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Sep 15, 2023 at 6:16 comment added ADL @Shri That case corresponds precisely to $\psi$ being an automorphism (as $u$ is fixed by $\psi^2$, and as it is a test word $\psi^2$ is an automorphism, and hence $\psi$ is an automorphism too by Hopficity of free groups). Different techniques will be needed for $\psi$ an automorphism. [The splitting of "automorphism" vs "non-automorphism" is really common in this area.]
Sep 15, 2023 at 5:35 comment added Shri Thank you for your answer. But what about when $u$ becomes a test word?
Sep 14, 2023 at 10:38 history answered ADL CC BY-SA 4.0