The metanorm and its influence on the group structure
The norm $N(G)$ of a group $G$ is the intersection of the normalizers of all subgroups of a group $G$ (see [\textit{R. Baer}, Compos. Math. 1, 254--283 (1934; Zbl 0009.15504)]). \textit{E. Schenkman} [Ill. J. Math. 4, 150--152 (1960; Zbl 0099.25104)] showed that $N(G)\leq \zeta_2(G)$, the second center of $G$. If $\mathfrak{X}$ is a group theoretical class, then the $\mathfrak{X}$-norm is the intersection of all the normalizers of the subgroups of $G$ which are not in $\mathfrak{X}$. The metanorm of $G$ is the subgroup $M(G)$ which is the $\mathfrak{X}$-norm in the case when $\mathfrak{X}$ is the class of abelian groups. Thus $M(G)$ is metahamiltonian-every nonabelian subgroup of $M(G)$ is normal in $M(G)$. Such groups were introduced by \textit{G. M. Romalis} and \textit{N. F. Sesekin} [Mat. Zap., Sverdl. 5, No. 3, 101--106 (1966; Zbl 0351.20020); ibid. 6, No. 3, 50--52 (1968; Zbl 0351.20021); ibid. 7, No. 3, 195--199 (1970; Zbl 0324.20036)]. Their results show that if $G$ is metahamiltonian, then $G'$ is a finite $p$-subgroup for some prime $p$, called the characteristic of $G$ in this paper. The main results are: Corollary 4.17: Let $G$ be a locally (polycyclic-by-finite) group whose metanorm $M$ is metabelian, but not nilpotent, and has characteristic $p$. Then $G$ is either metahamiltonian or $|M'|=p$. Theorem 5.2: Let $G$ be a locally finite group whose metanorm $M$ is not metabelian and let $p$ be the characteristic of $M$. Then $G$ contains an abelian subgroup of finite index dividing $p(p-1)(p^2-1)$. As usual with these authors there are a host of other interesting ideas rather too numerous to mention in a short review.
scientific article; zbMATH DE number 6894493
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