Finite groups with \(2p^m\) elements having the highest order are solvable. (Q1884514)

This research is related to Thompson's problem. In 1987, J. G. Thompson posed the following problem to the reviewer: For each finite group \(G\) and each integer \(d\geq 1\), let \(G(d)=\{x\in G\mid x^d=1\}\). Definition. \(G_1\) and \(G_2\) are of the same order type if and only if \(|G_1(d)|=|G_2(d)|\), \(\forall d=1,2,\dots\). Suppose \(G_1,G_2\) are finite groups of the same order type. Suppose also that \(G_1\) is solvable. Is it true that \(G_2\) is also necessarily solvable? (see \textit{W. Shi} [in Group theory, Proc. Conf., Singapore 1987, 531-540 (1989; Zbl 0657.20017)], and there is a printing error in Thompson's name in the paper). That is, whether or not the solvability of finite groups may be determined by their order types. From a special point of view, considering the number of the maximal order elements, the authors of this paper continue previous works [see \textit{C. Yang}, Chin. Ann. Math., Ser. A 14, No. 5, 561-567 (1993; Zbl 0829.20036), \textit{Y. Jiang}, ibid. 21, No. 1, 61-64 (2000; Zbl 0964.20008)] and prove that finite groups with \(2p^m\) elements of maximal order are solvable, where \(p>5\) is a prime and \(m\) is a positive integer.