Proposition
Let $G$ be a finite group and $|G|=pq$ where $p<q$ and $p,q$ are prime. Then, $G$ is not simple.
The questioneer of this page A finite group of order $pq$ cannot be simple. writes the proof of this proposition, but I wonder if that is correct.
Here is the overview of the proof.
By third sylow's theorem, $|\mathrm{syl}_p(G)|$ devides $q$. Since $q>p$ is prome, $|\mathrm{syl}_p(G)|=1.$ Let $P$ be the sylow-$p$ subgroup. Since $1<|P|<|G|$, $P$ is nontrivial subgroup of $G$. By the second sylow's theorem, $g^{-1} P g=P$ for all $g\in G$. Thus, $P$ is nontrivial and normal subgroup of $G$ and $G$ isn't simple.
In the last part, this proof says "By the second sylow's theorem, $g^{-1} P g=P$ for all $g$."
But the second sylow's theorem says that if $H$ and $K$ are $\mathrm{sylow-}p$ subgroups, then there exists $g\in G$ s.t. $g^{-1}Hg=K$.
Why can the proof say "all $g\in G$" ?
There exists only one $p$-Sylow subgroup,since the cardinal of $gPg^{-1}$ is $p$, it is a Sylow $p$-group and equal to $P$ since there exists only one group whose cardinal is $p$.