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Finite groups with many elements of the same order

Ryan McCulloch, Lee Tae Young

Abstract

We study a conjecture by Deaconescu on the solubility of finite groups with large number of elements of the same order k. We show that the original conjecture fails by presenting some counterexamples. By restricting to a fixed k, the conjecture may or may not hold depending on k. We prove that if k is a power of a prime other than 2 or 3, or if k = 2 or 3, then the conjecture holds, while it fails for many other choices of k including all multiples of 2 and 3 which are larger than 5. We also prove that for all k > 1, it is always possible to find a finite non-soluble group where at least 2/15 of the elements have order k.

Finite groups with many elements of the same order

Abstract

We study a conjecture by Deaconescu on the solubility of finite groups with large number of elements of the same order k. We show that the original conjecture fails by presenting some counterexamples. By restricting to a fixed k, the conjecture may or may not hold depending on k. We prove that if k is a power of a prime other than 2 or 3, or if k = 2 or 3, then the conjecture holds, while it fails for many other choices of k including all multiples of 2 and 3 which are larger than 5. We also prove that for all k > 1, it is always possible to find a finite non-soluble group where at least 2/15 of the elements have order k.
Paper Structure (6 sections, 26 theorems, 43 equations, 2 tables)

This paper contains 6 sections, 26 theorems, 43 equations, 2 tables.

Key Result

Theorem 1.1

(1) The conjecture holds when $k$ is a power of a prime $p>3$, and when $k=2$ and $3$. (2) The conjecture fails when $k>4$ is divisible by either $2$ or $3$. (3) The conjecture also fails for some $k$ which are not divisible by $2$ and $3$.

Theorems & Definitions (55)

  • Conjecture 1: M. Deaconescu notebook
  • Theorem 1.1
  • Theorem 1.2
  • Definition 1
  • Lemma 1
  • proof
  • Proposition 1
  • proof
  • Proposition 2
  • proof
  • ...and 45 more