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I just showed that if $G $ is a nonabelian group of order $pq $, $p <q $, then it has a non normal subgroup $K $ of index $q $. But now I want to show that $G $ is isomorphic to a subgroup of the normalizer in $S_q $ of the cyclic group generated by the cycle $(1~2~\cdots~q) $. I think I should let $G $ act on left cosets of $K $, but I don't see what to do next.

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$K$ is non-normal Sylow-$p$ subgroup of $G$, hence $G$ should contain more than one Sylow-$p$ subgroup.

As you said, let $G$ act on cosets of one Sylow-$p$ subgroups, say on $G/K$. The action is transitive (clear?). This gives a homomorphism from $G$ to $S_q$ as you mentioned.

What is kernel of this homomorphism? It contains those $g\in G$, such that $g$ fixes every coset of $K$, i.e. $gxK=xK$ for all $x\in G$, i.e. $g\in xKx^{-1}$ for all $x\in G$.

Thus, kernel of homomorphism contains the elements common in all conjugates of $K$, i.e. common in all Sylow-$p$ subgroup. What could be these common elements? Can you complete proof?

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