Exercise: 24 Section 4.3
Question: Let G be a group. The commutator subgroup, denoted G', is defined as the subgroup generated by all products x^-1y^-1xy for any x, y ∈ G. In other words,
G' = ⟨x^-1y^-1xy | x, y ∈ G⟩.
a) Prove that G' ⊴ G.
b) Prove (without using part a) that G' is a characteristic subgroup of G. (See Exercise 4.2.20)
c) Prove that G/G' is abelian.
Solution:
a) Let g ∈ G and x^-1y^-1xy ∈ G' be any one of the generators of the subgroup and consider the conjugation g(x^-1y^-1xy)g^-1 = (gx^-1y^-1)(xyg^-1) = (gx^-1y^-1g)(g^-1xyg^-1). Letting h = g^-1x and k = yg^-1 we see that this is equal to h^-1k^-1hk ∈ G'. By Theorem 4.2.8, we have G' ⊴ G.
b) Recall that a characteristic subgroup is a subgroup H that is fixed by any automorphism ψ (not just the inner automorphisms which are conjugation). Let ψ be any automorphism and x^-1y^-1xy ∈ G' be any generator of G'. Consider ψ(x^-1y^-1xy) = ψ(x^-1)ψ(y^-1)ψ(x)ψ(y) = ψ(x)^-1ψ(y)^-1ψ(x)ψ(y) = g^-1h^-1gh ∈ G'. Since any generator will end up back in G' and ψ is a homomorphism, this shows that ψ(G') ⊆ G'. Now, since ψ is an isomorphism, we know that in the finite (order of G' is finite) case, ψ(G') = G'. In the case when |G'| is infinite, we know that ψ(G') ⊆ G'. Now operating by ψ^-1 gets us, (G') ⊆ ψ^-1(G').
c) Consider the quotient G/G'. For any two element x, y ∈ G, the product x^-1y^-1xy ∈ G' ⇒ ̅x^-¹̅y^-¹̅x̅y = ̅1. Splitting up the elements we get ̅x^-¹̅y^-¹̅x̅y = ̅1. By moving the two inverses to the other side of the equation we get ̅x̅y = ̅y̅x, which shows that G/G' is abelian.