Free-by-cyclic groups are not relatively profinitely rigid
Abstract
For $m=5$ and for every $m \geq 7$ we construct families of pairwise non-isomorphic free-by-cyclic groups isomorphic to $F_{2m} \rtimes \mathbb{Z}$ having isomorphic profinite completions. This answers a question of Bridson & Reid from 2015.
Disclosure
“Acknowledgements. I wish to thank Martin Bridson and Alan Reid for encouragement and helpful discussions. Several steps in the proof were simplified with the use of GPT-5.6 (Sol) accessed via ChatGPT Pro. All words in this article were written by it (human) author. Date: July 28, 2026.”
PDF page 1
- Classification
- Proof ideas or individual proof-step assistance
- Multiplier
- 8
- Verified
Structural counts
Pages 10 pdf
Theorems 1 source
Lemmas 4 source
Propositions 3 source
Corollaries 0 source
Definitions 0 source
Displayed equations 19 source
Bibliography entries 17 source
Appendix pages 0 estimated
Count notes
- Source counts use the expanded primary TeX file freebycyclic-pr.tex.
- Appendix pages include the first PDF page with an explicit Appendix heading through the final page.