A subnormality criterion in finite groups related to character degrees
Suppose that \(A\) is a proper abelian subgroup of a finite group \(G\). Then, if \(\chi\in\text{Irr}(G)\), then \(\chi(1)\leq| G:A|\). One proves the following important assertion (Theorem A): If \(\chi(1)=| G:A|\), then \(A\) is subnormal in \(G\). The author shows that \(A\) need not be normal in \(G\). Note that even if \(A\) is normal in \(G\) and \(| G:A|\in\text{cd}(G)\), then \(G/A\) may be an arbitrary group. Theorem A cannot be extended to arbitrary nilpotent subgroups. The following theorem is an immediate consequence of Theorem A. Theorem B. Suppose that \(A\) is an abelian subgroup of \(G\). If \(| G:A|\in\text{cd}(G)\), then \(G_0=A^G\), the normal closure of \(A\) in \(G\), is a nilpotent group of central type whose center \(Z_0\) satisfies \(| A:Z_0|=| G_0:A|\). Moreover, \(C_G(Z_0)=G_0\) and \(| G:G_0|<| Z_0|\). The author poses the following Problem. Suppose the finite group \(G\) has a proper nilpotent subgroup \(H\) possessing a linear character \(\lambda\) such that \(\lambda^G\) is irreducible. Can \(G\) be a simple group? I think that this problem can be solved using the classification of finite simple groups. For example, if \(H\in\text{Syl}(G)\), then Chillag and Herzog proved that \(G\) in fact is not simple (moreover, in the last case, the Fitting subgroup of \(G>\{1\}\) is nonidentity).
scientific article; zbMATH DE number 1214104