(Frattini) Let \(K\) be a normal subgroup of a finite group \(G\), and let \(P\) be a Sylow \(p\)-subgroup of \(K\). Recall \(N_G(P)\) denotes the normalizer of \(P\) in \(G\). Show that \(G = KN_G(P)\).
Solution
1. For every \(g\in G\), there is \(k\in K\) with\(gPg^{-1}=kPk^{-1}\).
Normality gives \(gKg^{-1}=K\). Hence \(gPg^{-1}\) is a subgroup of \(K\). Conjugation preserves order, so it has order \(|P|\), the largest power of \(p\) dividing \(|K|\). It is therefore a Sylow \(p\)-subgroup of \(K\). Sylow conjugacy, applied inside the finite group \(K\), gives an element \(k\in K\) with the asserted equality [@DF04].
2. Every \(g\in G\) belongs to \(KN_G(P)\).
1. Then\[ (k^{-1}g)P(k^{-1}g)^{-1} =k^{-1}(gPg^{-1})k=P. \]Thus \(k^{-1}g\in N_G(P)\) and\(g=k(k^{-1}g)\in KN_G(P)\). This proves\(G\subseteq KN_G(P)\). The reverse inclusion holdsbecause both factors are subsets of the group \(G\) andtheir products lie in \(G\). Hence \(G=KN_G(P)\).