If $N\normal G$ and $P$ is Sylow in $N$, then $G=N_G(P)N$.Suggest a slogan
Let \(G\) be a finite group, let \(N\) be a normal subgroup of \(G\), let \(p\) be a prime, and let \(P\) be a Sylow $p$-subgroup of \(N\). Then \(G=N_G(P)N\), where \(N_G(P)\) is the normalizer of \(P\) in \(G\).
Let \(g\in G\). Since \(N\) is normal, \(gPg^{-1}\subseteq N\), and it is a Sylow \(p\)-subgroup of \(N\). By Sylow's second theorem in \(N\), there is \(n\in N\) with \(ngPg^{-1}n^{-1}=P\), so \(ng\in N_G(P)\). Hence \(g=n^{-1}(ng)\in NN_G(P)=N_G(P)N\), the last equality because \(N\) is normal.