Nam Hung Tran Nguyen

Abstract:  This post will demonstrate the strength of the notion of a fundamental group by indicating several well-known problems whose solutions can be found much easier thanks to the fundamental group.  1. Compute the (co)homology groups    The (co)homology groups of an arbitrary topological space X are defined complicatedly via the groups generated by all continuous maps from the standard simplices to X. Meanwhile, fundamental groups allow us to restrict ourselves to the space X alone. By Hurewicz theorem, which was named after the Polish mathematician Witold Hurewicz, in certain cases, we can compute the (co)homology groups via the fundamental groups, which simplify the set of mathematical objects that we need to consider.     Example: Consider the unorientable surface of genus g. Its fundamental group has the following representation: <c(1), c(2), ..., c(g) | c(1)^2c(2)^2...c(g)^2>. By Hurewicz theorem, its first homology group is the abelizatio...

1. What is a hopfian group?     There are (at least) two equivalent definitions of a hopfian group .  The first one, which was written in the book "Topics in Combinatorial Group Theory" (Gilbert Baumslag) is shown below.      The second one, which was written in "Combinatorial Group Theory" (Roger Lyndon and Paul Schupp), is as follows: a group G is called hopfian if every homomorphism from G onto G is an isomorphism. The equivalence between these two definitions can be explained rigorously as follows:    - If every homomorphism from G onto G is an isomorphism, for each normal subgroup N of G, consider the projection p : G -> G/N such that p(g) = gN for each g in G. If G is isomorphic to G/N, consider the isomorphism i : G/N -> G, it is obvious that the composition of p and i yields a homomorphism from G onto G. By our assumption, this composition must be an isomorphism and the projection is also an isomorphism, which means that N ...

Consider the following problem.  Problem (4b - Algebra for Group A or 5b - Algebra for Group B, National Student Mathematical Olympiad):   Consider the polynomial T(x) = x^3 - 3x. Prove that there doesn't exist any polynomial S(x) of degree 3 such that S(T(x)) has 9 distinct roots that form an arithmetic progression.  This problem can be solved combinatorially as follows.  Assume that there is a such polynomial S(x). Then, S(x) must have three distinct roots a, b, c and the 9 distinct roots of S(T(x)) consist of 3 roots of x^3 - 3x - a, 3 roots of x^3 - 3x - b and 3 roots of x^3 - 3x - c. Let the nine roots of S(T(x)) be A, A + d, A + 2d, ..., A + 8d, where d is a nonzero real number. Assume that:     i) The three roots of x^3 - 3x - a are A + a(1)d, A + a(2)d, A + a(3)d,     ii) Similarly, the three roots of x^3 - 3x - b are A + b(1)d, A + b(2)d, A + b(3)d and the three roots of x^3 - 3x - c are A + c(1)d, A + c(2)d, A + c(3)d.  M...