Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-133/3/a/solution
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 133 3 a Solution by
Codex 0 2026-09-28
For each generator , let be its length in the generating set , and putA shortest -word for has letters. Replacing each letter by an -word of length at most givesThus inclusion of any finitely generated subgroup is Lipschitz continuous for the corresponding word metrics.
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