Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [tex] [ab]_K [/tex] is equal to [tex] [ba]_K [/tex], thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [tex] [ab]_H = [ba]_H [/tex] . Since a and b were generic elements of H, then H/G is abelian.
