A group of order pq with a single subgroup of order p Given a group $G$ of order $pq$ (such that $p,q$ are primes and $p < q$) 
that have a single subgroup of order p (named $H$) 
prove that  $\forall h \in H , g\in G : ghg^{-1} = h$
 A: First note that $H$ is normal in $G$ since for all $g\in G$, the subgroup $g^{-1}Hg$ has $p$ elements.
Take an element $g\notin H$. If $g$ generates $G$, we are done. Else $g$ has order $q$. Take a generator $h$ of $H$. We want to show first that $gh=hg$.
To see this, consider the operation of $<\!g\!>$ on $H$ by conjugation: $x\mapsto g^{-1}xg$. This gives a homomorphism $<\!g\!>\to \mathrm{Sym}(H)$. But since $q$ does not divide the order of $\mathrm{Sym}(H)$ (which is $p!$), this map is not injective, so it must be the identity. 
Hence $gh=hg$. Now $gh$ has order $pq$.
Indeed, if $gh$ has order $p$, then $g^p=1$, which is not possible. On the other hand, if $gh$ has order $q$, then $h^{q-p}=1$, which is not possible.
So we see that $gh$ generates $G$.
A: By applying Sylow's 3rd theorem, we conclude that there is exactly one Sylow $p$ and exactly one Sylow $q$ subgroup and hence both of them are normal. Also their intersection is trivial since $p$ and $q$ are two distinct primes. So, $G$ is the direct product of $C_p$ and $C_q$  which is essentially $C_{pq}$
