TITLE: Groups of order $pq$ without using Sylow theorems
QUESTION [18 upvotes]: If $|G| = pq$,  $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems?

REPLY [2 votes]: This doesn't use any Sylow, as desired.
In general, if $|G|=pq$, with $p$ and $q$ distinct primes, then $G$ has center either trivial or the whole group (see here).
If $Z(G)$ is trivial, then the Class Equation reads:
$$pq=1+k_pp+k_qq \tag 1$$
where $k_i$ are the number of conjugacy classes of size $i$.  Now,  there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$)$^\dagger$. Since each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because by assumption $q\nmid p-1$).
Therefore, $(1)$ yields:
$$pq=1+k_pp+m'q(p-1) \tag 2$$
for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields:
$$p=n+m'(p-1) \tag 3$$
In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$). So, $1+k_pp=q$: but this is a contradiction, because $p\nmid q-1$ for $p>q$. So we are left with $G$ Abelian, and hence (Cauchy) cyclic.

$^\dagger$An element lies in a conjugacy class of size $q$ (respectively, $p$) if and only if its order is $p$ (respectively, $q$). This claim follows from the Orbit-Stabilizer Theorem and the fact that, for $g\in G\setminus\{e\}$, it is $\langle g\rangle=C_G(g)$.