On a Question of Stanojkovski: Abelian Subgroups of Breadth Two Distinguish Vaughan-Lee's Groups 8.5.14 and 8.5.15
Manuscript 14 October 2026 · Online 14 October 2026
Abstract
At the problem session of the Oberwolfach workshop Computational Group Theory (June 2025), M. Stanojkovski asked whether an invariant simpler than the order of the automorphism group can show that two groups \(G_1\) and \(G_2\) of order \(p^8\), class \(2\) and exponent \(p\) (\(p\) an odd prime) are not isomorphic. They are the groups 8.5.14 and 8.5.15 of Vaughan-Lee's list, whose presentations and automorphism group orders are due to Vaughan-Lee, and they are the one pair of \(5\)-generator groups of order \(p^8\) that the geometric invariants of O'Brien and Stanojkovski did not separate in their computations. For a group \(G\) of class \(2\) let \(\mathcal{A}(G)\) be the set of abelian subgroups \(A\ge G'\) with \(|A:G'|=p^2\) all of whose elements outside \(G'\) have centraliser of index \(p^2\). We prove that \(|\mathcal{A}(G_1)|=1\) and \(|\mathcal{A}(G_2)|=3\) for every odd prime \(p\) and every quadratic non-residue \(\omega\) in the presentation of \(G_1\); hence \(G_1\not\cong G_2\). The proof is uniform in \(p\): the elements of breadth two form, in the projective space of \(G/G'\), a plane and a cubic scroll; the members of \(\mathcal{A}(G)\) are the lines of this locus on which the commutator map vanishes; and the hypothesis on \(\omega\) enters only through \(t^2\ne\omega\). We also determine the commuting graph on the elements of breadth two, whose numbers of edges differ by \(p\), and we show that \(G_1\) depends only on the square class of \(\omega\) and is isomorphic to \(G_2\) when \(\omega\) is a square. The pair had already been separated without automorphism groups in the classification literature: Copetti's thesis of 2004, following Brahana's paper of 1951, attaches to the space of commutator forms a pencil of conics, which contains double lines only for 8.5.15. We show that \(|\mathcal{A}(G)|-1\) is the number of these double lines, so that \(|\mathcal{A}(G)|\) is a group-theoretic form of their invariant. Breadth and centralisers are classical notions; whether either invariant is simpler in the intended sense is for the proposers to judge. This is an unrefereed note.
Record
- Affiliation
- Mercury Software GmbH
- Contact
- [email protected] · GitHub
- Result
- Answer: yes, an elementary invariant separates the pair
- Categories
- math.GR · math.RA
- Manuscript
- 14 October 2026
- Online release
- 14 October 2026
- Version
- 1.0
- License
- Creative Commons Attribution 4.0 International
Files and verification
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Citation
Alper Ferudun, “On a Question of Stanojkovski: Abelian Subgroups of Breadth Two Distinguish Vaughan-Lee's Groups 8.5.14 and 8.5.15,” EulerSolve Research Papers, OWR-14299292-013, 2026. https://doi.org/10.5281/zenodo.23376497.
BibTeX
@misc{Ferudun2026Owr14299292013,
author = {Ferudun, Alper},
title = {On a Question of Stanojkovski: Abelian Subgroups of Breadth Two Distinguish Vaughan-Lee's Groups 8.5.14 and 8.5.15},
year = {2026},
howpublished = {EulerSolve Research Papers},
url = {https://eulersolve.org/papers/owr-14299292-013/},
doi = {10.5281/zenodo.23376497},
note = {OWR-14299292-013; unrefereed preprint}
}