A group \(G\) is called residually finite if for any nontrivial element \(g\) of \(G\), there exists a finite group \(K\) and a surjective homomorphism \(\rho: G \to K\) such that \(\rho(g)\) is a nontrivial element of \(K\).
Suppose \(G\) is a finitely generated residually finite group. Show that any surjective homomorphism from \(G\) to itself is an isomorphism.
Let \(I, J\) be connected open intervals such that \(I \cap J\) is a nonempty proper sub-interval of both \(I\) and\(J\). For instance, \(I = (0, 2)\) and \(J = (1, 3)\) form an example.
Let \(f\) (\(g\), resp.) be an orientation-preserving homeomorphism of the real line \(\mathbb{R}\) such that the set of points of \(\mathbb{R}\) which are not fixed by \(f\) (\(g\), resp.) is precisely \(I\) (\(J\), resp.).
Show that for large enough integer \(n\), the group generated by \(f^n, g^n\) is isomorphic to the group with the following presentation
\[ <a, b | [ab^{-1}, a^{-1}ba] = [ab^{-1}, a^{-2}ba^2] = id>. \]
The best solution was submitted by 김동률 (수리과학과 2015학번). Congratulations!
Here is his solution of problem 2019-12.