I recently tried to come up with my own proof of the following: Let $p$ and $q$ be primes with $p<q$. Prove that a non-abelian group $G$ of order $pq$ has a nonnormal subgroup of index $q$. I found the following proof from this link: Prove that a non-abelian group of order $pq$ ($p<q$) has a nonnormal subgroup of index $q$
Let $K$ be a subgroup of order $q$ of $G$. $K$ exists by Cauchy's Theorem. Then index$[G:K]=p$, the smallest prime dividing $|G|$. Now let $G$ act on the left cosets of $K$ by left-multiplication. The kernel of this action $C=core_G(K)$ is normal in $G$ and $G/C$ injects homomorphically in $S_p$. So $|G/C| \mid p!$. Since $p \lt q$, it follows that $|G/C|=p$, whence $K=C$, and $K$ is normal.
Now $G$ has a subgroup $H$ of order $p$. If it is normal, then we would have $H \cap K =1$ and $|HK|=|H||K|/|H\cap K|=pq$, so $G=HK$, and $G \cong H \times K \cong C_{pq}$, and $G$ would be abelian. So $H$ must be non-normal.
I understand that Cauchy's Theorem implies the existence of $K$, a group of prime order. Since the index $[G:K]=p$ is the smallest prime dividing $|G|$, $K$ is normal in $G$ (Dummit and Foote, page 120). But what's the purpose of the remaining paragraph when it's apparent that $core_G(K)=\bigcap_{g\in G} gKg^{-1}=K$ since $K$ is normal?
And I know that the 1st. isomorphism thm. implies that $|G/core_G(K)|\mid p!$. But I don't think the deduction 'Since $p \lt q$, it follows that $|G/core_G(K)|=p$' is valid: $|G/core_G(K)|=p$ is immediately followed by the fact that $core_G(K)=K$. However, the '$p<q$' seems to be crucial in the proof of the proposition. Just consider $Sym(3)$. It has order $6$, and has $3$ nonnormal subgroups of index $3$, but the only subgroup of index $2$ is normal. So how to appropriately employ the assumption '$p<q$' in the proof?
Another question regarding the second paragraph: 'Now $G$ has a subgroup $H$ of order $p$. If it is normal, then we would have $H \cap K =1$'. Why the 'If it is normal' if $H$ and $K$ have coprime orders implying that $H \cap K =1$?