Deaconescu's solubility conjecture for finite groups with many elements of the same order

Let kk be a fixed positive integer and let GG be a finite group. Suppose that at least half of the elements of GG have order kk. Deaconescu's conjecture. Then GG is soluble. The paper shows that the original conjecture is false by giving counterexamples. For fixed kk, it holds when kk is a power of a prime other than 22 or 33, and when k=2k=2 or 33, but fails for many other values of kk, including all multiples of 22 and 33 greater than 55.

References

Primary source

Ryan McCulloch and Lee Tae Young, “Finite groups with many elements of the same order”, arXiv:2602.19340 (2026).

Progress summary

Refreshed
Claimed solved

A 2026 preprint claims the conjecture is false in general, while proving it in several important special cases.

Deaconescu’s conjecture asserts that a finite group is soluble if at least half its elements have one fixed order. The 2026 preprint claims this unrestricted statement is false and identifies both surviving cases and broad families of counterexamples.

Known results

  • Wall proved the assertion for involutions (k=2k=2).
  • Liebeck and MacHale classified finite groups with at least half their elements of order 22.
  • Mann and Berkovich obtained further structural and classification results for k=2k=2.

2026 arXiv preprint, version 2

The revised preprint claims the conjecture holds for k=2k=2, k=3k=3, k=4k=4, and every prime power k=pak=p^a with p>3p>3, but fails for every k>4k>4 divisible by 22 or 33, as well as some other kk. It further claims non-soluble examples with at least 2/152/15 of elements of order kk, plus stronger quantitative bounds. These claims have no independent verification or published referee assessment in the retrieved sources.

Current status (as of September 2026): The unrestricted conjecture is claimed false, and the revised preprint claims complete classifications in several cases, but those claims remain unverified in the retrieved record.

Sources

Solutions 0

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