Prove non-ab $G$, $|G|=pq$, has $N\le G$ s.t. $[G:N]=q$, so $\exists\phi:G\rightarrowtail S_q$. Show $G\cong K\le N_{S_q}(\langle(12\dots q)\rangle)$

Question

Dummit and Foote -

Let $p$ and $q$ be primes with $p < q$. Prove that a non-abelian group $G$ of order $pq$ has a non normal subgroup of index $q$, so that there exists an injective homomorphism into $S_q$. Deduce that $G$ is isomorphic to a subgroup of the normalizer in $S_q$ of the cyclic group generated by the $q$-cycle $(1\ 2\ \dots\ q)$.

I've gotten as far as to see that $(1\ 2\ \dots\ q)$ is in this subgroup. And I'm blanking thereafter.

Attempt :

$q$ being the larger prime, any subgroup of order $q$ is normal and hall. A normal hall subgroup is unique. $Z(G)$ is $1$ or non abelian is contradicted. The class equation for $G$ is $1 + p + q(\dots)$. If a subgroup of index $q$ is normal, it would again be hall and then unique, contradicting $pq = 1 + p + q(\dots)$.

Let this be $H \le G$ of index $q$. $G$ acting on the cosets of $H$ has kernel the largest normal subgroup of $H$. Since $H$ not normal, and $|H|$ prime, the kernel is $1$ and $G/1 = G \cong$ a subgroup of $S_q$.

The generator $k$ of the unique normal subgroup of size $q$ cycles the cosets $1H, kH, \dots$ so the left regular presentation of $G$ has the cycle $(12\dots q)$ in it.

I'm stuck, can't see further. Pls help.