Let \(\left| G \right| = p^n q\) for some primes \(p > q\). Show that \(G\) contains a unique normal subgroup of index \(q\).
Solution
Let \(P\) be a Sylow \(p\)-subgroup of \(G\). Then \(|P|=p^n\), so \([G:P]=q\).
1. The number \(n_p\) of Sylow \(p\)-subgroups is \(1\).
By Sylow’s theorems, \[ n_p\mid q \qquad\text{and}\qquad n_p\equiv1\pmod p. \] Since \(q\) is prime, \(n_p\) is either \(1\) or \(q\). If \(n_p=q\), then \(q\equiv1\pmod p\), so \(p\) divides \(q-1\). But \(0<q-1<p\) because \(p>q\), which is impossible. Hence \(n_p=1\).
2. The subgroup \(P\) is normal and has index \(q\).
A unique Sylow subgroup is normal, so 1 gives \(P\trianglelefteq G\). Also \[ [G:P]=\frac{p^nq}{p^n}=q. \]
3. Any subgroup \(H\le G\) of index \(q\) equals \(P\).
If \([G:H]=q\), then \[ |H|=\frac{|G|}{q}=p^n. \] Thus \(H\) is a Sylow \(p\)-subgroup. By 1 the Sylow \(p\)-subgroup is unique, so \(H=P\).
4. Therefore \(G\) has a unique normal subgroup of index \(q\).
Existence follows from 2, and uniqueness from 3.