. Then he takes an element $g$ of order $p$ from $S_p$ and an element $h$ of order $q$ from $S_q$ which are possible to take since both $S_p$ and $S_q$ are cyclic of order $p$ and $q$ respectively. Now using the normality criterion of $S_q$ it follows that $ghg^{-1} \in S_q$. Since $S_q$ is a cyclic group generated by $h$ so $ghg^{-1} = h^a$ for some $0 \le a \le q-1$. At this place he claims that $a^p \equiv 1\ (\text {mod}\ q)$ as $g^p = e$. Though I don't understand why it should be true. Then the last few arguments go as follows $:$
So $a$ has order dividing $p$ in $(\Bbb Z/ q \Bbb Z)^*$ which is of order $q-1$. Since $p \nmid q-1$ it follows that $a \equiv 1\ (\text {mod}\ q)$. So $ghg^{-1} = h$ i.e. $g$ and $h$ commute with each other. Since order of $g$ and $h$ are relatively prime to each other it follows that $o(gh) = o(g)o(h) = pq,$ where $o(g)$ denotes the order of the element $g \in G$. This proves that $G$ is a cyclic group generated by the element $gh \in G$.
Can anybody please help me to point out why $a^p \equiv 1\ (\text {mod}\ q)$? Then it will be really helpful for me.
Thank you very much.
