Finite \(p\)-groups with some isolated subgroups (Q306522)

A subgroup \(H\) of a group \(G\) is \textit{isolated} if for each \(x\in G\) either \(x\in H\) or \(\langle x\rangle \cap H=\{1\}\).NEWLINENEWLINERecall that the Hughes subgroup \(H_p(G)\) of a \(p\)-group \(G\) is the subgroup of \(G\) generated by all elements of order different from \(p\).NEWLINENEWLINEThe author considers finite non-abelian \(p\)-groups with certain isolated subgroups. Here are some of the main results of the paper.NEWLINENEWLINETheorem 1. Let \(G\) be a non-abelian \(p\)-group of exponent strictly greater than \(p\), all of whose maximal abelian subgroups of exponent strictly greater than \(p\) are isolated in \(G\). Then, \(G\) has an abelian maximal subgroup \(A\) of exponent strictly greater than \(p\) such that \(A=H_p(G)\).NEWLINENEWLINETheorem 2. Let \(G\) be a non-abelian \(p\)-group of exponent strictly greater than \(p\), all of whose maximal abelian subgroups are isolated in their normalizers. Then, \(p>2\) and \(G\) has an abelian maximal subgroup \(A\) of exponent strictly greater than \(p\) such that \(A=H_p(G)\). Conversely, all such \(p\)-groups satisfy the assumptions of the theorem.NEWLINENEWLINETheorem 3. Let \(G\) be a non-abelian \(p\)-group of exponent strictly greater than \(p\), all of whose maximal abelian subgroups of exponent strictly greater than \(p\) are isolated in their normalizers. Then, \(G\) has an abelian subgroup \(A\) of exponent strictly greater than \(p\) and index \(p\) and \(A=H_p(G)\). Conversely, all such \(p\)-groups satisfy the assumptions of the theorem.