Let G be a group. An Abelianization of G is a pair (A, f), where A is an Abelian group and f: G --> A is a group homomorphism, which satisfies the following mapping property: given any Abelian group B and any group homomorphism h: G --> B, there is a unique group homomorphism h_a: A --> B such that h_a * f = h.
i) Explain why an Abelianization fo G is unique up to isomorphism.
ii) Construct an Abelianization of G as follows: Let G' be the subgroup fo G generated by all commutators [x,y] = xyz^-1 y^-1, where x, y are real members of G.. Show that G' is a normal subgroup. [Hint: it is enough (Why?) to show for each g is a real member of G that g[x,y]g^-1 is a real member of G'. Check that by a computation/]
ii) Show that if f: G --> B is a homomorphism from G to an Abelian group B, then f(G') = 1.
iv) Conclude that G/G' together with the natural projection mapping pi: G --> G/G' is an Abelianization of G.© BrainMass Inc. brainmass.com March 21, 2019, 3:20 pm ad1c9bdddf
This solution investigates the albelianization of a group in an attached PDF file. A .ps file is also attached, which is a postscript document.