A normal subgroup together with the normalizer of one of its Sylow subgroups generates the whole group.Suggest a slogan
Let \(G\) be a finite group, \(H\normal G\) a normal subgroup, \(p\) a prime, and \(P \in \Syl_p(H)\) a Sylow $p$-subgroup of \(H\). Then \(G=H\,N_G(P)\), and \([G: H]\) divides \(\abs{N_G(P)}\).
Let \(g\in G\). Since \(H\normal G\), \(gPg^{-1}\leq H\) is a Sylow \(p\)-subgroup of \(H\), so by Sylow's second theorem in \(H\) there is \(h\in H\) with \(hgPg^{-1}h^{-1}=P\). Then \(hg\in N_G(P)\), so \(g\in h^{-1}N_G(P)\subseteq H\,N_G(P)\). Hence \(G=H\,N_G(P)\), and by the second isomorphism theorem \(G/H\cong N_G(P)/(N_G(P)\cap H)\), so \([G:H]\) divides \(\abs{N_G(P)}\).