Math-2006 · mathematics · Group theory

Math-2006

Questions in Geometric Group Theory — Q 13.9

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

For a finite presentation $P=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_m\rangle$, write $G(P)$ for the presented group. Such a presentation is balanced because it uses equally many generators and relators. It presents a perfect group when $G(P)/[G(P),G(P)]=1$, equivalently when the exponent-sum matrix of its relators has determinant $1$ or $-1$.

Is there an effectively produced sequence $(P_j)_{j\geq1}$ of finite balanced presentations of perfect groups such that \[ T=\{j\geq1:G(P_j)=1\} \] is not a decidable subset of the positive integers? Effective production means an algorithm receives $j$ and returns the finite strings specifying $P_j$. Nondecidability means no algorithm halts on every $j$ and correctly tests membership in $T$.

The number of generators and the lengths of the relators may vary with $j$; the source does not impose a uniform bound on either. The perfectness promise applies to every member of the sequence, including the nontrivial groups, and prevents abelianization from deciding triviality in the intended examples.

This question concerns recognition of the trivial group across a family of presentations. It is different from the word problem inside a single group. A presentation being trivial can be verified by enumerating consequences until each generator is proved trivial, but the difficulty is obtaining an algorithm that also terminates on all the nontrivial members.

Short English statement

Can triviality be undecidable even within an effective sequence of balanced presentations of perfect groups?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.