Abstract

Let G a group and ω(G) be the set of element orders of G. Let k∈ω(G) and let sk be the number of elements of order k in G. Let nse(G)={sk∣k∈ω(G)}. In Khatami et al. and Liu's works, L3(2) and L3(4) are uniquely determined by nse(G). In this paper, we prove that if G is a group such that nse(G) = nse(U3(7)), then G≅U3(7).

Highlights

  • A finite group G is called a simple Kn-group if G is a simple group with |π(G)| = n

  • Thompson posed a very interesting problem related to algebraic number fields as follows

  • 5.43 ∉ ω(G); it follows that the Sylow 5-subgroup of G acts fixed-point-freely on the set of elements of order 43, and so |P5| | s43, which is a contradiction

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Summary

Introduction

Let T(G) = {(n, sn) | n ∈ ω(G) and sn ∈ nse(G)}, where sn is the number of elements with order n. All sporadic simple groups are characterizable by nse and order (see [5]). Comparing the sizes of elements of same order but disregarding the actual orders of elements in T(G) of Thompson’s problem, whether it can characterize finite simple groups? The set of element orders of G is denoted by ω(G). Let π(G) denote the set of prime p such that G contains an element of order p. Ln(q) denotes the projective special linear group of degree n over finite fields of order q. Un(q) denotes the projective special unitary group of degree n over finite fields of order q. The other notations are standard (see [9])

Some Lemmas
Proof of Theorem
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