On elements of order \(p\) in powerful \(p\)-groups.
Let \(p\) be a prime, \(\varepsilon_2=2\) and \(\varepsilon_p=1\) for \(p>2\). A \(p\)-group \(G\) is said to be powerful if \(G'\leq\mho_{\varepsilon_p}(G)\). Given a subset \(M\) in a \(p\)-group \(G\), let \(\Omega_{\{1\}}(M)\) be the set of all elements of order \(\leq p\) in \(M\). The authors give an example to show that for \(p=2\) there is a powerful \(p\)-group such that \(\Omega_1(G)\neq\Omega_{\{1\}}(G)\). The following results are proved. Theorem 1. For a powerful \(p\)-group \(G\) the following hold: (i) \(|\Omega_{\{1\}}(G)|=|G:\mho_1(G)|\). (ii) If \(x\in G\) is of order \(p\) and \(N\) is a powerful normal subgroup of \(G\) such that \(\langle x,N\rangle\) is powerful, then \(|\Omega_{\{1\}}(xN)|=|\Omega_{\{1\}}(N)|\). Corollary 2. The following statements are equivalent for a powerful \(p\)-group \(G\): (i) \(|\Omega_1(G)|\leq|G:\mho_1(G)|\). (ii) \(|\Omega_1(G)|=|G:\mho_1(G)|\). (iii) \(\exp(\Omega_1(G))=p\). (iv) There exists \(H\leq G\) of exponent \(p\) such that \(|H|\geq|G:\mho_1(G)|\). By analogy, one can define the set \(\Omega_{\{n\}}(M)\) as the set of all elements of orders \(\leq p^n\) in a subset \(M\) of a \(p\)-group \(G\). Is it true, in a powerful \(p\)-group \(G\), that the following equality \(|\Omega_{\{n\}}(G)|=|G:\mho_n(G)|\) holds for \(n>1\)?
scientific article; zbMATH DE number 2026913