Completely realizable groups (Q6607138)

Given a functor \(f\) on groups, a group \(G\) is \(f\)-realizable if there is a group \(H\) such that \(G \simeq f(H)\), and completely \(f\)-realizable if there is a group \(H\) such that \(G \simeq f(H)\) and every subgroup of \(G\) is isomorphic to \(f(H_{1})\) for some \(H_{1} \leq H\) and vice versa. Several new results related to this problem have been obtained for \(f=D\), where \(D(H)=H'\) is the derived subgroup of \(H\) (see [\textit{J. Araújo} et al., Isr. J. Math. 234, No. 1, 149--178 (2019; Zbl 1458.20029)]). In this case, \(H\) is called an integral of \(G\).\N\NIn the paper under review, devoted to finite groups, the authors determine completely \(\mathrm{Aut}\)-realizable groups (Theorem 2.2: A group is completely \(\mathrm{Aut}\)-realizable if and only if it is an elementary abelian \(2\)-group of rank at most \(3\)). They also study \(f\)-realizable groups for \(f \in \{ Z, F, M, D, \Phi \}\), where \(Z(G)\), \(F(G)\), \(M(G)\), \(D(G)\) and \(\Phi(H)\) denote the center, the Fitting subgroup, the Chermak-Delgado subgroup, the derived subgroup and the Frattini subgroup of the group \(G\), respectively.\N\NIn particular, the authors show that every abelian group is completely \(Z\), \(M\), \(\Phi\) and \(D\)-realizable (the last case is known since \textit{R. M. Guralnick} [Glasg. Math. J. 19, 159--162 (1978; Zbl 0377.20033)] showed that if \(A\) is an abelian group, then \(A \wr \mathbb{Z}_{2}\) is an integral of \(A\)). They further assert that a group \(G\) is completely \(F\)-realizable if and only if \(G\) is nilpotent (but the reviewer suspects that the case where \(G\) is a cyclic \(2\)-group should be excluded).