On relative normal complements in finite groups (Q5916303)

Let \((G,H,H_ 0,\pi)\) denote the following configuration: \(G\) is a finite group, \(H,H_ 0 \leq G\), \(H_ 0 \triangleleft H\) and \(\pi = \pi(H/H_ 0)\). Then \(H\) is said to have a relative normal complement in \(G\) if there exists \(G_ 0\triangleleft G\) such that \(G = HG_ 0\) and \(H \cap G_ 0 = H_ 0\). It is obvious that a necessary condition for \(H\) to have a relative normal complement in \(G\) is that: \((\overline{\text{A}}_ 0)\) If two \(\pi\)-elements \(x,y \in H\) are conjugate in \(G\), then \(xH_ 0\) and \(yH_ 0\) are conjugate in \(H/H_ 0\). The two main results of this paper are the following ones: Let \((G,H,H_ 0,\pi)\) satisfy condition \((\overline{\text{A}}_ 0)\). Then \(H\) has a relative normal complement in \(G\) if one of the following holds: i) \((| G:H|,| H:H_ 0|) = 1\) and \(\pi = \{p\}\). ii) \(G\) is \(\pi\)-solvable, \((| G:H|,| H|) = 1\) and \(| G:H|\) is a prime power.