Subgroups of index \(p\), \(p\) being the smallest prime dividing \(\left|G\right|\) (FGT 1A.1)

Let \(H\) be a subgroup of prime index \(p\) in the finite group \(G\), and suppose that no prime smaller than \(p\) divides \(\left|G\right|\). Prove that \(H \triangleleft G\).

Let \(\left|G\right| = pn\). The action of \(G\) on left cosets of \(H\) induces a homomorphism \(\varphi : G \rightarrow S_p\), whose kernel we shall denote \(K\). Since \(K \le H\), we have \(1 \le \left|K\right| \le n\), so that \(p \le \left|G/K\right| \le \left|G\right|\). Since \(p\) is the smallest prime dividing \(\left|G\right|\), it follows that \(\left|G/K\right|\) is a product of primes \(\ge p\). But \(\left|S_p\right| = p!\) is a product of primes \(\le p\), where the prime \(p\) appears just once. It can be seen that \(\left|G/K\right|\) divides \(p!\) only when \(n = 1\), in which case \(H = 1\) is surely normal. If it does not, we have a contradiction, since \(G/K\) is isomorphic to a subgroup of \(S_p\). This completes the proof.