Non-abelian groups of order pq, p and q prime (D&F 4.3.27)

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 \(q\), so that there exists an injective homomorphism into \(S_q\). Deduce that \(G\) is isomorphic to a subgroup of \(N_{S_q}(\left<(1\ 2\ \ldots\ q)\right>)\).

Note that \(Z(G) \neq G\) since \(G\) is non-abelian, and also \(Z(G)\) is neither \(p\) nor \(q\) since otherwise \(G/Z(G)\) would be cyclic. This forces \(Z(G) = 1\). Let \(g_1, g_2, \ldots, g_r\) be representatives of the conjugacy classes of \(G\) not in \(Z(G)\). For each \(g_i\), the subgroup \(G_i = \left<g_i\right>\) cannot have order \(pq\) since that means \(G\) is cyclic. Hence \(\left|G_i\right|\) is either \(p\) or \(q\). If \(\left|G_i\right| = p\), then \(G_i \le C_G(G_i) < G\), so \(\left|G : C_G(G_i)\right| = q\). Likewise if \(\left|G_i\right| = q\) then \(\left|G : C_G(G_i)\right| = p\). Hence the class equation of \(G\) is \(1 + mp + nq\), for \(m,n \in \mathbb{Z^+}\). It is clear that the value of any sum including the term \(1\) is either \(1\) or some \(k > p\). Thus any order-\(p\) subgroup cannot be expressed as a union of conjugacy classes, and so is nonnormal in \(G\).

By Cauchy's Theorem, there exists an order-\(p\) subgroup \(H\) and an order-\(q\) element \(x\). Since the only normal subgroup of \(G\) contained in \(H\) is the trivial subgroup, \(\ker \pi_H\) (the permutation representation of the action of \(G\) on \(G/H\)) is trivial, and so \(\pi_H\) is an injective homomorphism from \(G\) to \(S_{G/H} \cong S_q\). Now, \(G/H\) is cyclic since it has prime order. Since \(\left<x\right> \cap H = 1\) (which is up to the reader to justify because I don't feel like doing it), we can write \(G/H = \left<xH\right>\). In particular, \(\pi_H(x) = (H\ xH\ x^2H\ \ldots\ x^{q-1}H)\). Furthermore \(\left<x\right> \trianglelefteq G\) since it has index \(p\) (the smallest prime dividing \(G\)). This implies \(\left<\pi_H(x)\right> = \pi_H(\left<x\right>) \trianglelefteq \text{im}\ \pi_H\). By the First Isomorphism Theorem we can conclude that \(G \cong \text{im}\ \pi_H \le N_{S_{G/H}}(\left<\pi_H(x)\right>) = N_{S_{G/H}}(\left<(H\ xH\ x^2H\ \ldots\ x^{q-1}H)\right>)\).

(ok to be honest I looked up the solution for that last part)