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20.41 (2022)

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Suppose that $G$ is a non-cyclic residually finite group in which every subgroup of finite index (including the group itself) is defined by a single defining relation, while all infinite index subgroups are free. Is it true that $G$ is either free or isomorphic to the fundamental group of a compact surface?

See 7.36 for a negative solution of a similar question without the assumption on infinite index subgroups.

Contributor: D. Kielak

Progress

*Yes, it is true (H. Wilton, Preprint, 2024, https://arxiv.org/abs/2406.02121).

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