Kourovka Notebook Problem 21.45

Definitions and mathematical statement

A finite presentation of a group is an expression \[ G=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_s\rangle =F_m/\langle\!\langle r_1,\ldots,r_s\rangle\!\rangle \] with $m,s$ finite, where $F_m$ is free on the listed generators and the denominator is the normal subgroup generated by the relators. The group is simple if it is nontrivial and has no proper nontrivial normal subgroup. For a finitely generated group, let $d(G)$ be the smallest cardinality of a generating set, allowing arbitrary elements of $G$ as generators.

Does there exist an infinite simple group $G$ with a finite presentation and $d(G)>2$? Equivalently, one seeks a finitely presented infinite simple group such that \[ \langle a,b\rangle\ne G\qquad\text{for every }a,b\in G. \] The subgroup $\langle a,b\rangle$ includes products using inverses and may itself be infinite; it must simply be proper in every case.

The number of generators displayed in a chosen presentation is not necessarily minimal. Presenting a group with three or more named generators therefore does not prove the desired lower bound. A witness must rule out every alternative pair, while also proving infinitude, simplicity, and finite presentability. The question asks for some finite minimal generator number at least three; no particular number is prescribed. Infinite simple groups lacking finite presentations do not settle the problem even if they require more than two generators.

Short English statement

Does some finitely presented infinite simple group require at least three generators?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2554 (KOU-21.45), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, new problems, Problem 21.45. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.