Question

Modern Algebra optional standard: I can prove properties of homomorphisms Let G and H be groups, and let φ:G→H be a homomorphism. Suppose H is abelian. Show that aba⁻¹b⁻¹∈kerφ for all a,b∈G.

Answer 1

The statement aba^{-1}b^{-1} is in kerφ for all a, b ∈ G has been proven by showing the homomorphism φ applied to aba^{-1}b^{-1} equals the abelian group H's identity element, which is the definition of being in the kernel of φ.

The question pertains to proving a property of homomorphisms in the context of group theory within abstract algebra. Given a homomorphism φ:G→H where H is an abelian group, it's required to show that for any elements a, b in G, the element aba-1b-1 lies in the kernel of φ. To demonstrate this, recall that a group is abelian if its operation is commutative, meaning that for all x, y in H, xy = yx.

Now, we'll proceed as follows:

1. Apply φ to aba-1b-1 to get φ(aba-1b-1).
2. Since φ is a homomorphism, it preserves the group operation, leading to φ(a)φ(b)φ(a)-1φ(b)-1.
3. In an abelian group, φ(a)φ(b) = φ(b)φ(a), so the expression simplifies to the identity element of H, which denotes that aba-1b-1 is in kerφ.

Therefore, since the expression evaluates to the identity element, we have proven that aba-1b-1∈kerφ for all a, b ∈ G.