Finite groups with many elements of the same order

TL;DR

This paper investigates a conjecture about the solubility of finite groups with a majority of elements sharing the same order, providing counterexamples and conditions under which the conjecture holds or fails.

Contribution

It refutes the original conjecture, identifies specific cases where it holds, and determines bounds on element orders in non-soluble groups.

Findings

The conjecture fails in general, with counterexamples.

The conjecture holds when k is a prime power other than 2 or 3, or when k=2, 3, or 4.

For k=4, the sharp upper bound of element ratio in non-soluble groups is established.

Abstract

We study a conjecture by Deaconescu on the solubility of finite groups with claims that if more than half of the elements in a finite group has the same order $k$, then the group is soluble. 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 $k=2, 3$ or $4$, then the conjecture holds, while it fails for many other choices of $k$ including all multiples of $2$ and $3$ which are larger than $5$. For $k=4$ we also find the sharp upper bound of the ratio of elements of order $4$ in non-soluble groups. 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 $k$.