Final answer:
To demonstrate that H equals the preimage under f of the image of H, it is shown that elements of H are mapped to f[H] by f and that elements of f⁻¹(f[H]) must belong to H by the properties of homomorphisms and because {Ker}(f) is a subset of H.
Step-by-step explanation:
We are given a group G, a subset H of G, and a homomorphism f from G to another group. The statement {Ker}(f) ⊆ H indicates that the kernel of f, which is the set of elements in G that are mapped to the identity in the codomain by f, is a subset of H. We want to show that H=f⁻¹(f[H]), which means that H equals the preimage under f of the image of H under f.
To prove this, we'll show that every element in H is also in f⁻¹(f[H]) and every element in f⁻¹(f[H]) is also in H. This will demonstrate that the two sets are equal.
Let x be any element in H. Then f(x) is in f[H] because f maps elements of H to f[H]. Now, consider any y in G such that f(y) is in f[H]. Since f is a homomorphism, if f(x) = f(y), then f(x)^{-1}f(y) = e where e is the identity element in the codomain, meaning that x^{-1}y is in {Ker}(f). Because the kernel is a subset of H and H is a subset of G, x^{-1}y is also in H. As subgroup properties ensure closure under taking inverses and products, and since x is in H, xx^{-1}y = y must also be in H. Hence, every element y in f⁻¹(f[H]) is also in H establishing the equality.