Finite groups of the same type as Suzuki groups

For a finite group \(G\) and a positive integer \(n\), let \(G(n)\) be the set of all elements in \(G\) such that \(x^{n}=1\). The groups \(G\) and \(H\) are said to be of the same (order) type if \(G(n)=H(n)\), for all \(n\). The main aim of this paper is to show that if \(G\) is a finite group of...

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Bibliographic Details
Published in:arXiv.org 2016-05
Main Authors: Seyed Hassan Alavi, Daneshkhah, Ashraf, Hosein Parvizi Mosaed
Format: Article
Language:English
Subjects:
Group theory
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Summary:For a finite group \(G\) and a positive integer \(n\), let \(G(n)\) be the set of all elements in \(G\) such that \(x^{n}=1\). The groups \(G\) and \(H\) are said to be of the same (order) type if \(G(n)=H(n)\), for all \(n\). The main aim of this paper is to show that if \(G\) is a finite group of the same type as Suzuki groups \(Sz(q)\), where \(q=2^{2m+1}\geq 8\), then \(G\) is isomorphic to \(Sz(q)\). This addresses the well-known J. G. Thompson's problem (1987) for simple groups.
ISSN:2331-8422