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Borovik–Lubotzky–Myasnikov generalized Andrews–Curtis conjecture

Let GG be a finitely generated group, let w(G)w(G) be its weight, and let Gab=G/[G,G]G_{ab}=G/[G,G] be its abelianization. For n≥max⁡(w(G),2)n\geq\max(w(G),2), an ordered nn-tuple is normally generating when its components normally generate GG; two such tuples are Andrews–Curtis equivalent when they are related by finite compositions of Nielsen transformations and arbitrary conjugations of components. Borovik–Lubotzky–Myasnikov generalized Andrews–Curtis conjecture. Any two normally generating nn-vectors of GG are Andrews–Curtis equivalent if and only if their images in GabG_{ab} are Andrews–Curtis equivalent. The formulation isolates the possible obstruction arising from abelianization, and the paper proves this property for finitely generated soluble groups; the statement is presented as a generalized conjecture rather than as an unresolved assertion for every finitely generated group.

Sources & referencesView supporting material

Primary source

Luc Guyot, “On Andrews-Curtis conjectures for soluble groups”, arXiv:1612.06912 (2017).

abelianizationcombinatorial-group-theorysoluble-groups

Progress summary

Refreshed
Partially solved

The conjecture is proved for all finitely generated soluble groups, but the general finitely generated case remains open.

The Borovik–Lubotzky–Myasnikov generalized conjecture, also called GACC1\mathrm{GACC1}, asks whether Andrews–Curtis equivalence of normally generating tuples is completely determined by their images in the abelianization. The unrestricted question is whether every finitely generated group satisfies this property.

Known results

  • Finite groups satisfy GACC1\mathrm{GACC1}.
  • Finitely generated free soluble groups satisfy GACC1\mathrm{GACC1}.
  • Finitely generated groups whose maximal subgroups are normal satisfy GACC1\mathrm{GACC1}.
  • Certain direct products of nonabelian simple groups satisfy GACC1\mathrm{GACC1}.

2016 soluble-group theorem

The paper “On Andrews–Curtis conjectures for soluble groups” proves GACC1\mathrm{GACC1} for every finitely generated soluble group. Its results that some soluble Baumslag–Solitar groups fail GACC2\mathrm{GACC2} do not provide a counterexample to GACC1\mathrm{GACC1}; no general proof or counterexample was found.

Current status (as of August 2026): GACC1\mathrm{GACC1} is settled for finitely generated soluble groups and several other classes, while whether any finitely generated group fails it remains open.

Sources

Solutions 0

No solutions have been posted yet.