Question

Define the commutator subgroup Gᶜ to be the subgroup generated by all elements xyx⁻¹y⁻¹ with x,y∈G Such elements are called commutators. 22. (a) Show that the commutator subgroup is a normal subgroup. (b) Show that G/Gᶜ is abelian.

Answer 1

The commutator subgroup Gᶜ is shown to be normal by demonstrating that conjugation of its elements by any element of G remains within Gᶜ. Furthermore, G/Gᶜ is proven to be abelian by showing that the product of its cosets commutes.

Step-by-step explanation:

The commutator subgroup Gᶜ of a group G is the subgroup generated by all elements of the form xyx⁻¹y⁻¹, where x and y are elements of G. To show that Gᶜ is a normal subgroup, we need to show that gGᶜg⁻¹ = Gᶜ for any g in G. Take any element h = aba⁻¹b⁻¹ in Gᶜ and any g in G, we want to show ghg⁻¹ is in Gᶜ.

Note that ghg⁻¹ = gabg⁻¹a⁻¹gag⁻¹b⁻¹, which may be written as (ga)(gb)(ga)⁻¹(gb)⁻¹, where ga and gb are elements of G, thus showing that ghg⁻¹ is in Gᶜ and Gᶜ is normal.

To show that G/Gᶜ is abelian, consider any two cosets aGᶜ and bGᶜ in G/Gᶜ. Their product is (aGᶜ)(bGᶜ) = abGᶜ. But since Gᶜ contains all commutators, we can insert any element of Gᶜ to manipulate the product ab into ba, which means the product of cosets commutes. Consequently, G/Gᶜ is an abelian group.